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# Q-- A wire carrying a current i in X-Y plane along the curve y=sin(2Πx/λ). A magnetic field exists in the z-direction. Find magnitude of magnetic force on the portion of the wire between x=0 and x=λ. EXPLAIN
Grade:Upto college level
## 1 Answers
Chetan Mandayam Nayakar
312 Points
13 years ago
dF=idLxB= iB(∫dL)=-2πiBjˆ
therefore,magnitude=2πiB along negative y-axis
## ASK QUESTION
Get your questions answered by the expert for free
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > resghm Structured version Visualization version GIF version
Theorem resghm 18350
Description: Restriction of a homomorphism to a subgroup. (Contributed by Stefan O'Rear, 31-Dec-2014.)
Hypothesis
Ref Expression
resghm.u 𝑈 = (𝑆s 𝑋)
Assertion
Ref Expression
resghm ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → (𝐹𝑋) ∈ (𝑈 GrpHom 𝑇))
Proof of Theorem resghm
Dummy variables 𝑎 𝑏 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2820 . 2 (Base‘𝑈) = (Base‘𝑈)
2 eqid 2820 . 2 (Base‘𝑇) = (Base‘𝑇)
3 eqid 2820 . 2 (+g𝑈) = (+g𝑈)
4 eqid 2820 . 2 (+g𝑇) = (+g𝑇)
5 resghm.u . . . 4 𝑈 = (𝑆s 𝑋)
65subggrp 18258 . . 3 (𝑋 ∈ (SubGrp‘𝑆) → 𝑈 ∈ Grp)
76adantl 484 . 2 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → 𝑈 ∈ Grp)
8 ghmgrp2 18337 . . 3 (𝐹 ∈ (𝑆 GrpHom 𝑇) → 𝑇 ∈ Grp)
98adantr 483 . 2 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → 𝑇 ∈ Grp)
10 eqid 2820 . . . . 5 (Base‘𝑆) = (Base‘𝑆)
1110, 2ghmf 18338 . . . 4 (𝐹 ∈ (𝑆 GrpHom 𝑇) → 𝐹:(Base‘𝑆)⟶(Base‘𝑇))
1210subgss 18256 . . . 4 (𝑋 ∈ (SubGrp‘𝑆) → 𝑋 ⊆ (Base‘𝑆))
13 fssres 6518 . . . 4 ((𝐹:(Base‘𝑆)⟶(Base‘𝑇) ∧ 𝑋 ⊆ (Base‘𝑆)) → (𝐹𝑋):𝑋⟶(Base‘𝑇))
1411, 12, 13syl2an 597 . . 3 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → (𝐹𝑋):𝑋⟶(Base‘𝑇))
1512adantl 484 . . . . 5 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → 𝑋 ⊆ (Base‘𝑆))
165, 10ressbas2 16531 . . . . 5 (𝑋 ⊆ (Base‘𝑆) → 𝑋 = (Base‘𝑈))
1715, 16syl 17 . . . 4 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → 𝑋 = (Base‘𝑈))
1817feq2d 6474 . . 3 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → ((𝐹𝑋):𝑋⟶(Base‘𝑇) ↔ (𝐹𝑋):(Base‘𝑈)⟶(Base‘𝑇)))
1914, 18mpbid 234 . 2 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → (𝐹𝑋):(Base‘𝑈)⟶(Base‘𝑇))
20 eleq2 2899 . . . . . 6 (𝑋 = (Base‘𝑈) → (𝑎𝑋𝑎 ∈ (Base‘𝑈)))
21 eleq2 2899 . . . . . 6 (𝑋 = (Base‘𝑈) → (𝑏𝑋𝑏 ∈ (Base‘𝑈)))
2220, 21anbi12d 632 . . . . 5 (𝑋 = (Base‘𝑈) → ((𝑎𝑋𝑏𝑋) ↔ (𝑎 ∈ (Base‘𝑈) ∧ 𝑏 ∈ (Base‘𝑈))))
2317, 22syl 17 . . . 4 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → ((𝑎𝑋𝑏𝑋) ↔ (𝑎 ∈ (Base‘𝑈) ∧ 𝑏 ∈ (Base‘𝑈))))
2423biimpar 480 . . 3 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎 ∈ (Base‘𝑈) ∧ 𝑏 ∈ (Base‘𝑈))) → (𝑎𝑋𝑏𝑋))
25 simpll 765 . . . . 5 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → 𝐹 ∈ (𝑆 GrpHom 𝑇))
2615sselda 3943 . . . . . 6 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ 𝑎𝑋) → 𝑎 ∈ (Base‘𝑆))
2726adantrr 715 . . . . 5 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → 𝑎 ∈ (Base‘𝑆))
2815sselda 3943 . . . . . 6 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ 𝑏𝑋) → 𝑏 ∈ (Base‘𝑆))
2928adantrl 714 . . . . 5 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → 𝑏 ∈ (Base‘𝑆))
30 eqid 2820 . . . . . 6 (+g𝑆) = (+g𝑆)
3110, 30, 4ghmlin 18339 . . . . 5 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑎 ∈ (Base‘𝑆) ∧ 𝑏 ∈ (Base‘𝑆)) → (𝐹‘(𝑎(+g𝑆)𝑏)) = ((𝐹𝑎)(+g𝑇)(𝐹𝑏)))
3225, 27, 29, 31syl3anc 1367 . . . 4 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → (𝐹‘(𝑎(+g𝑆)𝑏)) = ((𝐹𝑎)(+g𝑇)(𝐹𝑏)))
335, 30ressplusg 16588 . . . . . . . 8 (𝑋 ∈ (SubGrp‘𝑆) → (+g𝑆) = (+g𝑈))
3433ad2antlr 725 . . . . . . 7 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → (+g𝑆) = (+g𝑈))
3534oveqd 7148 . . . . . 6 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → (𝑎(+g𝑆)𝑏) = (𝑎(+g𝑈)𝑏))
3635fveq2d 6648 . . . . 5 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → ((𝐹𝑋)‘(𝑎(+g𝑆)𝑏)) = ((𝐹𝑋)‘(𝑎(+g𝑈)𝑏)))
3730subgcl 18265 . . . . . . . 8 ((𝑋 ∈ (SubGrp‘𝑆) ∧ 𝑎𝑋𝑏𝑋) → (𝑎(+g𝑆)𝑏) ∈ 𝑋)
38373expb 1116 . . . . . . 7 ((𝑋 ∈ (SubGrp‘𝑆) ∧ (𝑎𝑋𝑏𝑋)) → (𝑎(+g𝑆)𝑏) ∈ 𝑋)
3938adantll 712 . . . . . 6 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → (𝑎(+g𝑆)𝑏) ∈ 𝑋)
4039fvresd 6664 . . . . 5 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → ((𝐹𝑋)‘(𝑎(+g𝑆)𝑏)) = (𝐹‘(𝑎(+g𝑆)𝑏)))
4136, 40eqtr3d 2857 . . . 4 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → ((𝐹𝑋)‘(𝑎(+g𝑈)𝑏)) = (𝐹‘(𝑎(+g𝑆)𝑏)))
42 fvres 6663 . . . . . 6 (𝑎𝑋 → ((𝐹𝑋)‘𝑎) = (𝐹𝑎))
43 fvres 6663 . . . . . 6 (𝑏𝑋 → ((𝐹𝑋)‘𝑏) = (𝐹𝑏))
4442, 43oveqan12d 7150 . . . . 5 ((𝑎𝑋𝑏𝑋) → (((𝐹𝑋)‘𝑎)(+g𝑇)((𝐹𝑋)‘𝑏)) = ((𝐹𝑎)(+g𝑇)(𝐹𝑏)))
4544adantl 484 . . . 4 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → (((𝐹𝑋)‘𝑎)(+g𝑇)((𝐹𝑋)‘𝑏)) = ((𝐹𝑎)(+g𝑇)(𝐹𝑏)))
4632, 41, 453eqtr4d 2865 . . 3 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎𝑋𝑏𝑋)) → ((𝐹𝑋)‘(𝑎(+g𝑈)𝑏)) = (((𝐹𝑋)‘𝑎)(+g𝑇)((𝐹𝑋)‘𝑏)))
4724, 46syldan 593 . 2 (((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) ∧ (𝑎 ∈ (Base‘𝑈) ∧ 𝑏 ∈ (Base‘𝑈))) → ((𝐹𝑋)‘(𝑎(+g𝑈)𝑏)) = (((𝐹𝑋)‘𝑎)(+g𝑇)((𝐹𝑋)‘𝑏)))
481, 2, 3, 4, 7, 9, 19, 47isghmd 18343 1 ((𝐹 ∈ (𝑆 GrpHom 𝑇) ∧ 𝑋 ∈ (SubGrp‘𝑆)) → (𝐹𝑋) ∈ (𝑈 GrpHom 𝑇))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 208 ∧ wa 398 = wceq 1537 ∈ wcel 2114 ⊆ wss 3912 ↾ cres 5531 ⟶wf 6325 ‘cfv 6329 (class class class)co 7131 Basecbs 16459 ↾s cress 16460 +gcplusg 16541 Grpcgrp 18079 SubGrpcsubg 18249 GrpHom cghm 18331 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2177 ax-ext 2792 ax-rep 5164 ax-sep 5177 ax-nul 5184 ax-pow 5240 ax-pr 5304 ax-un 7437 ax-cnex 10569 ax-resscn 10570 ax-1cn 10571 ax-icn 10572 ax-addcl 10573 ax-addrcl 10574 ax-mulcl 10575 ax-mulrcl 10576 ax-mulcom 10577 ax-addass 10578 ax-mulass 10579 ax-distr 10580 ax-i2m1 10581 ax-1ne0 10582 ax-1rid 10583 ax-rnegex 10584 ax-rrecex 10585 ax-cnre 10586 ax-pre-lttri 10587 ax-pre-lttrn 10588 ax-pre-ltadd 10589 ax-pre-mulgt0 10590 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3or 1084 df-3an 1085 df-tru 1540 df-ex 1781 df-nf 1785 df-sb 2070 df-mo 2622 df-eu 2653 df-clab 2799 df-cleq 2813 df-clel 2891 df-nfc 2959 df-ne 3007 df-nel 3111 df-ral 3130 df-rex 3131 df-reu 3132 df-rab 3134 df-v 3475 df-sbc 3752 df-csb 3860 df-dif 3915 df-un 3917 df-in 3919 df-ss 3928 df-pss 3930 df-nul 4268 df-if 4442 df-pw 4515 df-sn 4542 df-pr 4544 df-tp 4546 df-op 4548 df-uni 4813 df-iun 4895 df-br 5041 df-opab 5103 df-mpt 5121 df-tr 5147 df-id 5434 df-eprel 5439 df-po 5448 df-so 5449 df-fr 5488 df-we 5490 df-xp 5535 df-rel 5536 df-cnv 5537 df-co 5538 df-dm 5539 df-rn 5540 df-res 5541 df-ima 5542 df-pred 6122 df-ord 6168 df-on 6169 df-lim 6170 df-suc 6171 df-iota 6288 df-fun 6331 df-fn 6332 df-f 6333 df-f1 6334 df-fo 6335 df-f1o 6336 df-fv 6337 df-riota 7089 df-ov 7134 df-oprab 7135 df-mpo 7136 df-om 7557 df-wrecs 7923 df-recs 7984 df-rdg 8022 df-er 8265 df-en 8486 df-dom 8487 df-sdom 8488 df-pnf 10653 df-mnf 10654 df-xr 10655 df-ltxr 10656 df-le 10657 df-sub 10848 df-neg 10849 df-nn 11615 df-2 11677 df-ndx 16462 df-slot 16463 df-base 16465 df-sets 16466 df-ress 16467 df-plusg 16554 df-mgm 17828 df-sgrp 17877 df-mnd 17888 df-grp 18082 df-subg 18252 df-ghm 18332 This theorem is referenced by: ghmima 18355 resrhm 19537 reslmhm 19797 dimkerim 31031
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Prove recurrence relation via mathematical induction
$$T(n) = \begin{cases} 2 & \text{if } n = 2 \\ 2T(\frac{n}{2})+n & \text{if } n = 2^k \text{, for } k > 1 \end{cases}\\ \text{ } \\ \text{ } \\ \text{ } \\ \text{Prove } T(n) = n\lg(n) \text{ if } n = 2^k\text{, for } k > 1.$$
I am crawling through the "Introduction to Algorithms" textbook, in preparation for a future course, and this question has stumped me. I was able to understand proving algorithmic correctness using loop invariants, but I don't have a lot (any) experience in proofs. As a side-note, lg(n) is the binary logarithm of n. Also: The book did not make it explicit, but k is above 1 and an integer.
Here is what I have so far.
Base case
\begin{align} \text{Let } n = 2^k \text{ where } k &= 2 \text{ (minimal case)}\\ n &= 2^k = 4\\ 2T(\frac{n}{2}) + n &= n\lg n\\ 2T(2) + 4 &= 4\lg 4\\ 2*2 + 4 &= 4*2\\ 8 &= 8 \end{align}
Inductive hypothesis
$$T(2^{k+1}) = 2^{k+1}\lg{2^{k+1}}$$
Inductive step
$$2T(\frac{2^{k+1}}{2}) + 2^{k+1} = 2^{k+1}(k+1)$$
Beyond that, I cannot go. I would greatly appreciate any hints.
(I reiterate... this is not homework, I'm trying to solve this as a personal challenge)
Mentallic
Homework Helper
Now if $n=2^{k+1}$, then what does $2^{k+1}(k+1)$ equal to?
Last edited:
## 2^{k+1}(k+1) = k2^{k+1} + 2^{k+1} ##, via distribution
So...
$$2T(\frac{2^{k+1}}{2})+2^{k+1} = k2^{k+1} + 2^{k+1}\\ 2T(\frac{2^{k+1}}{2}) = k2^{k+1} \\$$
And also...
$$2T(\frac{2^{k+1}}{2}) = k2^{k+1} \\ 2T(2^k) = k2^{k+1} \\$$
If I divide by two...
$$T(2^k) = k2^k \\$$
This is neat, but it doesn't seem useful.
Edit: Wait... HOLY GOD
I understand now why some people do this for a living... the shock and elation at seeing the answer right before my eyes.
Now I just replace with my inductive hypothesis...
$$(2^k)\lg(2^k) = k2^k \\ (2^k)k = k2^k \\ \text{ } k2^k = k2^k \\$$
Last edited:
Mentallic
Homework Helper
That's fine, but not the usual systematic way of how we prove by induction.
Usually, you show the base case of k=1 or k=2 etc. is true. Then you assume the nth case to hold, that is you assume that what you are trying to prove is true:
$$T(n)=n\lg{n}, n=2^k$$
i.e.
$$T(2^k)=2^k\lg{2^k}=2^kk$$
This last expression here which is $T(2^k)=2^kk$ we assume to be true.
Now for the inductive hypothesis, we increment k by 1, which means we now test to see if
$$T(N)=N\lg{N}, N=2^{k+1}$$
(I'm using N here to denote that it's different to our previous n)
so we are trying to show that
$$T(2^{k+1})=2^{k+1}\lg\left({2^{k+1}}\right)$$
But we aren't assuming it's true. We're going to try to turn the left hand expression into the right hand expression using what we've been given.
By the recurrence relation,
$$T(2^{k+1})=2T\left(\frac{2^{k+1}}{2}\right)+2^{k+1} = 2T(2^k)+2^{k+1}$$
But from our assumption, $T(2^k)=2^kk$. Plugging this in gives us
$$T(2^{k+1})=2(2^kk)+2^{k+1}=2^{k+1}k+2^{k+1}=2^{k+1}(k+1)$$
What were we trying to show again? That
$$T(N)=N\lg{N}, N=2^{k+1}$$
So we want to express everything in terms of N.
$$T(2^{k+1})=T(N)$$
and since $\lg\left({2^{k+1}}\right)=k+1$ then
$$2^{k+1}(k+1)=2^{k+1}\lg\left({2^{k+1}}\right)=N\lg(N)$$
as required. Hence it's proven.
Ah, it took me a night's sleep to understand your response.. Today, while looking over my original solution (the one I posted before yours), I was bothered. It seems as if I was assuming something was true, and then trivially deriving it was true from that assumption (putting the cart before the horse).
Now I see how mathematical induction is supposed to work. You prove a base case, thereby clarifying that T(n) = n lg n, for the lowest n. Then you prove than n+1 (here, 2^(k+1) is true by rewriting it in terms of 2^k, which you've already proved in the base case.
Thank you for clarifying. I'm going to try proving a few recurrence relations like this... it seems now like a problem only of algebraic skill.
Mentallic
Homework Helper
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# Troubleshooting: 'colnames' Set on Object with Less Than 2-Dimensions
## Introduction
When working with R programming language, you may encounter an error message like this: "Error in colnames: 'colnames' requires a list or object with two or more dimensions". This error occurs when you try to set column names for a data frame that has less than two dimensions. In this article, we will discuss the causes of this error and provide step-by-step solutions to resolve it.
## Causes of the error
The error message "Error in colnames: 'colnames' requires a list or object with two or more dimensions" occurs when you try to set column names for a data frame that has less than two dimensions. This means that you have either provided an object that is not a data frame or you have provided a data frame that has only one dimension, such as a vector or a matrix.
## Solution
To resolve this error, you need to make sure that you are providing a data frame with at least two dimensions to the colnames function. Here are the steps to follow:
1. Check the dimension of your data frame using the dim function. If the dimension is less than two, then you need to convert it to a data frame with at least two dimensions.
``````# Create a vector
vec <- c(1, 2, 3)
# Convert the vector to a data frame with one column
df <- data.frame(vec)
# Check the dimension of the data frame
dim(df) # Output: [1] 3 1
# Convert the data frame to a matrix with one column
mat <- as.matrix(df)
# Convert the matrix to a data frame with two columns
df_new <- data.frame(mat)
# Check the dimension of the new data frame
dim(df_new) # Output: [1] 3 2
``````
1. Once you have a data frame with at least two dimensions, you can set the column names using the colnames function.
``````# Set column names for the data frame
colnames(df_new) <- c("Col1", "Col2")
``````
## FAQ
### Q1. What is a data frame in R?
A data frame is a two-dimensional array-like structure in R, where each column can have a different data type.
### Q2. Why do I get the error message "Error in colnames: 'colnames' requires a list or object with two or more dimensions"?
This error occurs when you try to set column names for a data frame that has less than two dimensions.
### Q3. How do I convert a vector to a data frame in R?
You can convert a vector to a data frame using the data.frame function.
``````# Create a vector
vec <- c(1, 2, 3)
# Convert the vector to a data frame with one column
df <- data.frame(vec)
``````
### Q4. How do I convert a matrix to a data frame in R?
You can convert a matrix to a data frame using the data.frame function.
``````# Create a matrix
mat <- matrix(1:6, nrow = 2)
# Convert the matrix to a data frame
df <- data.frame(mat)
``````
### Q5. How do I check the dimension of a data frame in R?
You can check the dimension of a data frame using the dim function.
``````# Create a data frame
df <- data.frame(Col1 = c(1, 2, 3), Col2 = c("A", "B", "C"))
# Check the dimension of the data frame
dim(df) # Output: [1] 3 2
``````
## Conclusion
In this article, we discussed the causes of the error message "Error in colnames: 'colnames' requires a list or object with two or more dimensions" and provided step-by-step solutions to resolve it. By converting a vector or matrix to a data frame with at least two dimensions, you can set column names using the colnames function without encountering this error.
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https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-2-section-2-2-the-multiplication-property-of-equality-exercise-set-page-110/29
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## Algebra: A Combined Approach (4th Edition)
$\frac{2}{3}$ y - 11 = -9 $\frac{2}{3}$ y = 2 y = 3
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Aptitude Tests 4 Me
Data Interpretation
194. In which year was the combined number of fatalities and serious injuries greatest?
(a) 2001 (b) 2002 (c) 2004 (d) 2005
195. Which year showed the greatest decrease in the number of serious injuries over the previous year?
(a) 2003 (b) 2004 (c) 2005 (d) 2006
196. What is the average (mean) number of fatalities over the eight years?
(a) 58 (b) 59.25 (c) 60.5 (d) 61
TOTAL
Detailed Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155
Passage Reading Verbal Logic Non Verbal Logic Numerical Logic Data Interpretation Reasoning Analytical Ability Basic Numeracy About Us Contact Privacy Policy Major Tests FAQ
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Free Version
Easy
# Subtracting Complex Numbers
Free
SATSTM-4QE1QZ
Simplify the following expression of complex numbers:
$$(4-5i)-(9+2i)$$
A
$-5–3i$
B
$-5–7i$
C
$5–3i$
D
$5–7i$
E
$-12i$
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### I'm Eight
Find a great variety of ways of asking questions which make 8.
### Let's Investigate Triangles
Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?
### Noah
Noah saw 12 legs walk by into the Ark. How many creatures did he see?
# Two-digit Targets
## Two-digit Targets
You have a set of the digits from $0$ - $9$.
Can you arrange these digits in the five boxes below to make two-digit numbers as close to the targets as possible? You may use each digit once only.
How will you know that your solution is as close to the targets as possible?
You could use this interactivity to try out your ideas.
This activity has been adapted from one of BEAM's free Maths of the Month resources, which unfortunately are no longer available.
Here is an alternative Flash version of the interactivity:
If you can see this message Flash may not be working in your browser
Please see http://nrich.maths.org/techhelp/#flash to enable it.
Full screen version
### Why do this problem?
This problem is a fantastic opportunity for learners to apply knowledge of place value and offers a context for learning and practising the relevant vocabulary (odd, even, multiple). The interactivity will help learners satisfy their curiosity in the sense of finding a 'better' solution, as it enables them to play around with the digits without having to commit anything to paper. This act of deciding whether one solution is better than another also provides a meaningful context in which to compare and order numbers.
### Possible approach
You could start by having two different digits and asking how the digits could be arranged to make the number that is the larger/smaller. Try this several times with different combinations of two digits. Could the digits be arranged to make a multiple of $5$? If not, why not?
Next you could introduce one of the criteria in the problem, such as making the largest possible even two-digit number. Alternatively, you could make up your own examples such as the largest even number or the nearest to $70$. You will need to establish whether $0$ can be used at the beginning of a number. This, in itself, can form an interesting discussion point. (The final decision itself does not matter - it is the reasons that are important, and the fact that the children feel as if it is their decision!)
Give pairs time to continue working on this problem, using the interactivity and/or a set of digit cards. Pairs may find this sheet useful for recording their thinking.
In the plenary, facilitate a general discussion on the strategies learners used to tackle the task and how they determined how 'good' a solution was.
### Key questions
What have you tried so far?
What do you know about odd numbers/even numbers/multiples of 5? How does this help you place the digits?
Can you create a 'better' solution? How will you know it is better?
### Possible extension
Some children might like the challenge of this four-digit version of the task.
### Possible support
Making the numbers one at a time with set of $0$ - $9$ digit cards, or using the interactivity, should help all children access this problem.
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## Conversion formula
The conversion factor from meters to yards is 1.0936132983377, which means that 1 meter is equal to 1.0936132983377 yards:
1 m = 1.0936132983377 yd
To convert 1.6 meters into yards we have to multiply 1.6 by the conversion factor in order to get the length amount from meters to yards. We can also form a simple proportion to calculate the result:
1 m → 1.0936132983377 yd
1.6 m → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 1.6 m × 1.0936132983377 yd
L(yd) = 1.7497812773403 yd
The final result is:
1.6 m → 1.7497812773403 yd
We conclude that 1.6 meters is equivalent to 1.7497812773403 yards:
1.6 meters = 1.7497812773403 yards
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 0.5715 × 1.6 meters.
Another way is saying that 1.6 meters is equal to 1 ÷ 0.5715 yards.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one point six meters is approximately one point seven five yards:
1.6 m ≅ 1.75 yd
An alternative is also that one yard is approximately zero point five seven two times one point six meters.
## Conversion table
### meters to yards chart
For quick reference purposes, below is the conversion table you can use to convert from meters to yards
meters (m) yards (yd)
2.6 meters 2.843 yards
3.6 meters 3.937 yards
4.6 meters 5.031 yards
5.6 meters 6.124 yards
6.6 meters 7.218 yards
7.6 meters 8.311 yards
8.6 meters 9.405 yards
9.6 meters 10.499 yards
10.6 meters 11.592 yards
11.6 meters 12.686 yards
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# Number 31020022331
### Properties of number 31020022331
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
738effa3b
Base 32:
ssevuhr
sin(31020022331)
-0.9517771595797
cos(31020022331)
0.30679021904618
tan(31020022331)
-3.1023712637867
ln(31020022331)
24.157898714585
lg(31020022331)
10.491642106122
sqrt(31020022331)
176125.01903761
Square(31020022331)
9.6224178541574E+20
### Number Look Up
Look Up
31020022331 which is pronounced (thirty-one billion twenty million twenty-two thousand three hundred thirty-one) is a very unique figure. The cross sum of 31020022331 is 17. If you factorisate the figure 31020022331 you will get these result 1327 * 23376053. The number 31020022331 has 4 divisors ( 1, 1327, 23376053, 31020022331 ) whith a sum of 31043399712. The number 31020022331 is not a prime number. 31020022331 is not a fibonacci number. The number 31020022331 is not a Bell Number. The figure 31020022331 is not a Catalan Number. The convertion of 31020022331 to base 2 (Binary) is 11100111000111011111111101000111011. The convertion of 31020022331 to base 3 (Ternary) is 2222001211122101221222. The convertion of 31020022331 to base 4 (Quaternary) is 130320323333220323. The convertion of 31020022331 to base 5 (Quintal) is 1002012111203311. The convertion of 31020022331 to base 8 (Octal) is 347073775073. The convertion of 31020022331 to base 16 (Hexadecimal) is 738effa3b. The convertion of 31020022331 to base 32 is ssevuhr. The sine of the number 31020022331 is -0.9517771595797. The cosine of the number 31020022331 is 0.30679021904618. The tangent of 31020022331 is -3.1023712637867. The square root of 31020022331 is 176125.01903761.
If you square 31020022331 you will get the following result 9.6224178541574E+20. The natural logarithm of 31020022331 is 24.157898714585 and the decimal logarithm is 10.491642106122. that 31020022331 is unique number!
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# Using divisibility tests, determine which of following numbers are divisible by 6: 12583
Harshit Singh
3 years ago
Dear Student
Since, the last digit of the number is 3. Hence, the number is not divisible by 2
By adding all the digits of the number, we get 19 which is not divisible by 3.
Hence, the numberis not divisible by 3
∴The number is not divisible by both 2 and 3. Hence, the number is not divisible by 6
Thanks
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# Thread: Trig substituion: what is the flaw in my logic?
1. ## Trig substituion: what is the flaw in my logic?
Hi all,
I have a general question about trig substitution. My calculus book says if you ever have an expression dealing with Sqrt(a^2 - x^2), you let x =a Sin(theta), and from a constructing your triangle, you get that Cos(theta) = Sqrt(a^2 - x^2). Why is WRONG to switch Sine and Cosine here? For example, you could x be the adjacent side, Sqrt(a^2-x^2) be the opposite side, and a be the hyp, and you get:
x = a Cos(theta)
Sqrt(1-x^2) = a Sin(Theta).
And the triangle is still correct. But I tried working a problem this way and got the wrong answer. What is the deal?
Also, at the end of the integral, when you solve for Theta in terms of x, does it matter whether you use Arctan, Arcsin, or Arccos?
Thanks,
Tyler
2. Originally Posted by james121515
Hi all,
I have a general question about trig substitution. My calculus book says if you ever have an expression dealing with Sqrt(a^2 - x^2), you let x =a Sin(theta), and from a constructing your triangle, you get that Cos(theta) = Sqrt(a^2 - x^2). Why is WRONG to switch Sine and Cosine here? For example, you could x be the adjacent side, Sqrt(a^2-x^2) be the opposite side, and a be the hyp, and you get:
x = a Cos(theta)
Sqrt(1-x^2) = a Sin(Theta).
And the triangle is still correct. But I tried working a problem this way and got the wrong answer. What is the deal?
Also, at the end of the integral, when you solve for Theta in terms of x, does it matter whether you use Arctan, Arcsin, or Arccos? Mr F says: What matters is substituting the correct expression for theta based on the substitution you made.
Thanks,
Tyler
If you make the substitution $x = a \sin \theta$ then clearly
from a constructing your triangle, you get that Cos(theta) = Sqrt(a^2 - x^2)
If you make the substitution $x = a \cos \theta$ then clearly
you could x be the adjacent side, Sqrt(a^2-x^2) be the opposite side, and a be the hyp, and you get:
x = a Cos(theta)
Sqrt(1-x^2) = a Sin(Theta).
Both substitutions lead to equivalent correct answers.
If you need more help, please post a concrete question and your solution to it (note: solution = working plus answer) that illustrates your problem with the substitution method.
3. sin is usually chosen so that $\frac{-\pi}{2}\leq {\theta}\leq \frac{\pi}{2}$. Because then $cos{\theta}>0$.
It is about the restriction on theta.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
You are not logged in.
## #26 2011-09-20 06:02:32
bobbym
Offline
### Re: maths product
Hi rstakits;
Best I can do is:
800 x
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #27 2011-10-24 01:34:54
fatty
Guest
### Re: maths product
#### remona wrote:
hey guy i need some1 to talk to so if u could ud be a great help.....if u could talk then just gimmi yr msn addi or email:D
[removed by moderator]
## #28 2011-10-24 06:35:27
bobbym
Offline
### Re: maths product
Hi fatty;
It is not wise to just go handing your email address out like that. The asker here might be legitimate but anyone on the internet can see it also. You are leaving yourself open to spammers and hackers as well as numerous bots.
You might have been unaware of that. If you put it back I will assume you are willing to chance it and I will not interfere further.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #29 2012-02-06 22:54:22
maria tan
Guest
### Re: maths product
The product of two number is 32 and thier quotient is 8.find the number.
## #30 2012-02-06 23:16:23
bobbym
Offline
### Re: maths product
Hi maria tan;
The two numbers are 16 and 2.
There are only 3 products that equal 32. 8 x 4 = 32, 2 x16 = 32 and 1 x 32 = 32. Only one of them has a quotient of 8. Welcome to the forum.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #31 2012-06-05 07:57:37
Umar
Guest
### Re: maths product
#### mathsyperson wrote:
Presumably it wants you to find 1x2x3x4x5x6x7x8x9x10x11x12x13x14x15, also known as "15!".
it actually doesn't equall fifteen it equals 3026144400. becuase you have to find out what 1*2 is and the product of that *3 and the product of that #4 and the product of that *5 and keep going until you get to the product of 15.
15! means 15 facturo or something like that which is basically the product of 15
## #32 2012-06-05 08:09:24
anonimnystefy
Real Member
Offline
### Re: maths product
15! is not the product of 15, it is the product of all natural numbers less than 15. Maybe that is what you meant?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
## #33 2012-06-05 08:44:51
bobbym
Offline
### Re: maths product
Hi Umar;
Welcome to the forum. You are correct in drawing attention to the miscalculation.
15! = 1307674368000 not 3026144400
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #34 2012-08-22 08:34:52
noelevans
Full Member
Offline
### Re: maths product
Hi rstakits!
Using N for "a number" that would be 800 times N, or 800N or 800xN where x means times.
If you have a collection of numbers like {3,10,5} then X{3,10,5}=1x3x10x5=150 would be the
product of the numbers in the collection. We in essence "scale up" the 1 by the numbers in the
set. If there are no numbers in the set like the set { } then we don't have anything to scale the
1 up by so we just get the 1. X{ }=1.
So this gives us a short notation for multiplying any collection of numbers we want. If all the
numbers in the set are the SAME then we have exponentiation. X{2,2,2}=1x2x2x2=2^3, which
is called "two cubed."
That may be more than you wanted, but maybe it will help somewhere along the line.
Have a stupendous day!
Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).
LaTex is like painting on many strips of paper and then stacking them to see what picture they make.
## #35 2012-08-22 08:42:31
noelevans
Full Member
Offline
### Re: maths product
Hi bobbym!
Finally! Patience works. I now have a signature. I'll probably change it to something else before
long since I just put something in the "slot" to see when it would finally appear.
Have a lovely day!
Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).
LaTex is like painting on many strips of paper and then stacking them to see what picture they make.
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• Create Account
## Quantifying Difference Between Two Images
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5 replies to this topic
### #1phyitbos Members
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Posted 03 October 2012 - 05:56 AM
Hi,
I am learning about genetic algorithms and my goal is to make a replica (similar at any rate) to the one shown in this YouTube video:
I need to be able to rate the "fitness" of my generated images. To do this, I need to compare them to a supplied picture, and was wondering what algorithm I should use to quantify the similarity (distance?) between the two images. I'm not familiar with algorithms in this category, so any insight would be very helpful! One that is simple to implement would be best, considering the goal of my project is the genetic algorithm and not comparing images, although a little learning never hurts
### #2Álvaro Members
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Posted 03 October 2012 - 06:29 AM
The first idea is to treat the images as vectors in R^(height*width), with one component per pixel, and use Euclidean distance. There are more sophisticated measures that are closer to what humans perceive as being similar. They are usually based on the wavelet transform.
You can do something like reduce the resolution of the images to several different scales and add the corresponding Euclidean distances with some weights (I think that's what you would get with some of the wavelet methods if you were to use Haar wavelets).
### #3turch Members
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Posted 03 October 2012 - 06:33 AM
The simplest way would be to
int weight = 0;
for each (testPixel)
{
weight += abs(originalPixel.r - testPixel.r); // assuming each channel [0,255];
weight += abs(originalPixel.g - testPixel.g);
weight += abs(originalPixel.b - testPixel.b);
}
The lower the weight the better the match.
Edited by turch, 03 October 2012 - 06:34 AM.
### #4PolyVox Members
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Posted 03 October 2012 - 08:49 AM
You might find the ImageMagick 'compare' tool useful. It has a number of metrics which you can use when comparing images: http://www.imagemagick.org/script/command-line-options.php#metric
### #5Lauris Kaplinski Members
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Posted 05 October 2012 - 03:53 PM
You need at least some idea, what kind of similarity you want to use as the basis:
• The easiest is simply sum color differences of individual pixels (like turch suggested)
• Another is to find euclidean distance between MxN vectors (like Alvaro suggested)
• Now you may want to punish big differences proportionally more - in that case you may want to square the absolute differences before before adding them
• If you want to give less weight to high frequencies you should downsample images before comparing
• If you are interested in human percievable difference, you should transform your image to YUV or similar colorspace
• If you want to add more importance to shapes you may want to apply some edge-detection
And so on
Lauris Kaplinski
First technology demo of my game Shinya is out: http://lauris.kaplinski.com/shinya
Khayyam 3D - a freeware poser and scene builder application: http://khayyam.kaplinski.com/
### #6L. Spiro Members
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Posted 06 October 2012 - 05:29 AM
The simplest way would be to
int weight = 0;
for each (testPixel)
{
weight += abs(originalPixel.r - testPixel.r); // assuming each channel [0,255];
weight += abs(originalPixel.g - testPixel.g);
weight += abs(originalPixel.b - testPixel.b);
}
The lower the weight the better the match.
This is not the way to go and can in fact be very hurtful to the actual result.
What you want is to use MSE (mean squared error) if your goal is simply to compare qualities between different images, or PSNR (peak signal-to-noise-ratio) if you want to discretely quantify the differences between your images (in other words, calculating the MSE is part of determining the PSNR, and a higher MSE always results in a lower dB (the measurement for PSNR), so if your goal is simply to compare images to determine which is closer to the original than another, MSE is all that is needed, whereas if you want to quantify exactly how close using dB, you should continue on and calculate the PSNR, although even still this part is optional).
For MSE (all you probably really need), the pseudo code would be:
error = 0
for ( all Y ) {
for ( all X ) {
error += (src(X, Y) - copy(X, Y)) * (src(X, Y) - copy(X, Y))
}
}
error /= (width * height)
This is applied to each channel separately, so you would have one error value for R, one for G, and one for B.
Then you apply perceptual weights to these error values based on how humans perceive light.
Common weights are:
(0.299f, 0.587f, 0.114f)
(0.212656f, 0.715158f, 0.072186f)
(0.3086f, 0.6094f, 0.082f)
Notice how they all add up to 1, and also how green is always weighted much higher than the rest. This is also why the 565 16-bit color format exists, because green is the most important color perceptually. For example, if it was instead red that was most important to us humans, the 16-bit color format would instead have been 655.
Your final total perceptually adjusted error would be (assuming weights of (0.212656f, 0.715158f, 0.072186f)):
redError * 0.212656f + greenError * 0.715158f + blueError * 0.072186f
This gets you a properly weighted perceptually adjusted error metric that allows you to determine how close you are to the original, and is simple to implement.
L. Spiro
Edited by L. Spiro, 06 October 2012 - 05:32 AM.
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You are reading an old version of the documentation (v1.2.1). For the latest version see https://matplotlib.org/stable/
# axes_grid example code: demo_axes_rgb.pyΒΆ
```import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1.axes_rgb import make_rgb_axes, RGBAxes
def get_demo_image():
from matplotlib.cbook import get_sample_data
f = get_sample_data("axes_grid/bivariate_normal.npy", asfileobj=False)
z = np.load(f)
# z is a numpy array of 15x15
return z, (-3,4,-4,3)
def get_rgb():
Z, extent = get_demo_image()
Z[Z<0] = 0.
Z = Z/Z.max()
R = Z[:13,:13]
G = Z[2:,2:]
B = Z[:13,2:]
return R, G, B
def make_cube(r, g, b):
ny, nx = r.shape
R = np.zeros([ny, nx, 3], dtype="d")
R[:,:,0] = r
G = np.zeros_like(R)
G[:,:,1] = g
B = np.zeros_like(R)
B[:,:,2] = b
RGB = R + G + B
return R, G, B, RGB
def demo_rgb():
fig = plt.figure(1)
fig.clf()
ax = fig.add_subplot(111)
ax_r, ax_g, ax_b = make_rgb_axes(ax, pad=0.02)
#fig.add_axes(ax_r)
#fig.add_axes(ax_g)
#fig.add_axes(ax_b)
r, g, b = get_rgb()
im_r, im_g, im_b, im_rgb = make_cube(r, g, b)
kwargs = dict(origin="lower", interpolation="nearest")
ax.imshow(im_rgb, **kwargs)
ax_r.imshow(im_r, **kwargs)
ax_g.imshow(im_g, **kwargs)
ax_b.imshow(im_b, **kwargs)
def demo_rgb2():
fig = plt.figure(2)
ax = RGBAxes(fig, [0.1, 0.1, 0.8, 0.8], pad=0.0)
#fig.add_axes(ax)
#ax.add_RGB_to_figure()
r, g, b = get_rgb()
kwargs = dict(origin="lower", interpolation="nearest")
ax.imshow_rgb(r, g, b, **kwargs)
ax.RGB.set_xlim(0., 9.5)
ax.RGB.set_ylim(0.9, 10.6)
for ax1 in [ax.RGB, ax.R, ax.G, ax.B]:
for sp1 in ax1.spines.values():
sp1.set_color("w")
for tick in ax1.xaxis.get_major_ticks() + ax1.yaxis.get_major_ticks():
tick.tick1line.set_mec("w")
tick.tick2line.set_mec("w")
return ax
demo_rgb()
ax = demo_rgb2()
plt.show()
```
Keywords: python, matplotlib, pylab, example, codex (see Search examples)
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Warning: Creating default object from empty value in /home/admin/public_html/includeTop.php on line 210
Product Markets >> Theory Of Consumer Choice
### Individual Demand Curve
Example 1:
• If we plot the demand schedule (in the 1st quadrant) we get the demand curve.
• Draw a graph of the information in the flollowing table, with price on the y-axis and quantity demanded (Qd) on the x-axis.
• Then plot each price-quantity combination as shown here, and we get the demand curve
• The demand curve has a negative (downward) slope.
• As price goes up, the quantity demanded goes down and vice-versa.
• So what the demand curve shows us is the price, and a specific quantity demanded at that price.
• For example, here at 6\$/pound, we demand 4000 pounds of this good (say coffee.)
• Here, at \$ 2.10 price, quantity demanded is 990 pounds. Then as the price keeps on increasing (say by 10 cents each time), the quantity demnaded keeps on dropping steadily. Join them just like above, and we get the demand curve. Thus the important point here is that at each price point (y-axis), there will be a definite Qd point (x-axis). Thus there is a 1:1 (one to one correspondence) between price and Qd. For one price we have one and only one specific Qd. Pick a price, any price. Then draw a horizontal line till it hits the demand curve. Then drop a vertical from there on the x-axis. This will give us the quantity demanded at that specific price.
Example 2:
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# Combining Meshes / SubMeshes
## Recommended Posts
I'm trying to wrap my head around submeshes so that I can built something similar to BABYLON.Mesh.MergeMeshes, but maintain separation via sub-meshes and materials via mutlimaterial
In this PG, http://www.babylonjs-playground.com/#28EUMN#21, I cannot seem to comprehend the relationship of indexStart and indexCount
The same goes for this PG, http://www.babylonjs-playground.com/#1FSS3J - how do I color both faces of the Y axis?
##### Share on other sites
Let's imagine you have 30 indices (10 faces).
IndexStart for the global object = 0
IndexCount for the gobal object = 30
Now let's say you want to create a submesh for face 1 to 3:
indexStart = 3 (index 0, 1 and 2 are for the first face)
indexCount = 3 * 3 (3 faces)
Does it makes sense?
##### Share on other sites
Somewhat. In the 2nd PG example, how would I go about making the positive Y face(s) green as well? I think this would help clear it up for me.
I expected this to work:
`var submesh1 = new BABYLON.SubMesh(0, 0, verticesCount, 0, 12, box);var submesh2 = new BABYLON.SubMesh(1, 0, verticesCount, 12, 12, box);var submesh3 = new BABYLON.SubMesh(2, 0, verticesCount, 24, 12, box);`
24 vertices & 36 indices total
Scratch that, I copied from another PG example and didn't notice it was using emissiveColor. It does show it working as expected now that it's diffuseColor: http://www.babylonjs-playground.com/#1FSS3J#1
The sphere one was tricky to me because of the numbers it used. I understand now the numbers depend on the segments used in the sphere. Here's a PG with even numbers! http://www.babylonjs-playground.com/#28EUMN#22
Thanks for helping me clear this up!
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Tuesday
May 3, 2016
# Homework Help: physics
Posted by shyne on Friday, September 28, 2012 at 10:00pm.
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 22.2° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 56.0 kg, and the coefficient of kinetic friction between the skis and the snow is 0.138. Calculate the magnitude of the force that the tow bar exerts on the skier.
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# Probability that a random univariate polynomial of degree $n$ is irreducible?
Fix a prime $q.$ Let $p(x) = \sum_{i=0}^{n} p_i x^i \in \mathbb{F}_q[x]$ be a polynomial of degree $n,$ and $p$ is monic $(p_n = 1.)$
What is the probability that $p$ is irreducible:
• Over $\mathbb{F}_q[x]$? What I know so far: In $\mathbb{F}_q[x],$ the number of irreducible polynomials of degree $n$ is $O(\frac{q^n}{n}).$ So the probability should be $O(1/n).$
• Same polynomial, $p(x)$ treated as a polynomial over $\mathbb{Z}[x]$?
Motivation: Fix a prime $q,$ pick a random monic poly $p$ of degree $n,$ such all that coefficients of $p$ < $q.$ Test $p$ for irreducibility over both $\mathbb{Z}$ and $\mathbb{F}_q.$ Let $Pr_q$ denote the probability that $p$ is irreducible over $\mathbb{F}_q$ and $Pr_z$ that $p$ is irreducible over $\mathbb{Z}.$ Can we say something about $Pr_q > Pr_z$ or vice versa? Or they are not comparable at all?
• How are you randomly choosing a polynomial over $\mathbb{Q}[x]$? (You don't mean $\mathbb{Z}_q$, by the way; you mean $\mathbb{F}_q$, and you can strengthen the result to $\Theta(1/n)$.) I'm not sure I understand your motivation: for fixed $p$, if $P$ is irreducible $\bmod p$ then it's irreducible, so for any reasonable notion of "random polynomial" the probability that $P$ is irreducible is at least the probability that it's irreducible $\bmod p$. May 26, 2011 at 21:03
• I've edited the question. Fix a prime $q.$ Pick random poly $p(x)$ where all coefficients are uniformly chosen from $\mathbb{Z}$ and bounded by $q.$
– user2468
May 26, 2011 at 21:24
• This may be useful. In this paper Arthur Benjamin and Curtis Bennett prove that, given two randomly chosen non-constant polynomials in $\mathbf{F}_q$, the probability that they are relatively prime is $1-\frac{1}{q}$. A polynomial is irreducible iff it is relatively prime to every polynomial of smaller, positive degree. So it might be possible to piece these probabilities together to get what you want. This may not be the best route to a solution, though.
– Jeff
May 26, 2011 at 21:33
• @Jeff: thanks. I will read the paper.
– user2468
May 26, 2011 at 21:37
• How are you treating $p$ as a polynomial over $\mathbb{Z}[x]$? Are you identifying $\mathbb{F}_q$ with $\{ 0, 1, ... q - 1 \}$? May 26, 2011 at 21:49
The question in your motivation is much easier to answer than the actual question! But first: the number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_q[x]$ is
$$\frac{1}{n} \sum_{d | n} \mu(d) q^{n/d} = \frac{1}{n} \left( q^n + O(\sqrt{n} q^{n/2}) \right)$$
by Möbius inversion, from which it follows that for fixed $n$ and $q \to \infty$ the probability that a monic polynomial of degree $n$ over $\mathbb{F}_q[x]$ is irreducible is $\frac{1}{n} + O(\sqrt{n} q^{-n/2})$. In particular it approaches $\frac{1}{n}$ exponentially fast.
If $p$ is lifted to a polynomial $\tilde{p} \in \mathbb{Z}[x]$, for example by identifying $\mathbb{F}_q$ with $\{ 0, 1, ... q-1 \}$, then if $p$ is irreducible, so is $\tilde{p}$. It follows that the probability that $\tilde{p}$ is irreducible is at least the probability that $p$ is irreducible.
To show strict inequality it suffices to exhibit a monic integer polynomial of degree $n$ which is irreducible but reducible $\bmod q$. There are many ways to do this (for $n \ge 2$ of course). For example, if $q \ge 5$ then $p(x) = x^n + (q - 2) x^{n-1} + 1$ is irreducible by Perron's criterion, but $p(1) = q$. If $q \ge 3$ and $\ell$ is a prime dividing $q - 1$, then $p(x) = x^n + \sum_{i=1}^{n-1} c_i \ell x^i + \ell$ is irreducible by Eisenstein's criterion, but by choosing the $e_i$ appropriately we can arrange to have $p(1) = q$ again. If $q = 2$ then I think one can use complex-analytic tricks but I don't know any good way to generate examples off the top of my head.
There are MO discussions about variants of your second question here and here. This is a hard circle of questions, and I am pretty sure the answer to the second problem as written depends strongly on how you identify elements of $\mathbb{F}_q$ with integers.
• It follows that the probability that p˜ is irreducible is at least the probability that p is irreducible. This statement, and the lifting answers my question.
– user2468
May 26, 2011 at 22:09
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# Single Moment Microphysics Model
The conversion rates among the moist hydrometeors are parameterized assuming that
$\frac{\partial N_{m}}{\partial D} = n_{m}\left(D_{m}\right) = N_{0m} exp \left(-\lambda _{m} D_{m}\right)$
where $$N_{0m}$$ is the intercept parameter, $$D_{m}$$ is the diameters, and
$\lambda_{m} = (\frac{\pi \rho_{m} N_{0m}}{q_{m}\rho})^{0.25}$
where $$\rho_{m}$$ is the density of moist hydrometeors. Assuming that the particle terminal velocity
$v_{m} \left( D_{m},p \right) = a_{m}D_{m}^{b_{m}}\left(\frac{\rho_{0}}{\rho}\right)^{0.5}$
The total production rates including the contribution from aggregation, accretion, sublimation, melting, bergeron process, freezing and autoconversion are listed below without derivation. For details, please refer to Yuh-Lang Lin et al (J. Climate Appl. Meteor, 22, 1065, 1983) and Marat F. Khairoutdinov and David A. Randall’s (J. Atm Sciences, 607, 1983). The implementation of microphysics model in ERF is similar to the that in the SAM code (http://rossby.msrc.sunysb.edu/~marat/SAM.html)
## Accretion
There are several different type of accretional growth mechanisms that need to be included; these describe the interaction of water vapor and cloud water with rain water.
The accretion of cloud water forms in either the dry or wet growth rate can be written as:
$Q_{gacw} = \frac{\pi E_{GW}n_{0G}q_{c}\Gamma(3.5)}{4\lambda_{G}^{3.5}}(\frac{4g\rho G}{3C_{D}\rho})^{0.5}$
The accretion of raindrops by accretion of cloud water is
$Q_{racw} = \frac{\pi E_{RW}n_{0R}\alpha q_{c}\Gamma(3+b)}{4\lambda_{R}^{3+b}}(\frac{\rho_{0}}{\rho})^{1/2}$
## The bergeron Process
The cloud water transform to snow by deposition and rimming can be written as
$Q_{sfw} = N_{150}\left(\alpha_{1}m_{150}^{\alpha_{2}}+\pi E_{iw}\rho q_{c}R_{150}^{2}U_{150}\right)$
## Autoconversion
The collision and coalescence of cloud water to from raindrops is parameterized as following
$Q_{raut} = \rho\left(q_{c}-q_{c0}\right)^{2}\left[1.2 \times 10^{-4}+{1.569 \times 10^{-12}N_{1}/[D_{0}(q_{c}-q_{c0})]}\right]^{-1}$
## Evaporation
The evaporation rate of rain is
$Q_{revp} = 2\pi(S-1)n_{0R}[0.78\lambda_{R}^{-2}+0.31S_{c}^{1/3}\Gamma[(b+5)/2]a^{1/2}\mu^{-1/2}(\frac{\rho_{0}}{\rho})^{1/4}\lambda_{R}^{(b+5)/2}](\frac{1}{\rho})(\frac{L_{v}^{2}}{K_{0}R_{w}T^{2}}+\frac{1}{\rho r_{s}\psi})^{-1}$
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# NCERT Solutions for Class 8 Maths Chapter 2 (Ex 2.2) Linear Equations in One Variable
## NCERT Solutions for Class 8 Chapter 2 Linear Equations in One Variable -Free PDF Download
Free PDF download of NCERT Solutions Maths Class 8 Solutions Chapter 2 – Linear Equations in One Variable solved by Expert Maths Teachers on CoolGyan.Org. All Chapter 2 – Linear Equations in One Variable Questions with Solutions for NCERT to help you to revise complete Syllabus and Score More marks.
Maths Revision Notes for Class 8
Chapter Name Linear Equations in One Variable Chapter Chapter 2 Exercise Exercise 2.2 Class Class 8 Subject Maths NCERT Solutions Board CBSE TEXTBOOK CBSE NCERT Category NCERT Solutions
# NCERT SOLVED
1. If you subtract from a number and multiply the result by you get What is the number?
Ans. Let the number be
According to the question,
[Multiplying both sides by 2]
[Adding to both sides ]
Hence, the required number is
2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth?
Ans. Let the breadth of the pool be m.
Then, the length of the pool = m
Perimeter =
154 =
[Dividing both sides by 2]
[Subtracting 2 from both sides]
[Dividing both sides by 3]
m
Hence, length of the pool =
= 50 + 2 = 52 m
And,breadth of the pool = 25 m.
3. The base of an isosceles triangle is cm. The perimeter of the triangle is cm. What is the length of either of the remaining equal sides?
Ans. Let each of equal sides of an isosceles triangle be cm.
Perimeter of a triangle = Sum of all three sides
[Subtracting from both the sides]
[Dividing both sides by 2]
cm
Hence, each equal side of an isosceles triangle is cm.
4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Ans. Sum of two number = 95
Let the first number be
then another number be .
According to the question,
[Subtracting 15 from both sides]
[Dividing both sides by 2]
Hence, the first number = 40
And another number = 40 + 15 = 55.
5. Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Ans. Let the two numbers be and
According to question,
[Dividing both sides by 2]
Hence, first number = = 45 and second number = = 27.
6. Three consecutive integers add up to 51. What are these integers?
Ans. Let the three consecutive integers be and
According to the question,
[Subtracting 3 from both sides]
[Dividing both sides by 3]
Hence, first integer = 16,
second integer = 16 + 1 = 17 and
third integer = 16 + 2 = 18.
7. The sum of three consecutive multiples of 8 is 888. Find the multiples.
Ans. Let the three consecutive multiples of 8 be and
According to question,
[Subtracting 24 from both sides]
[Dividing both sides by 3]
Hence, first multiple of 8 = 288,
second multiple of 8 = 288 + 8 = 296 and
third multiple of 8 = 288 + 16 = 304.
8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Ans. Let the three consecutive integers be and
According to the question,
[Subtracting 11 from both sides]
[Dividing both sides by 9]
Hence first integer = 7, second integer
= 7 + 1 = 8 and third integer = 7 + 2 = 9.
9. The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Ans. Let the present ages of Rahul and Haroon be years and years respectively.
According to condition,
[Subtracting 8 from both sides]
[Dividing both sides by 12]
Hence, present age of Rahul = = 20 years and
present age of Haroon = = 28 years.
10. The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Ans. Let the number of girls be
Then, the number of boys =
According to the question,
[Transposing to L.H.S. and 40 to R.H.S.]
[Dividing both sides by ]
Hence the number of girls = 20 and number of boys = 20 + 8 = 28.
11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Ans. Let Baichung’s age be years,
then Baichung’s father’s age = years
and Baichung’s grandfather’s age = years.
According to condition,
[Subtracting 84 from both sides]
[Dividing both sides by 3]
years
Hence, Baichung’s age = 17 years,
Baichung’s father’s age = 17 + 29 = 46 years
And Baichung’s grandfather’s age = 17 + 29 + 26 = 72 years.
12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Ans. Let Ravi’s present age be years.
After fifteen years, Ravi’s age = years.
Fifteen years from now, Ravi’s age = years.
According to question,
[Transposing to L.H.S.]
[Dividing both sides by 3]
years
Hence, Ravi’s present age be 5 years.
13. A rational number is such that when you multiply it by and add to the product, you get What is the number?
Ans. Let the rational number be
According to the question,
[Subtracting from both sides]
[Dividing both sides by 60]
Hence, the rational number is
14. Lakshmi is a cashier in a bank. She has currency notes of denominations Rs. 100, Rs. 50 and Rs. 10 respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is Rs. 4,00,000. How many notes of each denomination does she have?
Ans. Let number of notes be and
According to question,
[Dividing both sides by 400]
Hence, number of denominations of Rs. 100 notes = = 2,000
Number of denominations of Rs. 50 notes = = 3,000
Number of denominations of Rs. 10 notes = = 5000
Therefore, required denominations of notes of Rs. 100, Rs. 50 and Rs. 10 are 2000, 3000 and 5000 respectively.
15. I have a total of Rs. 300 in coins of denomination Re. 1, Rs. 2 and Rs. 5. The number of Rs. 2 coins is 3 times the number of Rs. 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Ans. Total sum of money = Rs. 300
Let the number of Rs. 5 coins be
number of Rs. 2 coins be and
number of Re. 1 coins be
According to question,
[Subtracting 160 from both sides]
[Dividing both sides by 7]
Hence, the number of coins of Rs. 5 denomination = 20
Number of coins of Rs. 2 denomination = = 60
Number of coins of Rs. 1 denomination = = 160 – 80 = 80
16. The organizers of an essay competition decide that a winner in the competition gets a prize of Rs. 100 and a participant who does not win, gets a prize of Rs. 25. The total prize money distributed is Rs. 3,000. Find the number of winners, if the total number of participants is 63.
Ans. Total sum of money = Rs. 3000
Let the number of winners of Rs. 100 be
And those who are not winners =
According to the question,
75x + 1575 – 1575 = 3000 – 1575 [Subtracting 1575 from both sides]
75x = 1425
75x75=14257575×75=142575 [Dividing both sides by 75]
Hence, the number of winner is 19.
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https://www.mytutor.co.uk/answers/8506/GCSE/Maths/Expand-and-simplify-x-6-x-6/
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Expand and simplify (x+6)(x-6)
The first step is to expand the expression. This can be done using the FOIL method. This means that we multiply the First, Outside, Inside and Last terms in the bracket.First: x . x = x2Outside: x . -6 = -6xInside: 6 . x = 6xLast: 6 . -6 = -36Now we have multiplied the brackets out, we need to add all the terms together.= x+ (-6x) + 6x - 36Now we simplify:= x2-36
Answered by Monika P. Maths tutor
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Shuffle - Maple Help
Statistics
Shuffle
apply random permutation to a data sample
Calling Sequence Shuffle(X, options)
Parameters
X - options - (optional) equation of the form inplace = truefalse; indicates whether to change the argument
Description
• The Shuffle command applies a random permutation to X.
• The first parameter X is a data sample - given as e.g. a Vector.
• An option inplace = true or inplace = false can be specified as a second argument. With inplace = true, Shuffle will change X in place if it is an Array or a Vector; with inplace = false, Shuffle will return a new (shuffled) copy of X. If X is a list, then Shuffle returns a new copy either way. The default is inplace = false.
The option inplace = true can be abbreviated to inplace.
Notes
• The Shuffle command accepts non-numeric data.
Examples
> $\mathrm{with}\left(\mathrm{Statistics}\right):$
> $A≔\mathrm{Array}\left(\left[a,b,c,d,e,f,g,h,i,j\right]\right)$
${A}{≔}\left[\begin{array}{cccccccccc}{a}& {b}& {c}& {d}& {e}& {f}& {g}& {h}& {i}& {j}\end{array}\right]$ (1)
> $\mathrm{A1}≔\mathrm{Shuffle}\left(A\right)$
${\mathrm{A1}}{≔}\left[\begin{array}{cccccccccc}{g}& {d}& {e}& {a}& {i}& {f}& {b}& {h}& {j}& {c}\end{array}\right]$ (2)
> $\mathrm{A2}≔\mathrm{Shuffle}\left(A\right)$
${\mathrm{A2}}{≔}\left[\begin{array}{cccccccccc}{c}& {e}& {j}& {g}& {a}& {f}& {b}& {h}& {i}& {d}\end{array}\right]$ (3)
> $\mathrm{A3}≔\mathrm{Shuffle}\left(A\right)$
${\mathrm{A3}}{≔}\left[\begin{array}{cccccccccc}{c}& {d}& {i}& {b}& {h}& {e}& {f}& {j}& {g}& {a}\end{array}\right]$ (4)
> $\mathrm{sort}\left(\mathrm{A1}\right)$
$\left[\begin{array}{cccccccccc}{a}& {b}& {c}& {d}& {e}& {f}& {g}& {h}& {i}& {j}\end{array}\right]$ (5)
> $B≔\mathrm{Array}\left(\left[\mathrm{seq}\left(i,i=1..10\right)\right]\right)$
${B}{≔}\left[\begin{array}{cccccccccc}{1}& {2}& {3}& {4}& {5}& {6}& {7}& {8}& {9}& {10}\end{array}\right]$ (6)
> $C≔\mathrm{Shuffle}\left(B\right)$
${C}{≔}\left[\begin{array}{cccccccccc}{10}& {3}& {5}& {1}& {2}& {6}& {9}& {4}& {8}& {7}\end{array}\right]$ (7)
> $\mathrm{OrderByRank}\left(A,C\right)$
$\left[\begin{array}{cccccccccc}{d}& {e}& {b}& {h}& {c}& {f}& {j}& {i}& {g}& {a}\end{array}\right]$ (8)
So far, $A$ itself has not changed. The inplace option changes this.
> $A$
$\left[\begin{array}{cccccccccc}{a}& {b}& {c}& {d}& {e}& {f}& {g}& {h}& {i}& {j}\end{array}\right]$ (9)
> $\mathrm{Shuffle}\left(A,\mathrm{inplace}\right):$$A$
$\left[\begin{array}{cccccccccc}{e}& {d}& {h}& {i}& {b}& {j}& {g}& {c}& {f}& {a}\end{array}\right]$ (10)
Note that the output of Maple's random number generator is reproducible after a restart. If this is undesirable, one can use the randomize command.
> $\mathrm{restart}$
> $\mathrm{with}\left(\mathrm{Statistics}\right):$
> $A≔\mathrm{Array}\left(\left[a,b,c,d,e,f,g,h,i,j\right]\right)$
${A}{≔}\left[\begin{array}{cccccccccc}{a}& {b}& {c}& {d}& {e}& {f}& {g}& {h}& {i}& {j}\end{array}\right]$ (11)
> $\mathrm{A1}≔\mathrm{Shuffle}\left(A\right)$
${\mathrm{A1}}{≔}\left[\begin{array}{cccccccccc}{g}& {d}& {e}& {a}& {i}& {f}& {b}& {h}& {j}& {c}\end{array}\right]$ (12)
> $\mathrm{restart}$
> $\mathrm{with}\left(\mathrm{Statistics}\right):$
> $A≔\mathrm{Array}\left(\left[a,b,c,d,e,f,g,h,i,j\right]\right)$
${A}{≔}\left[\begin{array}{cccccccccc}{a}& {b}& {c}& {d}& {e}& {f}& {g}& {h}& {i}& {j}\end{array}\right]$ (13)
> $\mathrm{randomize}\left(\right)$
${3368728095}$ (14)
> $\mathrm{A1}≔\mathrm{Shuffle}\left(A\right)$
${\mathrm{A1}}{≔}\left[\begin{array}{cccccccccc}{h}& {f}& {d}& {e}& {j}& {b}& {g}& {a}& {c}& {i}\end{array}\right]$ (15)
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# Richard Willey
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I recommend that you look at the following example from the file exchange http://www.mathworks.com/matlabcentral/fileexchange...
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In general when I hear the expression parametric bootstrap I think "parametric residual bootstrap" I'm attaching some simple ...
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How do I Regression Fit a SinWave to a dataset?
Here's some simple code that illustrates how to perform nonlinear regression using the 12a release of Statistics Toolbox. Not...
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Homework help.
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I'm attaching code that shows a couple different ways to solve your problem. I prefer the second option. The R^2 is slightly...
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Hi ZazOufUMl A "Poly31" model is different than the one that I suggested. (Poly31 will contain a number of cross terms). ...
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Hi there The "Poly54" fit type is specifying a 5th order polynomial in one direction and a 4th order polynomial in the other....
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If Pv and Pa can both be described as functions of a, you might be able to use a univariate generator to generate plausible valu...
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Multicolinearity/Regression/PCA and choice of optimal model (2nd try)
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From the sounds of things, the simplest thing to do would be to generate all of your random numbers OUTSIDE your loop. Start ...
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Surface Fitting Tool Coefficient Structure
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I have a function called FitIt on the file exchange that might prove useful. FitIt combines Local regression (to smooth yo...
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Get Y values from Curve Fitting Tool
For simplicity, this code is using a simple linear model. However, the same syntax with work for your nonlinear regression. ...
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Here's a simple example using the new regression functions in the 12a release. All you need to do is pass the appropriate weigh...
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Multi-parametric fit with matlab
Hi Miguel fitensemble is able to handle multiple independent variables. You should be able to use this with the data set tha...
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Multi-parametric fit with matlab
Hi Miguel From the sounds of things, your central problem is that you're unable to specify an equation that describes the rel...
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The SVM implementation in Bioinformatics Toolbox is limited to binary classification. The Naive Bayes classifier in Statistic...
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How to obtain Std of Coefficients from Curve Fitting
Hi George Conveniently, 12a also has a function call NonLinearModel %% Generate some data X = 2* pi*rand(100,1); ...
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A nutritionist is studying the effects of the nutrients folic acid, choline, and inositol. He has three types of food available, and each type contains the following amounts of these nutrients per ounce.
Type A Type B Type C
Folic acid (mg) 4 2 4
Choline (mg) 3 2 3
Inositol (mg) 3 2 4
(a) Find the inverse of the matrix below and use it to solve the remaining parts of this problem.
4 2 4
3 2 3
3 2 4
(b) How many ounces of each food should the nutritionist feed his laboratory rats if he wants their daily diet to contain 18 mg of folic acid, 15 mg of choline, and 17 mg of inositol?
Type A
oz
Type B
oz
Type C
oz
(c) How much of each food is needed to supply 10 mg of folic acid, 9 mg of choline, and 10 mg of inositol?
Type A
oz
Type B
oz
Type C
oz
(d) Will any combination of these foods supply 6 mg of folic acid, 6 mg of choline, and 7 mg of inositol?
THERE IS A COMBINATION OF FOODS GIVING THE REQUIRED SUPPLY.
THERE IS NO COMBINATION OF FOODS GIVING THE REQUIRED SUPPLY.
THIS CANNOT BE DETERMINED.
We have used the Zero-Product Property to solve algebraic equations. Matrices do not have this property. Let O represent the 2 × 2 zero matrix.
O =
0 0
0 0
FIND 2 × 2 MATRICES
A ≠ O and B ≠ O such that AB = O.
×
0 0
9 0
= O
CAN YOU FIND A MATRIX
A ≠ O such that A2 = O?
Sample Solution
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https://drserendipity.com/notes/notes_by_subjects/artificial_intelligence/computer-vision/1-introduction-to-computer-vision/1-5-convolutional-filters-and-edge-detection/4-low-pass-filters/
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# 4 – Low-pass Filters
Earlier you saw an example of noise in an image of San Francisco City Hall. This noise is generally seen as speckles or discoloration in an image and it doesn’t contain any useful information. It might even mess with processing steps such as an edge detection when high pass filters can amplify noise if it’s not removed first. The most common way to remove noise is by using a low pass filter. These filters block certain high frequency content and effectively blur or smooth the appearance of an image, and this reduces high frequency noise. An example where this is very useful is in medical images which typically have noise that’s produced either by the imagery machinery or by a moving human subject. Let’s take a closer look at this cross-sectional image of a human brain. We can clearly see the outline of a skull and brain but we also see a lot of sort of salt and pepper speckle, and this is high frequency noise. You can imagine if we applied a high pass filter with the goal of detecting edges, we would detect and amplify a lot of this spotty noise. We can reduce this noise by taking a kind of average between neighboring pixels so that there are not such big jumps in intensity especially in small areas. This averaging of pixels in space is equivalent to implementing a low pass filter that blocks high frequency noise. Let’s see an example of a common kernel that does this. The first and simplest is the averaging filter. It’s a three by three kernel that weights are sent to pixel and its surrounding pixels all the same. Low pass filters typically take an average and not a difference as high pass filters do. So their components should all add up to one. This will preserve the image brightness and make sure that it doesn’t get brighter or darker overall. But we can see that the components of this kernel add up to nine. So we need to normalize by dividing the total value of the kernel by nine. Then our total sum becomes one. Now let’s go back to our main image and see what a convolution looks like between the image F of XY and the kernel K. Here’s a zoomed in portion of the image. To perform convolution we put our three by three kernel on top of each pixel in our image. I’ll choose this dark one with a value of 40 as our center pixel. Then looking at all the values in this three by three square, we perform multiplication in pairs using the weights in our kernel to multiply all the pixel values. In this case, we’re multiplying all the values by one and summing them up. Our last step will be dividing by nine to normalize the image and get a corresponding output pixel value, 85. We can see that this is just an average of the center pixel and its surrounding neighbor pixels. Since the surrounding pixels are mostly brighter than the center pixel, the new output pixel value is brighter too. If we do this with all the pixels in this image, we’ll get an average to smoothed out image with fewer abrupt changes in intensity. This can be useful for blowing out noise or making a background area within a certain intensity range look more uniform. In fact, this sort of filter is even used in Photoshop to soften and blur parts of an image. Next, let’s see how to do this in code.
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Exam2 Fall2011
# Exam2 Fall2011 - BE 1300 Exam 2 11:45-1:10 PM NAME ID Quiz...
• Test Prep
• 10
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BE 1300 Exam 2, October 20, 2011, 11:45-1:10 PM 1 NAME ____________________ ID# _______________ Quiz Section (Time, Day, and TA's name): ______________________________________________ CLOSED BOOK/CLOSED NOTES/CELL PHONES OFF: Nothing except this exam, your calculator (devoid of programmed equations) and a pen may be on your desk or in your line-of-sight at any time. Do not use pencils. Any scratch work may be done on the reverse sides of this exam. However, credit for anything on the reverse side of the exam will not be given unless you specifically mention that the problem is continued on the back of the page. College policy states that anyone giving or receiving information during an exam will be given an immediate failing grade for the course. I understand and agree to abide by the above rules. SIGNATURE: ___________________________________ NO EXAM WILL BE GRADED WITHOUT A SIGNATURE. DO NOT PRINT. Scores: For use of graders only Problem One _______ Problem Two _______ Problem Three _______ Problem Four _______ Problem Five _______ Problem Six _______ TOTAL___________________
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BE 1300 Exam 2, October 20, 2011, 11:45-1:10 PM 2 You may need the some of following information and formulas: k = 8.62 10 -5 eV/(atom K). N A = 6.023 10 23 atoms/mol. a = 2 3/2 R for FCC unit cell. a = 4 R /3 1/2 for BCC unit cell. Atomic Packing Factor APF = V Sphere / V Cell . A C N V nA R = 8.314 J/mol-K. N v (number of vacancies) = N (number of total lattice sites) exp( Q v / kT ). Fick’s first law: dx dC D J . Special case of Fick’s second law: Dt x erf C C C C s x 2 1 0 0 . RT Q D D d exp 0 .
BE 1300 Exam 2, October 20, 2011, 11:45-1:10 PM 3
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BE 1300 Exam 2, October 20, 2011, 11:45-1:10 PM 4 Do Problem One and any four of Problems Two, Three, Four, Five, and Six. Each problem is worth 20 points.
• Fall '14
• Fick
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### What will happen to the value of the mexican peso
Assignment Help Finance Basics
##### Reference no: EM13288193
If Mexicans go on a spending spree and buy twice as much french perfume, japanese tvs, english sweaters, swiss watches, and italian wine, what will happen to the value of the mexican peso?
#### What will happen to the u.s. exchange rate
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A currency trader observes that in the spot exchange marker, 1 U.S. dollar can be exchanged for 3.50 Israeli shekels or for 104.00 Japanese yen. What is the cross-exchange r
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a. Calculate the companys disbursement float, collection float, and net float. b. How would you answer to part (a) change if the collected funds were available in one day in
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Xerox has an 8.75% semi-annual coupon bond that has a remaining maturity of 16 years. the bond is callable in three years at a price of \$1100. its current price is \$1250.
#### How much the difference in the value of the iras
Two brothers each open IRAs in 2009 and plan to invest \$3,000 per year for the next 30 years. John makes his first deposit on January 1, 2009, and will make all future depos
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A = [−3, 4) B = {x ∈ R : −4 < x ≤ 2 or x > 3}
A∪B={x∈ℝ:-4<x≤2, 4 | x>3} because A is a subset of B. I use “|” to mean “or”.
by Top Rated User (762k points)
A or B = { -3, -2, -1, 0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 11........ }
by Level 1 User (260 points)
reshown by
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# Knowledge Base
### ValueWhen() Function
The VALUEWHEN function can be used to calculate the open, high, low, or close price or indicator value when a certain technical condition occurs. It can also be used with the BarDate/StrDate function to get a value as of a specific date.
For example, to find out the closing price of a stock when the 50MA last crossed below the 200MA:
V1 = MA(BARS=50, CALC=Close) CrossesBelow MA(BARS=200, CALC=Close);
VALUEWHEN(V1)
To find the high price on the day of the cross:
V1 = MA(BARS=50, CALC=Close) CrossesBelow MA(BARS=200, CALC=Close);
VALUEWHEN(HIGH(), V1)
To find the value of the MA50 when they crossed:
V1 = MA(BARS=50, CALC=Close) CrossesBelow MA(BARS=200, CALC=Close);
VALUEWHEN(MA(BARS=50, CALC=Close), V1)
To find the RSI(14) value when they crossed:
V1 = MA(BARS=50, CALC=Close) CrossesBelow MA(BARS=200, CALC=Close);
VALUEWHEN(RSI(BARS=14), V1)
To find the value of the last 10 bar pivot high:
VALUEWHEN(HIGH(),PIVOT(MIN=10, TYPE=High))
To find the RSI value on a specific date, eg March 23rd 2020:
D1 = BARDATE()==STRDATE(DATE=2020-03-23);
VALUEWHEN(RSI(BARS=14), D1)
#### Please Wait!
Please wait... it will take a second!
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Coloring picture fuzzy graphs through their cuts and its computation
(1) * Isnaini Rosyida (Department of Mathematics, Universitas Negeri Semarang, Indonesia)
(2) Suryono Suryono (Department of Doctor of Information System, Universitas Diponegoro, Indonesia)
*corresponding author
Abstract
In a fuzzy set (FS), there is a concept of alpha-cuts of the FS for alpha in [0,1]. Further, this concept was extended into (alpha,delta)-cuts in an intuitionistic fuzzy set (IFS) for delta in [0,1]. One of the expansions of FS and IFS is the picture fuzzy set (PFS). Hence, the concept of (alpha,delta)-cuts was developed into (alpha,delta,beta)-cuts in a PFS where beta is an element of [0,1]. Since a picture fuzzy graph (PFG) consists of picture fuzzy vertex or edge sets or both of them, we have an idea to construct the notion of the (alpha,delta,beta)-cuts in a PFG. The steps used in this paper are developing theories and algorithms. The objectives in this research are to construct the concept of (alpha,delta,beta)-cuts in picture fuzzy graphs (PFGs), to construct the (alpha,delta,beta)-cuts coloring of PFGs, and to design an algorithm for finding the cut chromatic numbers of PFGs. The first result is a definition of the (alpha,delta,beta)-cut in picture fuzzy graphs (PFGs) where (alpha,delta,beta) are elements of a level set of the PFGs. Further, some properties of the cuts are proved. The second result is a concept of PFG coloring and the chromatic number of PFG based on the cuts. The third result is an algorithm to find the cuts and the chromatic numbers of PFGs. Finally, an evaluation of the algorithm is done through Matlab programming. This research could be used to solve some problems related to theories and applications of PFGs.
Keywords
Picture fuzzy graph; Coloring graph; Cuts; Chromatic number; Intuitionistic fuzzy set
DOI
https://doi.org/10.26555/ijain.v7i1.612
Article metrics
Abstract views : 952 | PDF views : 213
References
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[21] M. Sitara, M. Akram, and M. Riaz, “Decision-Making Analysis Based on q-Rung Picture Fuzzy Graph Structures,” J Appl Math Comput ., 2021, doi: 10.1007/s12190-020-01471-z.
[22] P. Mani, B. Vasudevan, and M. Sivaraman, “Shortest path algorithm of a network via picture fuzzy digraphs and its application,” Mater. Today Proc., 2021, doi: 10.1016/j.matpr.2020.12.006.
[23] M. Pal, S. Samanta, and G. Ghorai, “Coloring of Fuzzy Graph,” 2020, 1st ed., pp. 175–193, doi: 10.1007/978-981-15-8803-7_7.
[24] I. Rosyida, Widodo, C. R. Indrati, D. Indriati, and Nurhaida, “Fuzzy Chromatic Number of Union of Fuzzy Graphs: An Algorithm, Properties and its Application,” Fuzzy Sets Syst., vol. 384, pp. 115–131, 2020, doi: 10.1016/j.fss.2019.04.028.
[25] I. Rosyida, J. Peng, L. Chen, W. Widodo, C. R. Indrati, and K. A. Sugeng, “An Uncertain Chromatic Number of an Uncertain Graph Based on α-Cut Coloring,” Fuzzy Optim Decis Mak., vol. 17, pp. 103–123, 2018, doi: 10.1007/s10700-016-9260-x.
[26] I. Rosyida, Widodo, C. R. Indrati, and D. Indriati, “On construction of fuzzy chromatic number of cartesian product of path and other fuzzy graphs,” J. Intell. Fuzzy Syst., vol. 39, no. 1, pp. 1073–1080, 2020, doi: 10.3233/JIFS-191982.
[27] I. Rosyida, Widodo, C. R. Indrati, and K. A. Sugeng, “A New Approach for Determining Fuzzy Chromatic Number of Fuzzy Graph”, J. Intell. Fuzzy Syst., vol. 28, no. 5, pp. 2331–2341, 2015, doi: 10.3233/IFS-141521.
[28] S. Ismail Mohideen and M. . Rifayathali, “Coloring of intuitionistic fuzzy graph using (alpha,beta)-cuts.,” Int. Res. J. Math. Eng. IT, vol. 2, no. 12, pp. 14–26, 2015. Available at: Google Scholar.
[29] A. Prasanna, M. . Rifayathali, and S. Ismail Mohideen, “Strong intuitionistic fuzzy graphs,” Int. J. Latest Eng. Res. Appl., vol. 02, no. 08, pp. 163–169, 2017, doi: 10.1007/s41478-018-0102-9.
[30] M. Ismayil, “Domination in Picture Fuzzy Graphs,” in 5th International Conference on Mathematical Methods and Computation (ICOMAC - 2019), 2019, no. February, 20-21, pp. 205–210. Available at: Google Scholar.
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# How to Prepare Amortization Schedule in Excel
An amortization schedule shows the interest applied to a fixed interest loan and how the principal is reduced by payments. It also shows the detailed schedule of all payments so you can see how much is going toward principal and how much is being paid toward interest charges. This wikiHow teaches you how to create your own amortization schedule in Microsoft Excel.
Method 1
Method 1 of 2:
### Creating an Amortization Schedule Manually
1. 1
Open a new spreadsheet in Microsoft Excel.
2. 2
Create labels in column A. Create labels for your data in the first column to keep things organized. Here's what you should put in each cell: .
• A1: Loan Amount
• A2: Interest Rate
• A3: Months
• A4: Payments
3. 3
Enter the information pertaining to your loan in column B. Fill out cells B1-B3 with information about your loan. Leave B4 (the cell next to the Payments label) blank.
• The "Months" value should be the total number of months in the loan term. For example, if you have a 2-year loan, enter 24.
• The "Interest Rate" value should be a percentage (e.g., 8.2%).
4. 4
Calculate your payment in cell B4. To do this, click cell B4, and then type the following formula into the formula (fx) bar at the top of the sheet and then press Enter or Return: =ROUND(PMT(\$B\$2/12,\$B\$3,-\$B\$1,0), 2).
• The dollar signs in the formula are absolute references to make sure the formula will always look to those specific cells, even if it is copied elsewhere into the worksheet.
• The loan interest rate must be divided by 12, since it is an annual rate that is calculated monthly.
• For example, if your loan is for \$150,000 at 6 percent interest for 30 years (360 months), your loan payment will calculate out to \$899.33.
5. 5
Create column headers in row 7. You'll be adding some additional data to the sheet, which requires a second chart area. Enter the following labels into the cells:
• A7: Period
• B7: Beginning Balance
• C7: Payment
• D7: Principal
• E7: Interest
• F7: Cumulative Principal
• G7: Cumulative Interest
• H7: Ending Balance.
6. 6
Populate the Period column. This column will contain your payment dates. Here's what to do:
• Type the month and year of the first loan payment in cell A8. You may need to format the column to show the month and year correctly.
• Click the cell once to select it.
• Drag down from the center of the selected cell downward to cover all cells through A367. If this doesn't make all of the cells reflect the correct monthly payment dates, click the small icon with a lightning bolt on it at the bottom-right corner of the bottommost cell and make sure the Last Month option is selected.
7. 7
Fill out the other entries in cells B8 through H8.
• The beginning balance of your loan into cell B8.
• In cell C8, type =\$B\$4 and press Enter or Return.
• In cell E8, create a formula to calculate the loan interest amount on the beginning balance for that period. The formula will look like =ROUND(\$B8*(\$B\$2/12), 2). The single dollar sign creates a relative reference. The formula will look for the appropriate cell in the B column.
• In cell D8, subtract the loan interest amount in cell E8 from the total payment in C8. Use relative references so this cell will copy correctly. The formula will look like =\$C8-\$E8.
• In cell H8, create a formula to subtract the principal portion of the payment from the beginning balance for that period. The formula will look like =\$B8-\$D8.
8. 8
Continue the schedule by creating the entries in B9 through H9.
• Cell B9 should include a relative reference to the ending balance of the prior period. Type =\$H8 into B9 and press Enter or Return.
• Copy cells C8, D8 and E8 and paste them into C9, D9 and E9 (respectively)
• Copy H8 and paste it into H9. This is where the relative reference becomes helpful.
• In cell F9, create a formula to tabulate cumulative principal paid. The formula will look like this: =\$D9+\$F8.
• Enter the cumulative interest formula into G9 like this: =\$E9+\$G8.
9. 9
Highlight cells B9 through H9. When you rest the mouse cursor over the bottom-right part of the highlighted area, the cursor will turn to a crosshair.
10. 10
Drag the crosshair all the way down to row 367. This populates all the cells through row 367 with the amortization schedule.
• If this looks funny, click the small spreadsheet-looking icon at the bottom-right corner of the final cell and select Copy Cells.
Method 2
Method 2 of 2:
### Using an Excel Template
1. 1
Go to https://templates.office.com/en-us/loan-amortization-schedule-tm03986974. This is a free, downloadable amortization schedule template that makes it easy to calculate the total interest and total payments. It even includes the option to add extra payments.[1]
2. 2
3. 3
• The data in the template is there as an example—you'll be able to add your own data.
• If prompted, click Enable Editing so you can make changes to the workbook.
4. 4
Type the loan amount into the "Loan Amount" cell. It's in the "ENTER VALUES" section near the top-left corner of the sheet. To type it, just click the existing value (\$5000) and type your own amount.
• When you press Return or Enter (or click another cell), the amounts in the rest of the sheet will recalculate. This will happen each time you change a value in this section.
5. 5
Enter your annual interest rate. This goes into the "Annual interest rate" cell.
6. 6
Enter the duration of your loan (in years). This goes into the "Loan period in years" cell.
7. 7
Enter the number of payments you make per year. For example, if you make payments once per month, type 12 into the "Number of payments per year" cell.
8. 8
Enter the loan start date. This goes into the "Start date of loan" cell.
9. 9
Enter a value for "Optional extra payments." If you pay over the minimum amount due on your loan each pay period, enter that extra amount into this cell. If not, change the default value to 0 (zero).
10. 10
Enter the name of the loan issuer. The default value of the "LENDER NAME" blank is "Woodgrove Bank." Change this to your bank's name for your own reference.
11. 11
Save the worksheet as a new Excel file. Here's how:
• Click the File menu at the top-left and select Save As.
• Select a location on your computer or in the cloud where you'd like to store your schedule.
• Enter a name for the file. If the file type is not already set to "Excel Workbook (*.xlsx)," select that option from the drop-down menu (below the file name) now.
• Click Save.
## Community Q&A
Search
• Question
How can I change the currency?
The currency is dependent on the format of the cell with dollar amounts in them. Right click on the cell you want to change and click on "Format." Then choose "Currency" and put in the appropriate codes for the currency you choose.
• Question
How do I do a weekly payment loan?
You have to change the periods. Try this: Change "Months" in cell A3 to "Periods." Change the "Payments" formula in cell B4, changing the "12" (which represents months) to "52" which represents the number of weeks in a year. The new formula will look like this: "=ROUND(PMT(\$B\$2/52,\$B\$3,-\$B\$8,0), 2)." You'll also need to change your periods starting in cell A9 to add 7 days instead of 1 month. I would use this formula in cell A9: "=DATE(YEAR(A8),MONTH(A8),DAY(A8)+7." Lastly, copy your new formula down to the rest of the schedule so all the periods will be correct.
• Question
How do I change the formula for quarterly payments rather than monthly payments?
Wherever there is a field being populated with the =ROUND(...) formula, replace the "12" entry with a "4." Example: In cell E8, create a formula to calculate the loan interest amount on the beginning balance for that period. The formula will look like "=ROUND(\$B8*(\$B\$2/4), 2)"
200 characters left
## Tips
• If you do not receive a final ending balance of \$0.00, make sure you have used the absolute and relative references as instructed and the cells have been copied correctly.
⧼thumbs_response⧽
• You can now scroll to any period during the loan payment to see how much of the payment is applied to the principal, how much is charged as loan interest and how much principal and interest you have paid to date.
⧼thumbs_response⧽
Submit a Tip
All tip submissions are carefully reviewed before being published
Thanks for submitting a tip for review!
## Warnings
• This will only work for home loans that calculate monthly. If the loan is a car loan or a daily compounded loan, this will only give rough estimates for interest paid.
⧼thumbs_response⧽
## Things You'll Need
• Computer
• Microsoft Excel
• Loan details
Written by:
wikiHow Technology Writer
This article was co-authored by wikiHow staff writer, Nicole Levine, MFA. Nicole Levine is a Technology Writer and Editor for wikiHow. She has more than 20 years of experience creating technical documentation and leading support teams at major web hosting and software companies. Nicole also holds an MFA in Creative Writing from Portland State University and teaches composition, fiction-writing, and zine-making at various institutions. This article has been viewed 1,156,997 times.
Co-authors: 11
Updated: August 11, 2022
Views: 1,156,997
Categories: Microsoft Excel
Article SummaryX
2. Open the template in Excel.
3. Fill out your loan data in the "ENTER VALUES" section.
4. Type the bank name as the "LENDER NAME."
5. Save the worksheet as an Excel file.
Thanks to all authors for creating a page that has been read 1,156,997 times.
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# GMAT Tip of the Week: Percents Are Easy, Words Are Hard
Pop quiz: 1) Your restaurant bill came to exactly \$64.00 and you want to leave a 20% tip. How much do you leave? 2) You’re running a charity half-marathon and your fundraising goal is \$6000. You’ve raised \$3300. What percent of your goal have you reached? 3) Your \$20,000 investment is now worth \$35,000. By what percent has your investment increased in value?
If that was easy for you, good. It better have been. After all, you’re applying to graduate school and that’s maybe 6th grade math in three real-life contexts. Percents are not hard! But percent problems can be. And that’s what savvy GMAT test-takers need to learn:
On the GMAT, percent problems aren’t hard because of the numbers. They’re hard because of the words.
Consider two situations:
1) A band sells concert t-shirts online for \$20 each, and in California, web-based sales are subject to a 10% sales tax. How much does a California-based purchaser pay in sales tax after buying a t-shirt?
2) At a concert in California, a band wants to sell t-shirts for \$20. For simplicity’s sake at a cash-only kiosk, the band wants patrons to be able to pay \$20 even – hopefully paying with a single \$20 bill – rather than having to pay sales tax on top. If t-shirts are subject to a 10% tax on the sale price, and the shirts are priced so that the after-tax price comes to \$20, how much will a patron pay in sales tax after buying a t-shirt?
The first, quite clearly, should be \$2. Take 10% of the \$20 price and there’s your answer. And taking 10% is easy – just divide by 10, which functionally means moving the decimal point one place to the left and keeping the digits the same.
The second is not \$2, however, and the reason is critical to your preparation for percent questions above the 600 level on the GMAT: the percent has to be taken OF the proper value. Patrons will pay 10% OF the before-tax price, not 10% of the after-tax price. \$20 is the after-tax price (just as \$22 is the after-tax price in the first example…note that there you definitely did not take the 10% of the \$22 after-tax price!). So the proper calculation is:
Price + 10% of the Price = \$20
1.1(P) = 20
P = 20/1.1 = 18.18
So the price comes out to \$18.18, meaning that \$1.82 is the amount paid in tax.
While the calculation of 20/1.1 may have been annoying, it’s not “clever” or “hard” – the reason that many people will just say \$2.00 to both isn’t that they screwed up dividing \$20 by 1.1, but instead because they saw a percent problem with two numbers (10% and \$20) and just “calculated a percent.” That’s what makes the majority of GMAT percent problems tricky – they require an attention to detail, to precision in wording, for examinees to ensure that the (generally pretty darned easy) percent calculations are taking the percent of the proper value.
They’re logic puzzles that require a bit of of arithmetic, not simple arithmetic problems that just test your ability to divide by 10 absent critical thought. So as you approach GMAT percent problems, remember that the math should be the easy part. GMAT percent problems are often more about reading comprehension and logic than they are about multiplication and division.
By Brian Galvin.
# GMAT Tip of the Week: Your GMAT Verbal (Donald) Trump Card
The general consensus coming out of this week’s Democratic debate for the 2016 U.S. Presidency was this: the Democrats were quick to defend and agree with each other, particularly in contrast to the recent Republican debates in which the candidates were much more apt to attack each other.
The Democrats discussed, but the Republicans DEBATED, fiercely and critically. And – putting politics aside – one of the main issues on which those Republican candidates have attacked each other is “who is the more successful CEO/entrepreneur?” (And the answer to that? Likely Wharton’s finest: Donald “You’re Fired” Trump.)
So as you watch the political debates in between GMAT study sessions, keep this in mind: on the GMAT verbal section, you want to think more like a Republican candidate, and if possible you want to think like The Donald. Trump thinking is your Trump card: on GMAT verbal, you should attack, not defend.
Why?
Because incorrect answers are very easy to defend if that’s your mindset. They’re wrong because of a small (but significant) technicality, but to the “I see the good in all answer choices” eye, they’ll often look correct. You want to be in attack mode, critically eliminating answer choices and enjoying the process of doing so. Consider an example:
From 1998 to 2008, the amount of oil exported from the nation of Livonia increased by nearly 20% as the world’s demand soared. Yet over the same period, Livonia lost over 8,000 jobs in oil drilling and refinement, representing a 25% increase in the nation’s unemployment rate.
Which of the following, if true, would best explain the discrepancy outlined above?
A) Because of a slumping local economy, Livonia also lost 5,000 service jobs and 7,500 manufacturing jobs.
B) Several other countries in the region reported similar percentages of jobs lost in the oil industry over the same period.
C) Because of Livonia’s overvalued currency, most of the nation’s crude oil is now being refined after it has been exported.
D) Technological advancements in oil drilling techniques have allowed for a greater percentage of the world’s oil to be obtained from underneath the ocean floor.
E) Many former oil employees have found more lucrative work in the Livonia’s burgeoning precious metals mining industry.
The paradox/discrepancy here is that oil exports are up, but that jobs in oil drilling and refinement are down. What’s a Wharton-bound Trump to do here? Donald certainly wouldn’t overlook the word “Critical” in “Critical Reasoning.” Almost immediately, he’d be attacking the two-part job loss – it’s not that “oil jobs” are down, it’s that oil jobs in “drilling AND refinement” are down. Divide and conquer, he’d think, one of those items (either drilling or refinement) is bound to be a “lightweight” ready to be attacked.
Choice A is something that you could talk yourself into. “Hey, the economy overall is down, so it only makes sense that oil jobs would be down, too.” But think critically – you ALREADY know that the oil sector is not down. Oil exports are up 20% and global demand is soaring, so these oil jobs should be different. Critical thinking shows you that the general economy and this particular segment are on different tracks. Choice A does not explain the discrepancy.
Choice B is similar: if you’re looking for a reason to make it right, you might think, “See, it’s just part of what’s going on in the world.” But again, be critical. This is a bad answer, because it overlooks information you already have. Livonia’s oil exports are up, so absent a major reason that those exports are occurring without human labor, we don’t have a sound explanation.
Choice C hits on Trump’s “divide and conquer” attack strategy outlined above: if a conclusion to a Critical Reasoning problem includes the word “AND” there’s a very high likelihood that one of the two portions is the weak link. So fixate on that “and” and try to find which is the lightweight. Here you see that the oil is being exported from Livonia, but no longer being REFINED there. Those are the jobs that are leaving the country, and that explains why exports could be up with employment going down.
Choice D is tempting (statistically the most popular incorrect answer choice to this problem, with Trump-like polling numbers in the ~25% range). Why? Because you’re conditioned to think, “Oh, they’re losing jobs to technology.” So if you’re looking to find a correct answer without much critical thought and effort, this one shines like a beacon. But get more critical on the second half of the sentence: it’s not that technology makes it easier to obtain oil without human labor, it’s that technology is allowing for more drilling from the ocean. But that’s irrelevant, because, again, Livonia’s exports are up! So whether it’s Livonia getting that seafloor oil or other countries doing so, the fact remains that with oil exports up, you’d think that Livonia would have more jobs in oil, and this answer doesn’t explain why that’s not the case.
Here it pays to be critical all the way through the sentence: just because the first few words match what you think you might want to hear, that doesn’t mean that the entire statement is true. Think of this in Trump terms: Megyn Kelly might start a sentence with, “Mr. Trump, you’re arguably the most successful businessman of your generation,” (and you know Trump will love that) but if she follows that with, “But many would argue that your success was largely a result of your father’s money and that your manipulation of bankruptcy laws is unbefitting of an American president,” you know he’d be in attack mode immediately thereafter. Don’t fall in love with the first few words of an answer choice – stay ready to attack at a moment’s notice!
And choice E is similarly vulnerable to attack: yes some oil employees may have taken other jobs, but someone has to be doing the oil work. And if unemployment is up overall (as you know from the stimulus) then people are waiting to take those jobs, so the fact that some employees have left doesn’t explain why no one has filled those spots. When Donald Trump had to surrender his post as the star of The Apprentice, Arnold Schwarzenegger was ready to take his place; so, too, should unemployed members of the labor pool in Livonia be ready to take those oil jobs, absent a major reason why they wouldn’t, and choice E fails to present one.
Overall, your job on GMAT Verbal is to be as critical as possible. You’re there to debate the answer choices, not to defend or discuss them. As you read the conclusion of a Critical Reasoning problem, you want to be scanning for a “lightweight” word or phrase that makes it all the more vulnerable to attack. And as you read each answer choice, you shouldn’t be quick to see the good in the sentence, but instead you should be probing it to see where it’s weak and vulnerable to attack.
Let the answer choices view you as a bully – you’re not at the GMAT test center to make friends. Always be attacking, always be looking for words, phrases, or ideas that are an answer choice’s undoing. Trump logic is your Trump card, take joy from telling four of five answer choices “You’re Fired.”
By Brian Galvin.
# GMAT Tip of the Week: What To Do When The GMAT Gets All Netflix On You
Picture this: a friend texts you and asks, “Do you want to get a pizza and watch a movie after work?”. Do you find that odd at all?
But now picture this: that same friend asks, instead, “Do you want to get a pepperoni, mushroom and olive pizza with white sauce on thin crust from Domino’s and watch a Critically-Acclaimed Inspiring Underdog movie on Neflix after work?”. That’s strange, right? And why is that? Because it’s so specific.
Well, on the GMAT you’ll often see questions that ask for something oddly specific; “What is the value of x?” is pretty normal, but “What is the value of 6x – y?” is the equivalent of the specific pizza and odd Netflix category question. Why did they ask that? Often that’s a clue, and if you notice that clue it will help you better set up the problem. Consider this example:
Reflect on what this question is asking about. Not x. Not y. But to paraphrase Netflix, “a partially coefficiented combination of additive variables with a strong horizontal lead.” 6x – y. That’s oddly specific, so your first inclination should be, “Is there an easy way to get 6x – y?” as opposed to, “Let’s start solving for x” (which of course you can’t do here…that’s why E is a trap answer choice).
With that in mind, even if you’ve forgotten (or temporarily blanked on) some exponent rules, you should immediately be thinking, “I have 2x – how does that become 6x,” and, “Where does the subtraction come from?”.
The 6x, of course, comes from breaking 27 down into 3^3, so that you have (3^3)^2x, which then becomes 3^6x. And then with that, you have a fraction:
And that’s where the subtraction comes from. When you divide two exponents of the same base, you subtract the exponents, so now you have your 6x – y ready to go. Of course, from there, you need to get a base of 3 on the other side of the equation, so you can express 81 as 3^4, and now you know that 6x – y = 4, answer choice B.
Most importantly here, when the GMAT asks you an oddly-specific question in the vein of the oddly-specific Netflix category, you should seize on that specificity. Very frequently on the GMAT, it’s easier to solve for that oddly-specific combination of variables than it is to solve for any of the individual variables themselves!
On Problem Solving questions this can save you plenty of time, taking that extra few seconds to ask yourself how you’d arrive at that specific combination. On Data Sufficiency, this practice can be even more a matter of correct or incorrect. Data Sufficiency problems often give you sufficient information to arrive at the oddly-specific combination from the question stem, but insufficient information to determine any of the individual components. Imagine this problem as a Data Sufficiency problem:
Here, as you know from above, Statement 1 is sufficient, but if you go into the problem trying to solve for the variables individually, you’ll likely think that you need Statement 2 so that you can plug the value of y back into Statement 1 to supply the value of x. That way you’ll have the entire picture filled in: x = 1, y = 2, and 6x – y = 4.
But you don’t NEED Statement 2, so on a question like this the GMAT will punish you for not seeing that Statement 1 alone is sufficient. And it’s only sufficient because of that oddly-specific question stem. Check out this follow-up question (with a similar setup, but variables changed to a and b since the actual numbers will change):
Here you cannot use Statement 1 to get directly to the oddly-specific question stem. You can get to 4a – b = 4, but that doesn’t tell you about 6a – b. So here, the answer is C because you need Statement 2 so that you can solve for each variable individually.
More often than not, when the GMAT asks for an oddly-specific combination of variables it provides a way to arrive at it. So pay attention to the question itself: if it’s asking for something out of the ordinary or oddly specific, see that as a thinly-veiled clue that allows you to be the Confident GMAT Problem Solver With Excellent Think Like The Testmaker Skills En Route To A 700+ that you know you can be.
By Brian Galvin.
# GMAT Tip of the Week: Yogi Berra Teaches GMAT Sentence Correction
The world lost a legend this week with the passing of Yogi Berra, a New York Yankee and World War II hero. Yogi was universally famous – his name was, of course, the inspiration for beloved cartoon character Yogi Bear’s – but to paraphrase the man himself, those who knew him didn’t really know him.
As news of his passing turned into news reports summarizing his life, many were stunned by just how illustrious his career was: 18 All-Star game appearances (in 19 pro seasons), 10 World Series championships as a player, 3 American League MVP awards, part of the Normandy campaign on D-Day… To much of the world, he was “the quote guy” who also had been a really good baseball player. His wordsmithery is what we all remembered:
• Never answer an anonymous letter.
• It ain’t over ’til it’s over.
• It gets late early out here.
• Pair up in threes.
And his command (or butchering) of the English language is what you should remember as you take the GMAT. Yogi Berra famously “didn’t say some of the things I said” but he did, however inadvertently, have a lot to say about GMAT Sentence Correction:
Pronouns Matter
What’s funny about his quote, “Always go to other people’s funerals, otherwise they won’t come to yours”?
It’s the pronoun “they.” You know what Yogi means – go to other people’s funerals so that other people will come to yours. But in that sentence, the logical referent for “they” is “other people(‘s)”, and those other people have already been designated in the sentence as people who have already died. So the meaning is illogical: those same people cannot logically attend a funeral in the future. When you use a pronoun, it has to refer back to a specific noun. If that noun cannot logically do what the pronoun is said to be doing, that’s a Sentence Correction, illogical meaning problem.
What’s funny about his quote, “When you come to a fork in the road, take it”?
Again, it’s the pronoun, this time “it.” Since a fork in the road is a place where the road diverges into two paths, you can’t take “it” – you have to pick one path. And this is a good example of another sentence correction theme. In order to fix this thought (and the one above), there’s really not a pronoun that will work. “Them” has no logical referent (there’s only one fork) so the meaning is extremely important.
The only way to fix it is to change something prior in the sentence. Perhaps, “When you come to a turnoff on the road, take it,” or, “when the road presents a turn, take it.” On the GMAT, a pronoun error isn’t always fixed by fixing the pronoun – often the correct answer will change the logic that precedes the pronoun so that in the correct answer the previously-incorrect pronoun is correct.
Modifiers Matter
What’s funny about his quote, “Congratulations. I knew the record would stand until it was broken”?
Of course records stand until they’re broken, but in a grammatical sense Yogi’s primary mistake was his placement of the modifier “until it was broken.” What he likely meant to say is, “Until the record was broken, I thought it might stand forever.” That’s a perfectly logical thought, but we all laugh at the statement he actually made because the placement of the modifier creates a laughable meaning. So learn to spot similarly-misplaced modifiers by checking to make sure the language means exactly what it should.
Redundancy Is Funny (but sometimes has its place)
What’s funny about, “We made too many wrong mistakes,” and “It’s like déjà vu all over again”?
They’re redundant. A mistake is, by nature, something that went wrong. And déjà vu is the feeling that something happened before, so of course it’s “all over again.” Redundancy does come up on the GMAT, but as Yogi himself would point out, there’s a fine line between “redundant (and wrong)” and “a useful literary device”.
Take, for example, his famed, “It ain’t over ’til it’s over” quote. In a sports context, even though the word “over” is repeated, that sentence carries a lot of useful meaning: “when someone might say that the game is over, if there is still time (or outs) remaining there’s always a chance to change the result.” The world chuckles at this particular Yogi quote, but in actuality it’s arguably his most famous because, in its own way, it’s quite poignant.
What does that mean for you on the GMAT? Don’t prioritize redundancy as a primary decision point! GMAT Sentence Correction, by nature, involves plenty of different literary devices and sentence structures, and it’s extremely unlikely that you’ll feel like an expert on all of them.
Students often eliminate correct answers because they perceive redundancy, but a phrase like “not unlike” (a “not” next to an “un-“? That’s a redundant double-negative!) actually has a logical and important meaning (“not unlike” means “it’s not totally different from…there are at least some similarities,” whereas “like” conveys significantly more similarity). Rules for modifiers and pronouns are much more absolute, and you can get plenty of practice with those. Be careful with redundancy because, as Yogi might say, sometimes saying it twice is twice as good as saying it once.
“Baseball is ninety percent mental and the other half is physical.”
To paraphrase the great Yogi Berra, 90% of Sentence Correction is mental and the other half is grammatical. When he talked about baseball, he was talking about the physical tools – the ability to hit, run, throw, catch – as meaning substantially less than people thought, but the mental part of the game – strategy, mental toughness, stamina, etc. – being more important than people thought. The exact percentages, as his quote so ineloquently suggests, are harder to pin down and less important than the takeaway.
So heed Yogi’s advice as it pertains to Sentence Correction. Memorizing and knowing hundreds of grammar rules is “the other half” (or maybe 10%) of the game – employing good strategy (prioritizing primary Decision Points, paying attention to logical meaning, etc.) is the more-important-but-often-overlooked part of success. However eloquently or inelegantly Yogi Berra may have articulated his lessons, at least he made them memorable.
By Brian Galvin.
# GMAT Tip of the Week: 10 Must-Know Divisibility Rules For the GMAT (#3 Will Blow Your Mind!)
You clicked, didn’t you? You’re helpless when presented with an enumerated list and a teaser that at least one of the items is advertised to be – but probably won’t be – mind-blowing. (In this case it kind of is…if not mind-blowing, it’s at least very powerful). So in this case, let’s use click bait for good and enumerated lists to talk about numbers. Here are 10 important (and “BuzzFeedy”) divisibility rules you should know heading into the GMAT:
1) 1
1 may be the loneliest number but it’s also a very important number for divisibility! Every integer is divisible by 1, and the result of any integer x divided by 1 is just x (when you divide an integer by 1, it stays the same). On the GMAT, the fact that every integer is divisible by 1 can be quite important. For example, a question might ask (as at least one official problem does): Does integer x have any factors y such that 1 < y < x?
Because every integer is divisible by itself and 1, that question is really just asking, “Is x prime or not?” because if there is a factor y that’s between 1 and x, x is not prime, and there is not such a factor, then x is prime. That “> 1” caveat in the problem may seem obtuse, but when you understand divisibility by 1, you can see that the abstract question stem is really just asking you about prime vs. not-prime as a number property. The concept that all integers are divisible by 1 may seem basic, but keeping it top of mind on the GMAT can be extremely helpful.
2) 2
It takes 2 to make a thing go right…in relationships and on the GMAT! A number is divisible by 2 if that number is even (and a number is even if it’s divisible by 2). That means that if an integer ends in 0, 2, 4, 6, or 8, you know that it’s divisible by 2. And here’s a somewhat-surprising fact: the number 0 is even! 0 is divisible by 2 with no remainder (0/2 = 0), so although 0 is neither positive nor negative it fits the definition of even and should therefore be something you keep in mind because 0 is such a unique number.
The GMAT frequently tests even/odd number properties, so you should make a point to get to know them. Because any even number is divisible by 2 (which also means that it can be written as 2 times an integer), an even number multiplied by any integer will keep 2 as a factor and remain even. So even x even = even and even x odd = even.
3) 3
It’s been said that good things come in 3s, and divisibility rules are no exception! The divisibility rule for 3 works much like a magic trick and is one that you should make sure is top of mind on test day to save you time and help you unravel tricky numbers. The rule: if you sum the digits of an integer and that sum is divisible by 3, then that integer is divisible by 3. For example, consider the integer 219. 2 + 1 + 9 = 12 which is divisible by 3, so you know that 219 is divisible by 3 (it’s 3 x 73).
This rule can help you in many ways. If you were asked to determine whether a number is prime, for example, and you can see that the sum of the digits is a multiple of 3, you know immediately that it’s not prime without having to do the long division to prove it. Or if you had a messy fraction to reduce and noticed that both the numerator and denominator are divisible by 3, you can use that rule to begin reducing the fraction quickly. The GMAT tests factors, multiples, and divisibility quite a bit, so this is a critical rule to have at your disposal to quickly assess divisibility. And since 1 out of every 3 integers is divisible by 3, this rule will help you out frequently!
4) 4
Presidential Election and Summer Olympics enthusiasts, be four-warned! You already know the divisibility rule for 4: take the last two digits of an integer and treat them as a two-digit number, and if that’s divisible by 4 so is the whole number. So for 2016 – next year and that of the next presidential election and Brazil Olympics – the last two-digit number, 16, is divisible by 4, so you know that 2016 is also divisible by 4.
If you fail to see immediately that a number is divisible by 4 given that rule, fear not! Being divisible by 4 just means that a number is divisible by 2 twice. So if you didn’t immediately see that you could factor a 4 out of 2016 (it’s 504 x 4), you could divide by 2 (2 x 1008) and then divide by 2 again (2 x 2 x 504) and end up in the same place without too much more work.
5) 5
Who needs only 5 fingers to divide by 5? All of us – divisibility by 5 is so easy you should be able to do it with one hand tied behind your back! If an integer ends in 5 or 0 you know that it’s divisible by 5 (and we’ll talk more about what extra fact 0 tells you in just a bit…).
6) 6
Your favorite character from the hit 1990’s NBC sitcom “Blossom” is also an easy-to-use divisibility rule! Since 6 is just the product of 2 and 3 (2 x 3 = 6), if a number meets the divisibility rules for both 2 (it’s even) and 3 (the sum of the digits is divisible by 3) it’s divisible by 6. So if you need to reduce a number like 324, you might want to start by dividing by 6, instead of by 2 or 3, so that you can factor it in fewer steps.
7) 7
Ah, magnificent 7. While there is a “trick” for divisibility by 7, 7 occurs much less frequently in divisibility-based problems (as do other primes like 11, 13, 17, etc.), so 7 is a good place to begin to think about a strategy that works for all numbers, rather than memorizing limited-use tricks for each number. To test whether a large number, such as 231, is divisible by 7, find an obvious multiple of 7 nearby and then add or subtract multiples of 7 to see whether doing so will land on that number. For 231, you should recognize that a nearby multiple of 7 is 210 (you know 21 is 7 x 3, so putting a 0 on the end of it just means that 210 is 7 x 30). Then as you add 7s to get there, you go to 217, then to 224, then to 231. So in your head you can see that 231 is 3 more 7s than 7 x 30 (which you know is 210), so 231 = 7 x 33.
8) 8
8 is enough! As you saw above with 4s and 6s, when you start working with non-prime factors it’s often easier to just divide out the smaller prime factors one at a time than to try to determine divisibility by a larger composite number in one fell swoop. Since 8 = 2 x 2 x 2, you’ll likely find more success testing for divisibility by 8 by just dividing by 2, then dividing by 2 again, then dividing by 2 a third time. So for a number like 312, rather than working through long division to divide by 8, just divide it in half (156) then in half again (78) then in half again (39), and you’ll know that 312 = 39 x 8.
9) 9
While “nein” may be German for “no,” you should be saying “yes” to divisibility by nine! 9 shares a big similarity with 3 in that a sum-of-the-digits rule applies here too. If you sum the digits of an integer and that sum is a multiple of 9, the integer is also divisible by 9. So, for example, with the number 729, because 7 + 2 + 9 = 18, you know that 729 is divisible by 9 (it’s 81 x 9, which actually is 9 to the 3rd power).
10) 10
We’ve saved the best for last! If a number ends in 0, it’s divisible by 10, giving you a great opportunity to make the math easy. For example, a number like 210 (which you saw above) lets you pull the 0 aside and say that it’s 21 x 10, which means that it’s 3 x 7 x 10.
Working with 10s makes mental (or pencil-and-paper) math quick and convenient, so you should seek out opportunities to use such numbers in your calculations. For example, look at 693: If you add 7, you get to a number that ends in two 0s (so it’s 7 x 10 x 10), meaning that you know that 693 is divisible by 7 (it’s 7 away from an easy multiple of 7) *and* that it’s 7 x 99 because it’s one less 7 than 7 x 100. Because the GMAT rewards quick mental math, it’s a good idea to quickly check for, “If I have to add x to get to the nearest 0, then does that give me a multiple of x?” (297 is 3 away from 300, so you know that 297 = 99 x 3). And since 10 = 2 x 5, it’s also helpful sometimes to double a number that ends in 5 (try 215, which times 2 = 430) to see how many 10s you have (43). That tells you that 215 = 43 x 5 because 215 x 2 = 43 x (2 x 5). Working with 10s can make mental math extremely quick – we’d rate numbers that end in 0 a perfect 10!
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Small Numbers Lead to Big Scores
The last thing you want to see on your score report at the end of the GMAT is a small number. Whether that number is in the 300s (total score) or in the single-digits (percentile), your nightmares leading up to the test probably include lots of small numbers flashing on the screen as you finish the test. So what’s one of the most helpful tools you have to keep small numbers from appearing on the screen?
Have you found yourself on a homework problem or practice test asking yourself “can I just multiply these?”? Have you forgotten a rule and wondered whether you could trust your memory? Small numbers can be hugely valuable in these situations. Consider this example:
For integers x and y, 2^x + 2^y = 2^30. What is the sum x + y?
(A) 30
(B) 40
(C) 50
(D) 58
(E) 64
Every fiber of your being might be saying “can I just add x + y and set that equal to 30?” but you’re probably at least unsure whether you can do that. How do you definitively tell whether you can do that? Test the relationship with small numbers. 2^30 is far too big a number to fathom, but 2^6 is much more convenient. That’s 64, and if you wanted to set the problem up that way:
2^x + 2^y = 2^6
You can see that using combinations of x and y that add to 6 won’t work. 2^3 + 2^3 is 8 + 8 = 16 (so not 64). 2^5 + 2^1 is 32 + 2 = 34, which doesn’t work either. 2^4 + 2^2 is 16 + 4 = 20, so that doesn’t work. And 2^0 + 2^6 is 1 + 64 = 65, which is closer but still doesn’t work. Using small numbers you can prove that that step you’re wondering about – just adding the exponents – isn’t valid math, so you can avoid doing it. Small numbers help you test a rule that you aren’t sure about!
That’s one of two major themes with testing small numbers. 1) Small numbers are great for testing rules. And 2) Small numbers are great for finding patterns that you can apply to bigger numbers. To demonstrate that second point about small numbers, let’s return to the problem. 2^30 again is a number that’s too big to deal with or “play with,” but 2^6 is substantially more manageable. If you want to get to:
2^x + 2^y = 2^6, think about the powers of 2 that are less than 2^6:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
Here you can just choose numbers from the list. The only two that you can use to sum to 64 are 32 and 32. So the pairing that works here is 2^5 + 2^5 = 2^6. Try that again with another number (what about getting 2^x + 2^y = 2^5? Add 16 + 16 = 32, so 2^4 + 2^4), and you should start to see the pattern. To get 2^x + 2^y to equal 2^z, x and y should each be one integer less than z. So to get back to the bigger numbers in the problem, you should now see that to get 2^x + 2^y to equal 2^30, you need 2^29 + 2^29 = 2^30. So x + y = 29 + 29 = 58, answer choice D.
The lesson? When problems deal with unfathomably large numbers, it can often be quite helpful to test the relationship using small numbers. That way you can see how the pieces of the puzzle relate to each other, and then apply that knowledge to the larger numbers in the problem. The GMAT thrives on abstraction, presenting you with lots of variables and large numbers (often exponents or factorials), but you can counter that abstraction by using small numbers to make relationships and concepts concrete.
So make sure that small numbers are a part of your toolkit. When you’re unsure about a rule, test it with small numbers; if small numbers don’t spit out the result you’re looking for, then that rule isn’t true. But if multiple sets of small numbers do produce the desired result, you can proceed confidently with that rule. And when you’re presented with a relationship between massive numbers and variables, test that relationship using small numbers so that you can teach yourself more concretely what the concept looks like.
The best way to make sure that your GMAT score report contains big numbers? Use lots of small numbers in your scratchwork.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Eazy E Shows You How To Take Your Quant Score Straight Outta Compton And Straight To Cambridge
If you listened to any hip hop themed radio today, the day of the Straight Outta Compton movie premiere, you may have heard interviews with Dr. Dre. You almost certainly heard interviews with Ice Cube. And depending on how old school the station is there’s even a chance you heard from DJ Yella or MC Ren.
But on the radio this morning – just like on your GMAT exam – there was no Eazy-E. Logistically that’s because – as the Bone Thugs & Harmony classic “Tha Crossroads” commemorated – Eazy passed away about 20 years ago. But in GMAT strategy form, Eazy’s absence speaks even louder than his vocals on his NWA and solo tracks. “No Eazy-E” should be a mantra at the top of your mind when you take the GMAT, because on Data Sufficiency questions, choice E – the statements together are not sufficient to solve the problem – will not be given to you all that easily (Data Sufficiency “E” answers, like the Boyz in the Hood, are always hard).
Think about what answer choice E really means: it means “this problem cannot be solved.” But all too often, examinees choose the “Eazy-E,” meaning they pick E when “I can’t do it.” And there’s a big chasm. “It cannot be solved” means you’ve exhausted the options and you’re maybe one piece of information (“I just can’t get rid of that variable”) or one exception to the rule (“but if x is a fraction between 0 and 1…”) that stands as an obstacle to directly answering the question. Very rarely on problems that are above average difficulty is the lack of sufficiency a wide gap, meaning that if E seems easy, you’re probably missing an application of the given information that would make one or both of the statements sufficient. The GMAT just doesn’t have an incentive to reward you for shrugging your shoulders and saying “I can’t do it;” it does, however, have an incentive to reward those people who can conclusively prove that seemingly insufficient information can actually be packaged to solve the problem (what looks like E is actually A, B, C, or D) and those people who can look at seemingly sufficient information and prove why it’s not actually quite enough to solve it (the “clever” E).
So as a general rule, you should always be skeptical of Eazy-E.
Consider this example:
A shelf contains only Eazy-E solo albums and NWA group albums, either on CD or on cassette tape. How many albums are on the shelf?
(1) 2/3 of the albums are on CD and 1/4 of the albums are Eazy-E solo albums.
(2) Fewer than 30 albums are NWA group albums and more than 10 albums are on cassette tape.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked
(C) Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient
(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed
Statistically on this problem (the live Veritas Prep practice test version uses hardcover and paperback books of fiction or nonfiction, but hey it’s Straight Outta Compton day so let’s get thematic!), almost 60% of all test-takers take the Eazy-E here, presuming that the wide ranges in statement 2 and the ratios in statement 1 won’t get the job done. But a more astute examinee is skeptical of Eazy-E and knows to put in work! Statement 1 actually tells you more than meets the eye, as it also tells you that:
• 1/3 of the albums are on cassette tape
• 3/4 of the albums are NWA albums
• The total number of albums must be a multiple of 12, because that number needs to be divisible by 3 and by 4 in order to create the fractions in statement 1
So when you then add statement 2, you know that since there are more than 10 albums total (because at least 11 are cassette alone) so the total number could be 12, 24, 36, 48, etc. And then when you apply the ratios you realize that since the number of NWA albums is less than 30 and that number is 3/4 of the total, the total must be less than 40. So only 12, 24, and 36 are possible. And since the number of cassettes has to be greater than 10, and equate to 1/3 of the total, the total must then be more than 30. So the only plausible number is 36, and the answer is, indeed, C.
Strategically, being wary of Eazy-E tells you where to invest your time. If E seems too easy, that means that you should spend the extra 30-45 seconds seeing if you can get started using the statements in a different way. So learn from hip hop’s first billionaire, Dr. Dre, who split with Eazy long ago and has since seen his business success soar. Avoid Eazy-E and as you drive home from the GMAT test center you can bask in the glow of those famous Ice Cube lyrics, “I gotta say, today was a good day.”
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure tofind us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Jim Harbaugh Says Milk Does A GMAT Score Good
Someday when he’s not coaching football, playing with the Oakland Athletics, visiting with the Supreme Court, or Tweeting back and forth with Lil Wayne and Nicki Minaj, Jim Harbaugh should sit down and take the GMAT.
Because if his interaction with a young, milk-drinking fan is any indication, Harbaugh understands one of the key secrets to success on the GMAT Quant section:
Harbaugh, who told HBO Real Sports this summer about his childhood plan to grow to over 6 feet tall – great height for a quarterback – by drinking as much milk as humanly possible – is a fan of all kinds of milk: chocolate, 2%… But as he tells the young man, the ideal situation for growing into a Michigan quarterback is drinking whole milk, just as the ideal way to attend the Ross School of Business a few blocks from Harbaugh’s State Street office is to use whole numbers on the GMAT.
The main reason? You can’t use a calculator on the GMAT, so while your Excel-and-calculator-trained mind wants to say calculate “75% of 64” as 0.75 * 64, the key is to think in terms of whole numbers whenever possible. In this case, that means calling “75%” 3/4, because that allows you to do all of your calculations with the whole numbers 3 and 4, and not have to set up decimal math with 0.75. Since 64/4 is cleanly 16 – a whole number – you can calculate 75% by dividing by 4 first, then multiplying by 3: 64/4 is 16, then 16 * 3 = 48, and you have your answer without ever having to deal with messier decimal calculations.
This concept manifests itself in all kinds of problems for which your mind would typically want to think in terms of decimal math. For example:
With percentages, 25% and 75% can be seen as 1/4 and 3/4, respectively. Want to take 20%? Just divide by 5, because 0.2 = 1/5.
If you’re told that the result of a division operation is X.4, keep in mind that the decimal .4 can be expressed as 2/5, meaning that the divisor has to be a multiple of 5 and the remainder has to be even.
If at some point in a calculation it looks like you need to divide, say, 10 by 4 or 15 by 8 or any other type of operation that would result in a decimal, wait! Leaving division problems as improper fractions just means that you’re keeping two whole numbers handy, and knowing the GMAT at some point you’ll end up having to multiply or divide by a number that lets you avoid the decimal math altogether.
So learn from Jim Harbaugh and his obsession with whole milk. Whole milk may be the reasons that his football dreams came true; whole numbers could be a major reason that your business school dreams come true, too.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure tofind us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Behind the Scenes of Your GMAT Score
Among the most frequent questions we receive here at Veritas Prep headquarters (sadly, “How much am I allowed to tip my instructor?” is not one of them!) is the genre of “On my most recent practice test, I got X right and Y wrong and only Z wrong in a row… Why was my score higher/lower than my other test with A right and B wrong and C wrong in a row?” inquiries from students desperately trying to understand the GMAT scoring algorithm. We’ve talked previously in this space about why simply counting rights and wrongs isn’t all that great a predictor of your score. And perhaps the best advice possible relates to our Sentence Correction advice here a few months ago: Accept that there are some things you can’t change and focus on making a difference where you can.
But we also support everyone’s desire to leave no stone unturned in pursuit of a high GMAT score and everyone’s intellectual curiosity with regard to computer-adaptive testing. So with the full disclosure that these items won’t help you game the system and that your best move is to turn that intellectual curiosity toward mastering GMAT concepts and strategies, here are four major reasons that your response pattern — did you miss more questions early in the test vs. late in the test; did you miss consecutive questions or more sporadic questions, etc. — won’t help you predict your score:
1) The all-important A-parameter.
Item Response Theory incorporates three metrics for each “item” (or “question” or “problem”): the B parameter is the closest measurement to pure “difficulty”. The C parameter is essentially a measure of likelihood that a correct answer can be guessed. And the A parameter tells the scoring system how much to weight that item. Yes, some problems “count” more than others do (and not because of position on the test).
Why is that? Think of your own life; if you were going to, say, buy a condo in your city, you’d probably ask several people for their opinion on things like the real estate market in that area, mortgage rates, the additional costs of home ownership, the potential for renting it if you were to move, etc. And you’d value each opinion differently. Your very risk-loving friend may not have the opinion to value highest on “Will I be able to sell this at a profit if I get transferred to a new city?” (his answer is “The market always goes up!”) whereas his opinion on the neighborhood itself might be very valuable (“Don’t underestimate how nice it will be to live within a block of the blue line train”). Well, GMAT questions are similar: Some are extremely predictive (e.g. 90% of those scoring over 700 get it right, and only 10% of those scoring 690 or worse do) and others are only somewhat predictive (60% of those 700+ get this right, but only 45% of those below 700 do; here getting it right whispers “above 700” whereas before it screams it).
So while you may want to look at your practice test and try to determine where it’s better to position your “misses,” you’ll never know the A-values of any of the questions, so you just can’t tell which problems impacted your score the most.
2) Content balancing.
OK, you might then say, the test should theoretically always be trying to serve the highest value questions, so shouldn’t the larger A-parameters come out first? Not necessarily. The GMAT values balanced content to a very high degree: It’s not fair if you see a dozen geometry problems and your friend only sees two, or if you see the less time-consuming Data Sufficiency questions early in the test while someone else budgets their early time on problem solving and gets a break when the last ten are all shorter problems. So the test forces certain content to be delivered at certain times, regardless of whether the A-parameter for those problems is high or low. By the end of the test you’ll have seen various content areas and A-parameters… You just won’t know where the highest value questions took place.
3) Experimental items.
In order to know what those A, B, and C parameters are, the GMAT has to test its questions on a variety of users. So on each section, several problems just won’t count — they’re only there for research. And this can be true of practice tests, too (the Veritas Prep tests, for example, do contain experimental questions). So although your analysis of your response pattern may say that you missed three in a row on this test and gotten eight right in a row on the other, in reality those streaks could be a lot shorter if one or more of those questions didn’t count. And, again, you just won’t know whether a problem counted or not, so you can’t fully read into your response pattern to determine how the test should have been scored.
4) Item delivery vs. Score calculation.
One common prediction people make about GMAT scoring is that missing multiple problems in a row hurts your score substantially more than missing problems scattered throughout the test. The thinking goes that after one question wrong the system has to reconsider how smart it thought you were; then after two it knows for sure that you’re not as smart as advertised; and by the third it’s in just asking “How bad is he?” In reality, however, as you’ve read above, the “get it right –> harder question; get it wrong –> easier question” delivery system is a bit more nuanced and inclusive of experimentals and content balancing than people think. So it doesn’t work quite like the conventional wisdom suggests.
What’s more, even when the test delivers you an easier question and then an even easier question, it’s not directly calculating your score question by question. It’s estimating your score question-by-question in order to serve you the most meaningful questions it can, but it calculates your score by running its algorithm across all questions you’ve seen. So while missing three questions in a row might lower the current estimate of your ability and mean that you’ll get served a slightly easier question next, you can also recover over the next handful of questions. And then when the system runs your score factoring in the A, B, and C parameters of all of your responses to “live” (not experimental) questions, it doesn’t factor in the order in which those questions were presented — it only cares about the statistics. So while it’s certainly a good idea to get off to a good start in the first handful of problems and to avoid streaks of several consecutive misses, the rationale for that is more that avoiding early or prolonged droughts just raises your degree of difficulty. If you get 5 in a row wrong, you need to get several in a row right to even that out, and you can’t afford the kinds of mental errors that tend to be common and natural on a high-stakes exam. If you do manage to get the next several right, however, you can certainly overcome that dry spell.
In summary, it’s only natural to look at your practice tests and try to determine how the score was calculated and how you can use that system to your advantage. In reality, however, there are several unseen factors that affect your score that you just won’t ever see or know, so the best use of that curiosity and energy is learning from your mistakes so that the computer — however it’s programmed — has no choice but to give you the score that you want.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Kanye West’s Everything I Am Teaches Critical Reasoning
“Everything I’m not made me everything I am,” says Kanye West in his surprisingly-humble track Everything I Am. And while, unsurprisingly, much of what he’s talking about is silencing his critics, he might as well be rapping about making you an elite critic on Critical Reasoning problems. Because when it comes to some of the most challenging Critical Reasoning problems on the GMAT, everything they’re not makes them everything they are. Which is a convoluted way of saying this:
On challenging Strengthen and Assumption questions, the correct answer often tells you that a potential flaw with the argument is not true.
Everything that’s not true in that answer choice, then, makes the conclusion substantially more valid.
Consider this argument, for example:
Kanye received the most votes for the “Best Hip Hop Artist” award at the upcoming MTV Video Music Awards, so Kanye will be awarded the trophy for Best Hip Hop Artist.
If this were the prompt for a question that asked “Which of the following is an assumption required by the argument above?” a correct answer might read:
A) The Video Music Award for “Best Hip Hop Artist” is not decided by a method other than voting.
And the function of that answer choice is to tell you what’s not true (“everything I’m not”), removing a flaw that allows the conclusion to be much more logically sound (“…made me everything I am.”) These answer choices can be challenging in context, largely because:
1) Answer choices that remove a flaw can be difficult to anticipate, because those flaws are usually subtle.
2) Answer choices that remove a flaw tend to include a good amount of negation, making them a bit more convoluted.
In order to counteract these difficulties, it can be helpful to use “Everything I’m not made me everything I am” to your advantage. If what’s NOT true is essential to the conclusion’s truth, then if you consider the opposite – what if it WERE true – you can turn that question into a Weaken question. For example, if you took the opposite of the choice above, it would read:
The VMA for “Best Hip Hop Artist” is decided by a method other than voting.
If that were true, the conclusion is then wholly unsupported. So what if Kanye got the most votes, if votes aren’t how the award is determined? At that point the argument has no leg to stand on, so since the opposite of the answer directly weakens the argument, then you know that the answer itself strengthens it. And since we’re typically all much more effective as critics than we are as defenders, taking the opposite helps you to do what you’re best at. So consider the full-length problem:
Editor of an automobile magazine: The materials used to make older model cars (those built before 1980) are clearly superior to those used to make late model cars (those built since 1980). For instance, all the 1960’s and 1970’s cars that I routinely inspect are in surprisingly good condition: they run well, all components work perfectly, and they have very little rust, even though many are over 50 years old. However, almost all of the late model cars I inspect that are over 10 years old run poorly, have lots of rust, and are barely fit to be on the road.
Which of the following is an assumption required by the argument above?
A) The quality of materials used in older model cars is not superior to those used to make other types of vehicles produced in the same time period.
B) Cars built before 1980 are not used for shorter trips than cars built since then.
C) Manufacturing techniques used in modern automobile plants are not superior to those used in plants before 1980.
D) Well-maintained and seldom-used older model vehicles are not the only ones still on the road.
E) Owners of older model vehicles take particularly good care of those vehicles.
First notice that several of the answer choices (A, B, C, and D) include “is not” or “are not” and that the question stem asks for an assumption. These are clues that you’re dealing with a “removes the flaw” kind of problem, in which what is not true (in the answer choices) is essential to making the conclusion of the argument true. Because of that, it’s a good idea to take the opposites of those answer choices so that instead of removing the flaw in a Strengthen/Assumption question, you’re introducing the flaw and making it a Weaken. When you do that, you should see that choice D becomes:
D) Well-maintained and seldom-used older model vehicles ARE the only ones still on the road.
If that’s the case, the conclusion – “the materials used to make older cars are clearly superior to those used in newer ones” – is proven to be flawed. All the junkers are now off the road, so the evidence no longer holds up; you’re only seeing well-working old cars because they’re the most cared-for, not because they were better made in the first place.
And in a larger context, look at what D does ‘reading forward’: if it’s not only well-maintained and seldom-driven older cars on the road, then you have a better comparison point. So what’s not true here makes the argument everything it is. But dealing in “what’s not true” can be a challenge, so remember that you can take the opposite of each answer choice and make this “Everything I’m Not” assumption question into a much-clearer “Everything I Am” Weaken question.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Shake & Bake Your Way To Function Success
For many of us, cooking a delicious homemade meal and solving a challenge-level GMAT math problem are equally daunting challenges. So many steps, so many places to make a mistake…why can’t there be an easier way? Well, the fine folks at Kraft foods solved your first problem years ago with a product called “Shake and Bake.” You take a piece of chicken (your input), stick in the bag of seasoning, shake it up, bake it, and voila – you have yourself a delicious meal with minimal effort. So gourmet level cooking is now nothing to fear…but what about those challenging GMAT quant problems?
You’re in luck. Function problems on the GMAT are essentially Shake and Bake recipes. Consider the example:
f(x) = x^2 – 80
If that gets your heart rate and stress level up, you’re not alone. Function notation just looks challenging. But it’s essentially Shake and Bake if you dissect what a function looks like.
The f(x) portion tells you about your input. f(x) = ________ means that, for the rest of that problem, whatever you see in parentheses is your “input” (just like the chicken in your Shake and Bake).
What comes after the equals sign is the recipe. It tells you what to do to your input to get the result. Here f(x) = x^2 – 80 is telling you that whatever your input, you square it and then subtract 80 and that’s your output. …And that’s it.
So if they ask you for:
What is f(9)? (your input is 9), then f(9) = 9^2 – 80, so you square 9 to get 81, then you subtract 80 and you have your answer: 1.
what is f(-4)? (your input is -4), then f(4) = 4^2 – 80, which is 16 – 80 = -64.
What is f(y^2)? (your input is y^2), then f(y^2) = (y^2)^2 – 80, which is y^4 – 80.
What is f(Rick Astley), then your input is Rick Astley and f(Rick Astley) = (Rick Astley)^2 – 80.
It really doesn’t matter what your input is. Whatever the test puts in the parentheses, you just use that as your input and do whatever the recipe says to do with it. So for example:
f(x) = x^2 – x. For which of the following values of a is f(a) > f(8)?
I. a = -8
II. a = -9
III a = 9
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
While this may look fairly abstract, just consider the inputs they’ve given you. For f(8), just put 8 wherever that x goes in the “recipe” f(x) = x^2 – x:
f(8) = 8^2 – 8 = 64 – 8 = 56
And then do the same for the three other possible values:
I. f(-8) means put -8 wherever you see the x: (-8)^2 – (-8) = 64 + 8 = 72, so f(-8) > f(8).
II. f(-9) means put -9 wherever you see the x: (-9)^2 – (-9) = 81 + 9 = 90, so f(-9) > f(8)
III. f(9) means put a 9 wherever you see the x: 9^2 – 9 = 81 – 9 = 72, so f(9) > f(8), and the answer is E.
Ultimately with functions, the notation (like the Shake and Bake ingredients) is messy, but with practice the recipes become easy to follow. What goes in the parentheses is your input, and what comes after the equals sign is your recipe. Follow the steps, and you’ll end up with a delicious GMAT quant score.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Oh Thank Heaven For Seven-Eleven
A week after the Fourth of July, a lesser-known but certainly-important holiday occurs each year. Tomorrow, friends, is 7-11, a day to enjoy free Slurpees at 7-11 stores, to roll some dice at the craps table, and to honor your favorite prime numbers. So to celebrate 7-11, let’s talk about these two important prime numbers.
Checking Whether A Number Is Prime
Consider a number like 133. Is that number prime? The first three prime numbers (2, 3, and 5) are easy to check to see whether they are factors (if any are, then the number is clearly not prime):
2 – this number is not even, so it’s not divisible by 2.
3 – the sum of the digits (an important rule for divisibility by three!) is 7, which is not a multiple of 3 so this number is not divisible by 3.
5 – the number doesn’t end in 5 or 0, so it’s not divisible by 5
But now things get a bit trickier. There are, in fact, (separate) divisibility “tricks” for 7 and 11. But they’re relatively inefficient compared with a universal strategy. Find a nearby multiple of the target number, then add or subtract multiples of that target number. If we want to test 133 to see whether it’s divisible by 7, we can quickly go to 140 (you know this is a multiple of 7) and then subtract by 7. That’s 133, so you know that 133 is a multiple of 7. (Doing the same for 11, you know that 121 is a multiple of 11 – it’s 11-squared – so add 11 more and you’re at 132, so 133 is not a multiple of 11)
This is even more helpful when, for example, a question asks something like “how many prime numbers exist between 202 and 218?”. By finding nearby multiples of 7 and 11:
210 is a multiple of 7, so 203 and 217 are also multiples of 7
220 is a multiple of 11, so 209 is a multiple of 11
You can very quickly eliminate numbers in that range that are not prime. And since none of the even numbers are prime and neither are 205 and 215 (the ends-in-5 rule), you’re left only having to check 207 (which has digits that sum to a multiple of 3, so that’s not prime), 211 (more on that in a second), and 213 (which has digits that sum to 6, so it’s out).
So that leaves the process of testing 211 to see if it has any other prime factors than 2, 3, 5, 7, and 11. Which may seem like a pretty tall order. But here’s an important concept to keep in mind: you only have to test prime factors up to the square root of the number in question. So for 211, that means that because you should know that 15 is the square root of 225, you only have to test primes up to 15.
Why is that? Remember that factors come in pairs. For 217, for example, you know it’s divisible by 7, but 7 has to have a pair to multiply it by to get to 217. That number is 31 (31 * 7 = 217). So whatever factor you find for a number, it has to multiply with another number to get there.
Well, consider again the number 211. Since 15 * 15 is already bigger than 211, you should see that for any number bigger than 15 to be a factor of 211, it has to pair with a number smaller than 15. And as you consider the primes up to 15, you’re already checking all those smaller possibilities. That allows you to quickly test 211 for divisibility by 13 and then you’re done. And since 211 is not divisible by 13 (you could do the long division or you could test 260 – a relatively clear multiple of 13 – and subtract 13s until you get to or past 211: 247, 234, 221, and 208, so 211 is not a multiple of 13. Therefore 211 is prime.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Raising Your Data Sufficiency Accuracy From 33% To 99%
You’re looking at a Data Sufficiency problem and you’re feeling the pressure. You’re midway through the GMAT Quantitative section and your mind is spinning from the array of concepts and questions that have been thrown at you. You know you nailed that tricky probability question a few problems earlier and you hope you got that last crazy geometry question right. When you look at Statement 1 your mind draws a blank: whether it’s too many variables or too many numbers or too tricky a concept, you just can’t process it. So you look at Statement 2 and feel relief. It’s nowhere near sufficient, as just about anyone even considering graduate school would know immediately. So you smile as you cross off choices A and D on your noteboard, saying to yourself: “Good, at least I have a 33% chance now.”
You’re better than that.
Too often on Data Sufficiency problems, people are impressed by giving themselves a 33% or even 50% chance of success. Keep in mind that guessing one of three remaining choices means you’re probably going to get that problem wrong! And of more strategic importance is this: if one statement is dead obvious, you haven’t just raised your probability of guessing correctly – you should have just learned what the question is all about!
If a Data Sufficiency statement is clearly insufficient, it’s arguably the most important part of the problem.
Consider this example:
What is the value of integer z?
(1) z represents the remainder when positive integer x is divided by positive integer (x – 1)
(2) x is not a prime number
Many examinees will be thrilled here to see that statement 2 is nowhere near sufficient, therefore meaning that the answer must be A, C, or E – a 33% chance of success! But a more astute test-taker will look closer at statement 2 and think “this problem is likely going to come down to whether it matters that x is prime or not” and then use that information to hold Statement 2 up to that light.
For statement 1, many will test x = 5 and (x – 1) = 4 or x = 10 and (x – 1) = 9 and other combinations of that ilk, and see that the result is usually (always?) 1 remainder 1. But a more astute test-taker will see that word “prime” and ask themselves why a prime would matter. And in listing a few interesting primes, they’ll undoubtedly check 2 and realize that if x = 2 and (x – 1) = 1, the result is 1 with a remainder of 0 – a different answer than the 1, remainder 1 that usually results from testing values for x. So in this case statement 2 DOES matter, and the answer has to be C.
Try this other example:
What is the value of x?
(1) x(x + 1) = 2450
(2) x is odd
Again, someone can easily skip ahead to statement 2 and be thrilled that they’re down to three options, but it pays to take that statement and file it as a consideration for later: when I get my answer for statement 1, is there a reason that even vs. odd would matter? If I get an odd solution, is there a possible even one?
Factoring 2450 leaves you with consecutive integers 49 and 50, so x could be 49 and therefore odd. But is there any possible even value for x, a number exactly one smaller than (x + 1) where the product of the two is still 2450? There is: -50 and -49 give you the same product, and in that case x is even. So statement 2 again is critical to get the answer C.
And the lesson? A ridiculously easy statement typically holds much more value than “oh this is easy to eliminate.” So when you can quickly and effortlessly make your decision on a Data Sufficiency statement, don’t be too happy to take your slightly-increased odds of a correct answer and move on. Use that statement to give you insight into how to attack the other statement, and take your probability of a correct answer all the way up to “certain.”
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Talking About Equality
If you’ve ever struggled with algebra, wondered which operations you were allowed to perform, or been upset when you were told that the operation you just performed was incorrect, this post is for you. Algebra is all about equality.
What does that mean?
Consider the statement:
8 = 8
That’s obviously not groundbreaking news, but it does show the underpinnings of what makes algebra “work.” When you use that equals sign, =, you’re saying that what’s on the left of that sign is the exact same value as what’s on the right hand side of that sign. 8 = 8 means “8 is the same exact value as 8.” And then as long as you do the same thing to both 8s, you’ll preserve that equality. So you could subtract 2 from both sides:
8 – 2 = 8 – 2
And you’ll arrive at another definitely-true statement:
6 = 6
And then you could divide both sides by 3:
6/3 = 6/3
And again you’ve created another true statement:
2 = 2
Because you start with a true statement, as proven by that equals sign, as long as you do the exact same thing to what’s on either side of that equals sign, the statement will remain true. So when you replace that with a different equation:
3x + 2 = 8
That’s when the equals sign really helps you. It’s saying that “3x + 2” is the exact same value as 8. So whatever you do to that 8, as long as you do the same exact thing to the other side, the equation will remain true. Following the same steps, you can:
Subtract two from both sides:
3x = 6
Divide both sides by 3:
x = 2
And you’ve now solved for x. That’s what you’re doing with algebra. You’re taking advantage of that equality: the equals sign guarantees a true statement and allows you to do the exact same thing on either side of that sign to create additional true statements. And your goal then is to use that equals sign to strategically create a true statement that helps you to answer the question that you’re given.
Equality applies to all terms; it cannot single out just one individual term.
Now, where do people go wrong? The most common mistake that people make is that they don’t do the same thing to both SIDES of the inequality. Instead they do the same thing to a term on each side, but they miss a term. For example:
(3x + 5)/7 = x – 9
In order to preserve this equation and eliminate the denominator, you must multiply both SIDES by 7. You cannot multiply just the x on the right by 7 (a common mistake); instead you have to multiply everything on the right by 7 (and of course everything on the left by 7 too):
7(3x + 5)/7 = 7(x – 9)
3x + 5 = 7x – 63
Then subtract 3x from both sides to preserve the equation:
5 = 4x – 63
Then add 63 to both sides to preserve the equation:
68 = 4x
Then divide both sides by 4:
17 = x
The point being: preserving equality is what makes algebra work. When you’re multiplying or dividing in order to preserve that equality, you have to be completely equitable to both sides of the equation: you can’t single out any one term or group. If you’re multiplying both sides by 7, you have to distribute that 7 to both the x and the -9. So when you take the GMAT, do so with equality in mind. The equals sign is what allows you to solve for variables, but remember that you have to do the exact same thing to both sides.
Inequalities? Well those will just have to wait for another day.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAC Announces Two New GMAT Policies, Both In Your Favor!
MBA applicants, your path to submitting a score report that you can be proud of just got a bit smoother. In an announcement to test-takers today, GMAC revealed two new policies that each stand out as particularly student-friendly:
1) As of July 19, 2015, the designation “C” will no longer appear on score reports to designate a cancelled score. This couples nicely with the recent change to the GMAT’s cancellation policy allowing you to preview your score before you decide whether to “keep” or cancel it. So as of July 19, there is zero risk that anyone but you will ever notice that you had a bad test day (unless, of course, you decide to publish that score). Even better, this policy also applies to previously cancelled scores (not just to tests taken after July 19). If you submit a score report to a school now, and you have multiple test sittings in which you cancelled your score, business schools will never know it.
What does this policy mean for you? For one, you can feel markedly less pressure when you take the GMAT, as a bad score only has to be your business. There is no downside! Furthermore, you can feel confident selecting an aggressive timeline for your GMAT test date, as even if you do not perform to your goals at worst-case that attempt is “an expensive but very authentic practice test.” While in the vast majority of cases, that C never felt to schools like a Scarlet Letter, the stigma in students’ minds was often enough to inspire fear on test day and anxiety in the admissions process. Fear no more!
2) Effective immediately, you only need to wait 16 days (as opposed to 31) before retaking the GMAT. With that waiting period now cut just about in half, you have a few terrific advantages:
• If you decide you need to retake the exam, you can stick on your study regiment just 2.5 more weeks to polish up those last few concepts and you’ll take the test while everything is still fresh and you’re still in “game shape.”
• You’re significantly less likely to end up in limbo between “set the test date that maximizes your chance for success on THAT test” and “set the test date that gives you the safety net of one more try if the first one doesn’t go so well.” That month-plus between administrations made for tricky decisions for applicants in the past. Now you have that much more flexibility when choosing a date to get the test and a backup plan in before your applications are due.
Is there a downside? GMAC wouldn’t likely be as aggressive with the 16-day waiting period if it didn’t have the capacity to allow more GMAT administrations in the busy season, but there is a chance that the ~3 weeks leading up to the major application deadlines could get crowded at test centers. To have your pick of test dates for both your first shot and your backup, you may want to consider taking the GMAT 6 (and maybe 3) weeks before you and others need the score as opposed to 3 weeks and “immediately” before you need it.
Are you getting ready for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Dave Chappelle’s Friend Chip Teaches Data Sufficiency Strategy
“Officer, I didn’t know I couldn’t do that,” Dave Chappelle’s friend, Chip, told a police officer after being pulled over for any number of reckless driving infractions. In Chappelle’s famous stand-up comedy routine, he mocks the audacity of his (privileged) friend for even thinking of saying that to a police officer. But that’s the exact type of audacity that gets rewarded on Data Sufficiency problems, and a powerful lesson for those who, like Dave in the story, seem more resigned to their plight of being rejected at the mercy of the GMAT yet again.
How does Chip’s mentality help you on the GMAT? Consider this Data Sufficiency fragment:
Is the product of integers j, k, m, and n equal to 1?
(1) (jk)/mn = 1
The approach that most students take here involves plugging in numbers for j, k, m, and n and seeing what answer they get. Knowing that jk = mn (by manipulating the algebra in statement 1) they may pick combinations:
1 * 8 = 2 * 4, in which case the product is 64 and the answer is no
2 * 5 = 1 * 10, in which case the product is 100 and the answer is no
And so some will, after picking a series of arbitrary number choices, claim that the answer must be no. But in doing so, they’re leaving out the possibilities:
1 * 1 = 1 * 1, in which case the product jkmn = 1*1*1*1 = 1, so the answer is yes
-1 * 1 = -1 * 1, in which case the product is also 1, and the answer is yes
And here’s where Chip Logic comes into play: in any given classroom, when the two latter sets of numbers are demonstrated, at least a few students will say “How are we allowed to use the same number twice? No one told us we could do that?”. And the best response to that is Chip’s very own: “I didn’t know I COULDN’T do that.” Since the problem didn’t restrict the use of the same number twice (to do so they might say “unique integers j, k, m, and n”), it’s on you to consider all possible combinations, including “they all equal 1.” Data Sufficiency tends to reward those who consider the edge cases: the highest or lowest possible number allowed, or fractions/decimals, or negative numbers, or zero. If you’re going to pick numbers on Data Sufficiency questions, you have to think like Chip: if you weren’t explicitly told that you couldn’t, you have to assume that you can.
So on Data Sufficiency problems, when you pick numbers, do so with a sense of entitlement and audacity. Number-picking is no place for the timid – your job is to “break” the obvious answer by finding allowable combinations that give you a different answer; in doing so, you can prove a statement to be insufficient. So as you chip away at your goal of a 700+ score, summon your inner Chip. When it comes to picking numbers, “I didn’t know I couldn’t do that” is the mentality you need to know you can use.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Making the Most of Your Mental Stamina
One of the most fascinating storylines during the current 2015 NBA Finals is that of LeBron James’ workload and stamina. Responsible for such a huge percentage of Cleveland’s offense and a key component of the team’s necessarily suffocating defense, James needs to parcel out his energy usage much like an endurance athlete does in the Tour de France or Ironman World Championships. And it’s fascinating to watch as he slowly walks the ball up the court (killing time to shorten the game and also buying valuable seconds of rest before initiating the offense) and nervously watches his teammates lose ground while he takes his ~2 minute beginning-of-the-4th rest on the bench. At the final buzzer of each game he looks exhausted but thus far has been exhaustedly-triumphant twice. And watching how he handles his energy can teach you valuable lessons about how to manage the GMAT.
At the end of your GMAT exam you will be exhausted. But will you be exhaustedly triumphant? Here are 5 things you can do to help you tiredly walk out of the test center with a championship smile:
1) Practice The Way You’ll Play
The GMAT is a long test. You’ll be at the test center for about 4 hours by the time you’re done, and even during those 8-minute section breaks you’ll be hustling the whole time. Think of it this way: with a 30-minute essay, a 30-minute Integrated Reasoning section, a 75-minute Quant section and a 75-minute Verbal section, you’ll be actively answering questions for 3 hours and 30 minutes – a reasonable time for someone your age to complete a marathon (and well more than an hour off the world record). If you were training for a marathon, you wouldn’t stop your workouts after an hour or 90 minutes each time; at the very least you’d work up to where you’re training for over two hours at least once a week. And the same is true of the GMAT. To have that mental stamina to stay focused on a dense Reading Comprehension passage over 3 hours after you arrived at the test center, you need to have trained your mind to focus for 3+ hours at a time. To do so:
-Take full-length practice tests, including the AWA and IR sections.
-Practice verbal when you’re tired, after a long day of work or after you’ve done an hour or more of quant practice
-Make at least one 2-3 hour study session a part of your weekly routine and stick to it. Work can get tough, so whether it’s a Saturday morning or Sunday afternoon, pick a time that you know you can commit to and go somewhere (library, coffee shop) where you know you’ll be able to focus and get to work.
2) Be Ready For The 8-Minute Breaks
Like LeBron James, you’ll have precious few opportunity to rest during your “MBA Finals” date with the GMAT. You have an 8-minute break between the IR section and the Quant section and another 8-minute break between the Quant section and the Verbal section…and that’s it. And those breaks go quickly, as in that 8 minutes you need to check out with the exam proctor to leave the room and check back in to re-enter. A minute or more of your break will have elapsed by the time you reach your locker or the restroom…time flies when you’re on your short rest period! So be ready:
-Have a plan for your break, knowing exactly what you want to accomplish: restroom, water, snack. You shouldn’t have to make many energy-draining decisions during that time; your mind needs a break while you refresh your body, so do all of your decision-making before you even arrive at the test center.
-Practice taking 8-minute breaks when you study and take practice tests. Know how long 8 minutes will take and what you can reasonably accomplish in that time.
3) Use Energy Wisely
If you’re watching LeBron James during the Finals you’ll see him take certain situations (if not entire plays) off, conserving energy for when he has the opportunity to sprint downcourt on a fast break or when he absolutely has to get out on a ready-to-shoot Steph Curry. For you on the GMAT, this means knowing when to stress over calculations on quant or details on reading comprehension. Most students simply can’t give 100% effort for the full test, so you may need to consider:
-On this Data Sufficiency problem, do you need to finish the calculations or can you stop early knowing that the calculations will lead to a sufficient answer?
-As you read this Reading Comp passage, do you need to sweat the scientific details or should you get the gist of it and deal with details later if a question specifically asks for them?
-With this Geometry problem, is it worth doing all the quadratic math or can you estimate using the answer choices? If you do do the math, are you sure that it will get you to an answer in a reasonable amount of time?
Sometimes the answer is “yes” – if it’s a problem that you know you can get right, but only if you grind through some ugly math, that’s a good place to invest that energy. And other times the answer is “no” – you could do the work, but you’re not so sure you even set it up right and the numbers are starting to look ugly and you usually get these problems wrong, anyway. Practice is the key, and diagnosing how those efforts have gone on your practice tests. You might not have enough mental energy to give all the focus you’d like all day, so have a few triggers in there that will help you figure out which battles you can lose in an effort to win the war.
4) Master The “Give Your Mind A Break” Problems
Some GMAT problems are extremely abstract and require a lot of focus and ingenuity. Others are very process-driven if you know the process – among those are the common word problems (weighted averages, rate problems, Venn diagram problems, etc.) and straightforward “solve for this variable” algebra problem solving problems. If you’ve put in the work to master those content-driven problems, they can be a great opportunity to turn your brain off for a few minutes while you just grind out the necessary steps, turning your mind back on at the last second to double check for common mistakes.
This comes down to practice. If you recognize the common types of “just set it up and do the work” problems, you’ll know them when you see them and can relax to an extent as you perform the same steps you have dozens of times. If you recognize the testmaker’s intent on certain problems – in an “either/or” SC structure, for example, you know that they’re testing parallelism and can quickly eliminate answers that don’t have it; if a DS problem includes >0 or <0, you can quickly look for positive/negative number properties with the “usual suspects” that indicate those things – you can again perform rote steps that don’t require much mental heavy lifting. The test is challenging, but if you put in the work in practice you’ll find where you can take some mental breaks without getting punished.
The verbal section comes last, and that’s where focus can be the hardest as you face a barrage of problems on a variety of topics – astronomy, an election in a fake country, a discovery about Druid ruins, comparative GDP between various countries, etc. A verbal section will include thousands of words, but only a couple hundred are really operative words upon which correct answers hinge. So be proactive as you read verbal problems. That means:
-Scan the answer choices for obvious decision points in SC problems. If you know they’re testing verb tense, for example, then you’re looking at the original sentence for timeline and you don’t have to immediately focus on any other details. On many questions you can get an idea of what you’re reading for before you even start reading.
-Let details go on RC passages. Your job is to know the general author’s point, and to have a good idea of where to find any details that they might ask about. But in an RC passage that includes a dozen or more details, they may only ask you about one or two. Worry about those details when you’re asked for them, saving mental energy by never really stressing the ones that end up not mattering at all.
-Read the question stem first on CR problems. Before you read the prompt, know your job so that you know what to look for. If you need to weaken it, then look for the flaw in the argument and focus specifically on the key words in the conclusion. If you need to draw a conclusion, your energy needs to be highest on process-of-elimination at the answers, and you don’t have to stress the initial read of the prompt nearly as much.
Know that the GMAT is a long, exhausting day, and you won’t likely get out of the test center without feeling completely wiped out. But if you manage your energy efficiently, you can use whatever energy you have left to triumphantly raise that winning score report over your head as you walk out.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: No Calculator? No Problem.
For many GMAT examinees, the realization that they cannot use a calculator on the GMAT quantitative section is cause for despair. For most of your high school career, calculators were a featured part of the math curriculum (what TI are we up to now?); nowadays you almost always have Microsoft Excel a click away to perform those calculations for you.
But remember: it’s not that YOU don’t get to use a calculator on the GMAT quant section. It’s that NO ONE gets to use a calculator. And that creates the opportunity for a competitive advantage. If you know that the GMAT doesn’t include “calculator problems” – the testmakers know that you don’t get a calculator, too, so they create questions that savvy examinees can find efficient ways to solve by hand (or head) or estimate – then you can use that to your advantage, looking for “clean” numbers to calculate and saving calculations until they’re truly necessary. As an example, consider the problem:
A certain box contains 14 apples and 23 oranges. How many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples?
(A) 3
(B) 6
(C) 14
(D) 17
(E) 20
If you’re well-versed in “non-calculator” math, you should recognize a couple things as you scan the problem:
1) The 23 oranges represent a prime number. That’s an ugly number to calculate with in a non-calculator problem.
2) 70% is a very clean number, which reduces to 7/10. Numbers that end in 0 don’t tend to play well with or come from double-digit prime numbers, so in this problem you’ll need to “clean up” that 23.
3) The 14 apples are pretty nicely related to the 70%. 14*5 = 70, and 14 = 7 * 2, where 70% is 7/10. So in sum, the 14 is a pretty “clean” number you’re working with to find a relationship that includes that also “clean” 70%. And the 23 is ugly.
So if you wanted to plug in numbers here to see how many oranges should be removed, keep in mind that your job is to get that 23 to look a lot cleaner. So while the Goldilocksian conventional methodology for backsolving is to “start with the middle number, then determine whether it’s correct, too big, or too small,” if you’re preparing for non-calculator math you should quickly see that with answer choice C of 14, that would give you 14 apples and 9 (which is 23-14) oranges), and you’re stuck at that ugly number of 23 as your total number of pieces of fruit. So your goal should be to find cleaner numbers to calculate.
You might try choice A, 3, which is very easy to calculate (23 oranges minus 3 = 20 oranges left), but a quick scan there would show that that’s way too many oranges (still more oranges than apples). So the other number that can clean up the 23 oranges is 17 (choice D), which would at least give you an even number (23 – 17 = 6). Because you’re now dealing with clean numbers (14, 6, and 70%) it’s worth doing the full calculation to see if choice D is really correct. And since 14 apples out of 20 total pieces of fruit is, indeed, 70%, you know that D is correct.
Now, if you follow these preceding paragraphs step-by-step, they should look just as long and unwieldy as the algebra or some traditional backsolving. But to an examinee seasoned in non-calculator math, finding “clean numbers worth testing” is more about the scan than the process. You should know that Odd + or – Odd = Even, but that Odd + or – Even is Odd. So with an even “fixed” number of 14 apples and an odd “changeable” number of 23 oranges, an astute GMAT test-taker looking to save time would probably eschew plugging in C first and realize that it’s just not going to be correct. Then another scan of numbers shows that only 3 and 17 are odd and prone to becoming “clean” when subtracted from the prime 23, so D should start looking tempting within seconds.
Note: this strategy isn’t for everyone or for every problem, but for those shooting for the 700s it can be extremely helpful to develop enough “number fluency” that you can save time not-testing numbers that you can see don’t have a real chance. On a non-calculator test that typically involves “clean” (even, divisible by 10, etc.) numbers, quickly recognizing which numbers will result in good, clean, non-calculator math is a very helpful skill.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Snoop Dogg Keeps Your Data Sufficiency Ability Out Of Limbo
Whenever you’re picking numbers on a Data Sufficiency problem, you have to keep one image in your mind: Snoop Dogg at a limbo contest. How will that help you master Data Sufficiency? How can the Doggfather help you beat the Testmaker? Well think about the two questions that Snoop would be asking himself constantly at such a contest:
1) How high can I get? (Snoop’s general state of mind)
2) How low can I go? (Because you know Snoop’s in it to win it)
And that mindset is absolutely crucial in a Data Sufficiency number-picking situation. On these problems, the GMAT Testmaker knows your tendencies well: you’re predisposed to picking numbers that are easy to work with. Consider an example like:
If x is a positive integer less than 30, what is the value of x?
(1) When x is divided by 3 the remainder is 2.
(2) When x is divided by 5 the remainder is 2
On this problem, most can quite quickly eliminate statement 1, as x could be 5, 8, 11, 14, 17, 20, 23, 26, or 29. Typically your quick-thinking methodology will have you look at 3, then add the remainder of 2 (producing 5), then start looking at other multiples of 3 and doing the same (6 + 2 gives you 8, 9 + 2 gives you 11, and so on).
And similarly you can apply that logic to statement 2 and eliminate that pretty quickly. The obvious first candidate is 7 (add the remainder of 2 to 5), and then you should see the pattern: 7, 12, 17, 22, and 27 are your options.
So when you look at these quick lists and see that the only place they overlap is 17 (17/5 is 3 remainder 2 and 17/3 is 5 remainder 2), you might opt for C.
But where does Snoop Dogg’s Limbo Contest come in? Look at the range they gave you: a POSITIVE INTEGER (so anything > 0) LESS THAN 30 (so anything <30). So when you combine those, your range is 0 < x < 30. Then ask yourself:
*How high can you get? Well, on either list you’ve gotten as close to 30 as possible. The next possible number on the first list (5, 8, 11, 14, 17, 20, 23, 26, 29…) is 32, but they tell you that x is less than 30 so you can’t get that high. And the next possible number on the second list (7, 12, 17, 22, 27…) is also 32, but again you’re not allowed to get that high. So you’ve definitely answered that question well.
*How low can you go? On this one, you haven’t yet exhausted the lower limit. Look at the patterns on those lists – on the first one, all numbers are 3 apart but you started at 5. If you move down 3, you get to 2 (2, 5, 8…). And 2/3 is 0 remainder 2, so 2 is a legitimate number on that list, a positive integer that leaves a remainder of 2 when divided by 3. And on the second list, you started at 7 and kept adding 5s. Move 5 spots to the left and you’re again at 2, which does leave a remainder of 2 when divided by 5. So upon closer examination, this problem has two solutions: 2 and 17.
The GMAT does a masterful job of setting ranges that test-takers don’t exhaust, and that’s where the Snoop Limbo mentality comes into play. If you’re always asking yourself “how high can I get and how low can I go?” you’ll force yourself to consider all available options. So for example, if the test were to tell you that:
x^2 < 25 –> This doesn’t just mean that x is less than 5 (how high can you get) it also means that x is greater than -5 (how low can you go)
x is a positive three-digit integer –> make sure you try 100 (how low can you go) and 999 (how high can you get)
x > 0 –> You might want to start with 1, but make sure you consider fractions like 1/2 and 1/8, too (how low can you go? all the way to 0.00000….0001), and try a number in the thousands or millions too (how high can you get?) since most people will just test easy-reference numbers like 1, 2, 5, and 10. A massive number might react differently.
In triangle ABC, angle ABC measures greater than 90 degrees –> remember that “how high can you get” is capped by the fact that the three angles have to add to 180, but this obtuse angle can get up even above 179 (how high can you get?)
x is a nonnegative integer –> the smallest integer that’s not negative is 0, not 1! How low can you go? You’d better check 0.
3 < x < 5 –> it doesn’t have to be 4, as x could be 3.0000000001 or 4.99999999
So keep Snoop’s Limbo Contest in mind when you pick numbers on Data Sufficiency problems. Don’t just pick the easiest numbers to plug in or the first few numbers that come to mind. The GMAT often plays to the edge cases, so always ask yourself how high you can get and how low you can go.
(and for our readers who prefer East Coast rap to West Coast rap, feel free to substitute this with the “Biggie (how big a number can you use) Smalls (how small a number can you use)” method and you can end up with a notoriously big score).
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: 5 Common Quant Section Mistakes That You Must Avoid
Much of your GMAT preparation will focus on “more” – learning more content, memorizing more rules, feeling more comfortable with the test format, and ultimately getting more questions right. But might impact your score more than “more” is your emphasis on “less” (or “fewer”). Feeling less anxiety, taking less time on tricky problems, having to guess less than in your previous attempts, and this ever-important concept:
Making fewer mistakes.
On an adaptive test like the GMAT, making silly mistakes on problems that you should get right can be devastating to your score. Not only do you get that question wrong, but now you’re being served easier questions subsequent to that, with an even more heightened necessity of avoiding silly mistakes there. So you should make a point to notice the mistakes you make on practice tests so that you’re careful not to make them again. Particularly under timed pressure in a high-stress environment we’re all susceptible to making mistakes. Here are 5 of the most common so that you can focus on making fewer of these:
If someone asked you to pick a number 1-10, you might pick 5 or 6, or maybe you’d shoot high and pick 9 or low and pick 2. But you probably wouldn’t respond with 9.99 or 3 and 1/3. We tend to think in terms of integers unless told otherwise. Similarly, if someone asked “what number, squared, gives you 25” you’d immediately think of 5, but it might take a second to think of -5. We tend to think in terms of positive numbers unless told otherwise.
On the GMAT, a major concept you’ll be tested on is your ability to consider all relevant options (an important skill in business). So before you lock in your answer, ask yourself whether you considered: positive numbers (which you naturally will), negative numbers, fractions/nonintegers, zero, the biggest number they’d let you use, and the smallest number they’d let you use.
An easy way for the GMAT testmaker to chalk up a few more incorrect answers on the problem is to include an extra valuable or an extra step. For example, if a problem asked:
Given that x + y = 8 and that x – y = 2, what is the value of y?
You might quickly use the elimination method for systems of equations, stacking the equations and adding them together:
x + y = 8
x – y = 2
2x = 10
x = 5
But before you pick “5” as your answer, reconsider the question – they made it convenient to solve for x, but then asked about y. And in doing so, they baited several test-takers into picking 5 when the answer is 2. Make sure you always ask yourself whether you’ve answered the right question!
3) Multiplying/dividing variables across inequalities.
By the time you take the test you should realize that if you multiply or divide both sides of an inequality by a NEGATIVE number, you have to flip the sign. -x > 5 would then become x < -5. But the testmakers also know that you’re often trained mentally to only employ that rule when you see the negative sign, –
To exploit that, they may get you with a Data Sufficiency question like:
Is a > 5b?
(1) a/b > 5
And many people will simply multiply both sides of statement 1’s equation by b and get to an ‘exact’ answer: a > 5b. But wait! Since you don’t know whether b is positive or negative, you cannot perform that operation because you don’t know whether you have to flip the sign. When you see variables and inequalities, make sure you know whether the variables are negative or positive!
4) Falling in love with the figure.
On geometry questions, you can only rely on the figure’s dimensions as fairly-reliable measurements if: One, it’s a Problem Solving question (you can never bring in anything not explicitly provided on a DS problem); and, two, if the figure does not say “not drawn to scale”. But if it’s a Data Sufficiency problem *or* if the figure says not drawn to scale, you have to consider various ways that the angles and shapes could be drawn. Often times people will see a “standard” triangle with all angles relatively similar in measure (around 60 degrees, give or take a few), and then base all of their assumptions on their scratchwork triangle of the same dimensions. But wait – if you’re not told that one of the angles could be, say, 175 degrees, you could be dealing with a triangle that’s very different from the one on the screen or the one on your scratchwork. Don’t get too beholden to the first figure you see or draw – consider all the options that aren’t prohibited by the problem.
5) Forgetting that a definitive “no” answer to a Data Sufficiency question means “sufficient.”
Say you saw the Data Sufficiency prompt:
Is x a prime number?
1) x = 10! + y, where y is an integer such that 1 < y < 10
Mathematically, you should see that since every possible value of y is a number that’s already contained within 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, whatever y is the new number x will continue to be divisible by. For example, if y = 7, then you’re taking 10!, a multiple of 7, and adding another 7 to it, so the new number will be a multiple of 7.
Therefore, x is not a prime number, so the answer is “no.” But here’s where your mind can play tricks on you. If you see that “NO” and in your mind associate that with “Statement 1 — NO”, you might eliminate statement 1 when really statement 1 *is* sufficient. You can guarantee that answer that x is not prime, so even though the answer to the question is “no” the statement itself is “positive” in that it’s sufficient.
So be careful here – if you get a definitive “NO” answer to a statement, don’t cross it out or eliminate it!
Remember, a crucial part of your GMAT study plan should be making fewer mistakes. While you’re right to seek out more information, more practice problems, and more skills, “fewer” is just as important on a test like this. Make fewer of the mistakes above, and your score will take you more places.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Cedric the Entertainer becomes Cedric the GMAT Instructor
“They hope. We wish.”
In his classic routine from The Original Kings of Comedy, Cedric the Entertainer talks about the way that two different types of people view confrontation.
Some people hope that there’s no confrontation, worrying all the while that there might be.
Others – including Cedric himself – “wish a would” start some conflict. (Note: Kanye West borrowed this sentiment years later in a lyric for “The Good Life”)
On the GMAT, you want to be on Cedric’s team. Many test-takers go into the reasoning-based exam hoping that they don’t see too much Testmaker trickery, but those poised to score 700+ – the Original Kings of Calm on the test – wish the testmaker would. They’ve prepared to check negative numbers and nonintegers on Data Sufficiency. They’ve prepared to double-check their inferences on Critical Reasoning and Reading Comprehension questions to make sure they “must be true” (correct) and not just “probably true.” They’ve prepared to go back to the question on Problem Solving to make sure that the variable they solved for is the one that the question asked about. They’ve tracked the silly and recurring mistakes that they made in practice and they wish the test tries to sneak that by them on test day.
Why?
A few reasons. For one, any mistake you’ve made more than once in practice is something that you know is going to be difficult for people. By being ready for it, you’re poised to get “cheap” difficulty points (so to speak) when it’s really not that hard. If a question asks:
Starting with a full 12-gallon tank of gas, D.L. drove 225 miles getting 45 miles per gallon of gas burned. How much gas was left in D.L.’s tank at the end of the trip?”
You WANT them to ask about the gas that’s LEFT OVER (7 gallons) and include the amount of gas that was USED (5 gallons) as a trap answer. The math is pretty pedestrian, but that little twist – that you’ll solve for the amount used and then have to take just one more step to finish the problem, subtracting that 5 gallons used from the 12 you started with – will ensure that at least 20% more people get that problem wrong for just not reading carefully or from being in a hurry to finish the math and move on. You want to see those silly little trap answers there because they add difficulty (and therefore points) to your test without being truly “hard.”
Another reason is that there’s nothing more confidence-building than catching the GMAT trying to beat you with a silly trick that you’re more than prepared for. That’s Cedric’s point about concert tickets; sometimes it’s not sitting in great seats that makes you feel truly big-time, it’s being able to prove to someone else that you’ve earned the right to sit in them. That’s why Cedric wishes a would sit in his seats; he wants that pure satisfaction that comes from being justified in kicking them out! That adds happiness and satisfaction to the whole show. Similarly, when you catch the GMAT trying to trick you with a trap you saw coming from a mile away, that’s a huge confidence boost for the rest of the test. And that’s the ultimate point of this post – you can’t go into the test fearful of falling for traps. If that’s your mindset – “I really hope the GMAT doesn’t trick me into forgetting about zero” – then even if you catch that and save your answer, it can breed more stress. In a Data Sufficiency format, that could look like:
What is the value of x?
(1) 8x = x^2
(2) x is not a positive number
But on Cedric’s team – I wish the GMAT would try to sneak numbers like negatives, fractions, and zero past me – that same discussion looks like this (in bold because, well, it’s a bolder way of thinking):
What is the value of x?
(1) 8x = 8^2
<Cedric’s discussion with self: Man I know you want me to say 8 but that’s easy. I think x has to be 8 but I think you may be trying to trick me, GMAT. I’m too quick for that; I’m a grown-ass man dawg. We ain’t through here, you hear me?.>
(2) x is not a positive number
<Cedric’s discussion with self: There you go, always talking in code like that. x is not a positive number…you didn’t say it was negative so what’s the difference there. It’s zero; you don’t think I know that? So I see what you’re doing…I knew you’d try to throw zero at me. 8x = x^2 above? Anything times 0 is 0 so 0 is that second answer up top; I knew it wouldn’t be that easy. Statement 1 isn’t sufficient because of 0 and 8 and statement 2 says it can’t be 8. That’s C, dawg, as in you can’t C me easy like that. What do you have up next there Einstein?>
The real difference? Cedric’s mindset uses his knowledge that the GMAT will hit you with common traps as confidence. He knows it’s coming and he’s happy when he does see it, and catching those traps just breeds more confidence since he knows he’s better than the test and handling at least some of it’s difficulty with ease. The other mindset – even if it leads to a right answer on a particular question – breeds fear and anxiety, and those qualities can take a toll on future questions. By the time you take the GMAT you know what common traps it’s setting for you, so be confident when you see and avoid them! Like in this example:
x and y are consecutive integers such that x > y. What is the absolute value of y?
(1) The product xy is 20.
(2) x is a prime number.
Have you summoned your inner Cedric? Statement 1 begs you to say “oh, well if x is greater than y and they’re consecutive integers that multiply to 20, it’s 4 and 5 and x is the big one so y = 4. But wait – don’t you wish they’d try to throw a trick at you? Are you ready for it when it comes?
Statement 2 looks to just confirm what you saw before. Yep, x = 5 in statement 1, and if you take statement 2 alone it’s nowhere near sufficient. So what’s Cedric thinking? He wishes that the test would try to hit him with some of the low-level trickery it so often does. The test likes nonintegers? No, those don’t apply since the question says that x and y are indeed integers. The test likes 0? That doesn’t really apply either for statement 1 since 0 times anything can’t equal 20. But the GMAT also likes negative numbers, and you were wishing they’d try to get you with those. What other consecutive integers multiply to 20? -4 and -5. And in that case which is the smaller one (again, x > y)? That’s right, -5. So while the amateur might pick A thinking that the absolute value of y has to be 4, you can answer confidently like Cedric in the clip above:
“That’s right. Fo *and* five.”
Statement 1 is not sufficient alone, but statement 2 guarantees that the numbers have to be positive, so the answer is C. And since you wished the GMAT would try to get you with that positive/negative trick, you were looking for it, you answered correctly, and you confidently moved on to the next problem knowing that you’re on a roll.
On the GMAT, don’t hope they don’t try to make it difficult with those tricks that got you in practice. Wish they would make it difficult with those tricks because you’re confident you won’t fall for them again. They hope; you wish.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Thinking Out Loud With Drake, Nicki, and Wayne
It’s the hottest song in the country with a beat you just can’t get out of your head. Which is a good thing, because as you go in to take the GMAT you’d be well served to heed some of the lessons that Drake, Nicki Minaj, and Lil Wayne weave into the latest single from Nicki’s album.
The beat itself, with the steady bass line followed by the singsongy “You know…,” is a positive affirmation in and of itself. You DO know. You know how to solve these problems. You know that if you can’t make sense of the question you can often find clues in the answer choices. You know that just getting started and writing down what “you know…” is often the key to lessening anxiety and getting the prompt into an actionable format. You know.
But the master message in this track is the way that the three most prominent rappers in the game start each verse:
“Thinking out loud…”
Why is that important to you, the GMAT test-taker? Because that’s the way that the greatest test-takers start GMAT problems, too.
“Thinking out loud…”
Thinking out loud on the GMAT means having a conversation with yourself about the problem. It means staying relaxed and getting your thoughts together before you panic about the challenge of the problem. It means understanding that many problems won’t have an obvious set of steps that you can begin right away; they’ll require you to start loose and take account of your assets and the strategies in your toolkit. One of the keys to success on the GMAT is thinking out loud.
Check the rapgenius.com annotation for why Drizzy/Nicki/Weezy start each verse that way: All three MCs start their verses with some variation of “I’m thinkin’ out loud,” lending the song a breezy, friends-in-the-booth feeling. The recording was probably not a casual meeting, at all, but they’re good at sounding relaxed.
For that reason alone, thinking out loud is important for you. Their recording wasn’t a casual meeting – as they go on to say in all their lyrics they’re some of the wealthiest and most sought-after people on the planet, so that meeting was a big deal – but they were able to approach it like it was. Similarly your GMAT is, indeed, a big deal, but casual, calm problem solving is the name of the game. Teaching yourself to think out loud – “so I know that x and y must be positive but z could be either positive or negative…” – is a great way to get your mind thinking calmly and proactively as opposed to the all-too-commmon reactive mode of “I don’t even know where to start.”
But thinking out loud isn’t just a psychological tool, it’s also a tactical tool. Tricky GMAT problems are notorious for forcing you to see your assets from different angles before you can package them in a way to solve the problem. Too often students are looking for “the way” to do a problem when really they should be looking for “a way”. Which seems like a trivial difference but going in with the mindset that there may be several ways to solve the problem allows you to be flexible and see assets, not liabilities. Consider the example:
If side AB measures 3 and side BC measures 4, what is the length of line segment BD?
(A) 7/5
(B) 9/5
(C) 12/5
(D) 18/5
(E) 23/5
While many will rush into an abyss of Pythagorean Theorem, thinking out loud can show you a calm, proactive way to do this.
“Thinking out loud…I know that it’s a right triangle so if AB = 3 and BC = 4, it’s a 3-4-5 and side AC is 5. And as much as I want side AC to be cut in half by point D I don’t think I can do that. There are three different right triangles so I could go nuts with Pythagorean Theorem but that’s a lot of work. Thinking out loud, I also know that the perimeter is 3 + 4 + 5 and the area is 1/2(base)(height) so that’s 1/2 (3)(4) = 6. But what can I do with that?
Thinking out loud…the answer choices are all divided by 5…why do they all look like that? The only 5 in the problem so far is the 5 that’s side AC. Why would I multiply or divide by that?
Thinking out loud…BD is definitely going to be smaller than 4 because there’s no way it’s longer than side BC. So it can’t be E. But what else do I know about BD? It’s perpendicular to side AC, and AC is 5 and that’s that 5 in the denominator. Thinking out loud…what if I drew the triangle so that AC was on the bottom and not on the side? Then BD would be the height of triangle ABC and AC would be the base…but wait, I already know the area is 6, so that area 1/2 (side BD)(5) has to be 6, which means that side BD has to be 12/5, answer choice C.”
The takeaway here is that almost no one sees the area relationship with side BD right away, and that’s okay. The key to working on problems like these is staying loose and filling in unknowns. You can’t simply do math on paper and follow a set of steps…you need to do some thinking out loud and talk to yourself as you solve. For each of Drake, Nicki, and Wayne the phrase “thinking out loud” is followed by a wild description of how much money they have. Follow that “thinking out loud” philosophy and you’ll be on a similar pace with the help of an elite MBA.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Writing the AWA Without Engaging Your Brain
Writing a Friday GMAT Tip of the Week post on a tight deadline is a lot like writing the AWA essay in 30 minutes.
30 minutes is not a lot of time, many say, and because an effective essay needs to be well-organized and well-written it is therefore impossible to write a 30-minute essay.
Let’s discuss the extent to which we disagree with that conclusion, in classic AWA style.
In the first line of a recent blog post, the author claimed that writing an effective AWA essay in 30 minutes was impossible. That argument certainly has at least some merit; after all, an effective essay needs to show the reader that it’s well-written and well-organized. But this argument is fundamentally flawed, most notably because the essay doesn’t need to “be” well-written as much as it needs to “appear” well-written. In the paragraphs that follow, I will demonstrate that the conclusion is flawed, and that it’s perfectly possible to write an effective AWA essay in 30 minutes or less.
Most conspicuously, the author leans on the 30-minute limit for writing the AWA essay, when in fact the 30 minutes only applies to the amount of time that the examinee spends actually typing at the test center. In fact, much of the writing can be accomplished well beforehand if the examinee chooses paragraph and sentence structures ahead of time. For this paragraph, as an example, the transition “most conspicuously” and the decision to refute that claim with “in fact” were made long before I ever stopped to type. So while the argument has merit that you only have 30 minutes to TYPE the essay, you actually have weeks and months to have the general outline written in your mind so that you don’t have to write it all from scratch.
Furthermore, the author claims that the essay has to be well-written. While that’s an ideal, it’s not a necessity; if you’ve followed this post thus far you’ve undoubtedly seen a number of organizational cues beginning and then transitioning within each paragraph. However, once a paragraph’s point has been established the reader is likely to follow the point even if it’s a hair out of scope. Does this sentence add value? Maybe not, but since the essay is so well-organized the reader will give you the benefit of the doubt.
Moreover, while the author is correct that 30 minutes isn’t a lot of time, he assumes that it’s not sufficient time to write something actually well-written. Since the AWA is a formulaic essay – like this one, you’ll be criticizing an argument that simply isn’t sound – you can be well-prepared for the format even if you don’t see the prompt ahead of time. Knowing that you’ll spend 2-3 minutes finding three flaws in the argument, then plug those flaws into a template like this, you have the blueprint already in place for how to spend that time effectively. Therefore, it really is possible to write a well-written AWA in under 30 minutes.
As discussed above, the author’s insistence that 30 minutes is not enough time to write an effective AWA essay lacks the proper logical structure to be true. The AWA isn’t limited to 30 minutes overall, and if you’ve prepared ahead of time the 30 minutes you do have can go to very, very good use. How do I know? This blog post here took just under 17 minutes…
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Prepare for the GMAT Using the Study Plan Rule of Thirds
Here on the first Friday of April, we’ve officially ended the first quarter of the year and fiscal reports are streaming in. But who’s in a hurry to finish 2015?
We’re still firmly entrenched in the first third of the year, and if 2015 is the year that you plan to conquer the GMAT you’re in luck. Why?
Because your GMAT study plan should include three phases:
1) Learn
One of the most common mistakes that GMAT studiers make is that they forget that they need to learn before they can execute. Are you keeping an eye on the stopwatch on every question you complete? Are you taking multiple practice tests in your first month of GMAT prep? Have you uttered the phrase “how could I ever do this in two minutes???”? If so, you’re probably not paying nearly enough attention to the learning phase. In the learning phase you should:
• Review core skills related to the GMAT by DOING them and not just by trying to memorize them. You were once a master of (or maybe a B-student at) factoring quadratics and identifying misplaced modifiers and completing long division. Retrain your mind to do those things well again by practicing those skills.
• Learn about the GMAT question types and the strategies that will help you attack them efficiently. For this you might consider a prep course or self-study program, or you can always start by reviewing prep books and free online resources.
• Take as much time as you need to complete and learn from problems. You’ll learn a lot more from struggling through a problem in six minutes than you will from taking two minutes, giving up, and then reading the typewritten solution in the back of the book. Let yourself learn! Again, it’s critical to learn by doing – by actively engaging with problems and talking yourself into understanding – than it is to try to memorize your way to success. The stopwatch is not your friend in the first third of your preparation!
• Embrace mistakes and keep a positive attitude. The GMAT is a hard test; most people struggle with unfamiliar question formats (Data Sufficiency, anyone?) and challenging concepts (without a calculator, too). Recognize that it will take some time to learn/re-learn these skills, and that making mistakes and thinking about them is one of the best ways to learn.
2) Practice
Regardless of how you’ve studied, you’ll need to complete plenty of practice to make sure you’re comfortable implementing those strategies and using those skills on test day. Once you’ve developed a good sense of what the GMAT is testing and how you need to approach it, it’s time to spend a few weeks devouring practice problems. Among the best sources include:
In this phase, you can start concerning yourself with the stopwatch a little and you’ll want to identify weaknesses and common mistakes so that you can emphasize those. Particularly with GMAT verbal, the more official problems you see the more you develop a feel for the style of them, so it’s important to emphasize practice not just for the conscious skills but also for those subconscious feelings you’ll get on test day from having seen so many ways they’ll ask you a question.
3) Execute
Before you take the GMAT you should have taken several practice tests. Practice tests will help you:
• Work on pacing and develop a sense for how much time you’ll need to complete each section. From there you can develop a pacing plan.
• Determine which “silly” mistakes you tend to make under timed pressure and exam conditions, and be hyperaware of them on test day.
• Develop the kind of mental stamina you’ll need to hold up under a 4-hour test day. Verbal strategies can be much easier to employ in a 60-minute study session than at the end of a several-hour test! Make sure that at least a few times you take the entire test including AWA and IR for the first hour.
• Continue to see new problems and hone your skills.
While it’s not a terrible idea to take a practice test early in your study regimen and another partway through the Practice phase, most of your tests should come toward the end of your study process. Why? Because the learning and practice phases are so important. You can’t execute until you’ve developed the skills and strategies necessary to do so, and you won’t do nearly as effective a job of gaining and practicing those if you’re not allowing yourself the time and subject-by-subject focus to learn with an open mind.
So be certain to let yourself learn with a natural progression via the GMAT Study Rule of Thirds. Learn first; then focus on practice; then emphasize execution via practice tests. Studying in thirds is the best way to ensure that you get into a school that’s your first choice.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Kanye West Teaches You How To Live The Data Sufficiency Good Life
Welcome back to Hip Hop Month in the GMAT Tip of the Week space, where we know precisely why you want an MBA: so you can live some of the good life. You want a better job with a higher salary and better benefits. You want to invest big chunks of that higher salary to create passive income that brings you even more money per year. And if they hate then let ’em hate and watch the money pile up. Welcome to the Good Life.
Few are living the good life better than the author/performer of “The Good Life,” Kanye West-Kardashian. And while it may seem ironic for “The College Dropout” to provide the best advice for getting into a top graduate school, the way Kanye describes the Good Life provides you with critical advice for obtaining the good life via a high quant score on the GMAT. When you’re practicing Data Sufficiency, pay attention to Yeezy as he says:
“So I roll through good. Y’all pop the trunk; I pop the hood. (Ferrari)
If she got the goods, and she got that ass, I got to look. (Sorry)”
How does this lyric relate to Data Sufficiency? We’ll translate.
As you’re rolling through a standard Data Sufficiency problem, it’s quite common to make your decision on statement 1 alone (pop the trunk) and then on statement 2 alone (pop the hood). And since time is of the essence, you do so quickly (Ferrari). For example, you might see the problem:
Is yz > x?
(1) y > x/z
And then quickly think to yourself “if I take the given statement and multiply both sides by z, I get a direct answer: yz > x, so that’s sufficient. Now let’s look at statement 2 alone because the answer must be A or D.”
But if you’re on your way to the Good Life, you need to play the Data Sufficiency game at a higher level, and that level may be a little different from the status quo. (“50 told me go ‘head switch the style up…”) So read on:
“If she got the goods” refers to the other statement. If “the other statement” seems fairly obvious on its own, most of us will see that as very, very good. We can quickly make our determination, eliminate the last answer choice or two, and move on. But wait:
“And she got that ass, I got to look,” of course, refers to statements having “that assistance.” For example, if statement 2 in this problem were to say:
(2) z < 0
Knowing that z is negative is “that ass(istance)”. It’s clearly insufficient on its own (what about y and x?), but in giving you the goods that z is negative it’s assisting you in avoiding a catastrophic mistake. In statement 1, you multiplied both sides of an INEQUALITY by a variable, z. But statement 2 tells you that the variable is negative, which means that simply multiplying by z without flipping the sign – or at least considering that the sign might need to be flipped – was a mistake. You had to consider negative/positive there – if z were positive, you just multiply; if it were negative, you’d multiply and flip. And since you didn’t know what sign z took when you assessed statement 1 alone, statement 1 actually was not sufficient. You need statement 2’s ass(istance), so the answer is C.
And that’s where Kanye’s lyric is so important. “IF she got that ass(istance), I got to look (Sorry)” means that, while the standard operating procedure for Data Sufficiency is to adhere strictly to: 1 alone, then forget; 2 alone, then forget; if nether was sufficient alone then try them together, that strategy leaves some valuable points on the table. If statement 2 gives you information that you hadn’t considered when you assessed statement 1, you’ve got to look at how that new piece of information would have impacted your decision. Did you need to know that or not? And although this new strategic element may contradict the easy process-of-elimination that helped you learn Data Sufficiency in the first place (Sorry), it’s critical if you’re going to live the 700+ good life – difficult Data Sufficiency is structured to reward those who see the potential for clues in the question stem and in the “other” statement, those who leverage assets that may not be readily apparent to the average test taker.
Note that sometimes that new piece of information is unnecessary. For example, if the question were instead:
Is yz = x?
(1) y > x/z
(2) z < 0
You actually don’t need to know the sign. When you use statement 1 alone and multiply both sides by z, you either get yz > x (if z is positive) or yz < x (if z is negative). It’s either greater than or less than with no room for equals, so you don’t need the sign. So statement 2 isn’t always necessary, but if it appears to give assistance you’ve got to look – you have to at least consider whether it’s important, because that’s where the GMAT has set up the difficulty. On the most difficult problems, the GMAT will tend to reward those who can leverage all available information to think critically and make a good decision, so it pays to at least take a fairly-obvious-on-its-own statement and look at it in the context of the other statement, just in case.
So learn from Yeezy (who in classic yz > x form is a much greater instructor than Xzibit) and remember – the easier statement’s always got the goods, so on the chance that it’s got that ass(istance), you’ve got to look. The good life: it feels like Palo Alto, it feels like Cambridge, it feels like Fontainebleu. If you’ve got a passion for flashing that acceptance letter, when a Data Sufficiency statement looks too obvious on your own, ask yourself what would Yeezus do.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Slow Motion Is Better Than No Motion
Welcome back to Hip Hop Month in the GMAT Tip of the Week space, where 3-13 isn’t just a day to honor Eminem’s group “Three and a Third” from 8 Mile (we’ll save that for 10/3). It’s also Common’s birthday, so what better day to let one of the most intellectual rappers in the game help you take your game toward his South Side neighborhood (Chicago-Booth isn’t all that far away) or, we suppose, to the North Side and Kellogg?
Now, while you’re in the thick of the quant section looking for an un-Common-ly high score, the only Common lyric in your head is probably “Go!”. But particularly when you get to dense word problems, you’ll likely have more success if you heed his advice from the beginning and the refrain from “The Food“:
Slow motion better than no motion.
What’s Common trying to tell you about how to approach the quant section? Essentially this: most examinees hurry through their initial read of a problem, taking ~20 seconds to read the entire paragraph prompt, only to get to the question mark, sigh, and go back to the top to get started. That’s “no motion” on your first 20 seconds – which, if you’re holding to an average of 2 minutes per problem, is almost 17% of the time you have to get it done.
What should you do? Slow motion, which is better than no motion. What does that mean? Start writing and thinking while you read. For example, consider this problem:
Working in a South Side studio at a constant rate, Kanye can drop a full-length platinum LP in 5 weeks. Working at his own constant rate, Common can drop a full-length platinum LP in x weeks. If the two emcees work together at their independent rates, they can drop a full-length platinum compilation LP in 2 weeks. Assuming no efficiency is lost or gained from working together, how many weeks would it take Common, working alone, to drop a full-length platinum LP?
(A) 3 and 1/3 weeks
(B) 3 weeks
(C) 2 and 1/2 weeks
(D) 2 and 1/3 weeks
(E) 2 weeks
Now, while your instinct may be to Go! and speed through your initial read of this rate problem, remember: slow motion (is) better than no motion. As you read each sentence, you should start jotting down variables and relationships so that by the time you get to the question mark you have actionable math on your noteboard and you don’t have to read the question all over again to get started. You should be thinking:
Working in a South Side studio at a constant rate, Kanye can drop a full-length platinum LP in 5 weeks.
Rate (K) = 1 album / 5 weeks
Working at his own constant rate, Common can drop a full-length platinum LP in x weeks.
Rate (C) = 1 album / x weeks
If the two emcees work together…
I’m adding these rates, so their combined rate is 1/5 + 1/x
…they can drop a full-length platinum compilation LP in 2 weeks.
And they’re giving me the combined rate of 1 album / 2 weeks, so 1/5 + 1/x = 1/2
Assuming no efficiency is lost or gained from working together, how many weeks would it take Common, working alone, to drop a full-length platinum LP?
I’m using that equation to solve for Common’s time, so I’m solving for x.
Now by this point, that slow motion has paid off – your equation is set, your variable is assigned, and you know what you’ve solving for. Your job is to solve for x, so:
1/5 + 1/x = 1/2, so let’s get the x term on its own:
1/x = 1/2 – 1/5. and we can combine the two numeric terms by finding a common denominator of 10:
1/x = 5/10 – 2/10
1/x = 3/10, and from here you have options but let’s cross multiply:
10 = 3x, so divide both sides by 3 to get x alone:
10/3 = x, and that doesn’t look like the answer choices so let’s convert to a mixed number: 3 and 1/3 (there’s that number again), for answer choice A.
What’s the real lesson? It’s like Common says: slow motion (is) better than no motion, so you should read just a little slower but have some scratchwork to show for your initial read of the prompt. If you can:
-assign variables
-jot down relationships or equations
-write down which variable the answer wants
You’ll have a lot more to show for your initial 30 seconds with each problem, and you’ll find that you solve problems much more quickly this way because you have less wasted time. So heed Common’s uncommon wisdom (which is really just common sense): the best way to Go is to remember that slow motion > no motion.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: LLC Reasoning
In a time-honored tradition here at Veritas Prep, March is Hip Hop Month in the GMAT Tip of the Week space, and no cutting-edge hip hop blog in 2015 would be complete without mentioning the hottest thing from this year’s Grammys:
LL Cool J
Really? It’s been more than 20 years since LL had to start an album with the phrase “don’t call it a comeback; I’ve been here for years.” And yet as America’s favorite award show host and the star of either NCIS or CSI (all we know is that the man loves initials and acronyms), LL Cool J remains a household name in a young man’s game. Which should draw attention to his rather unique moniker:
LL Cool J
Which, of course, stands for Ladies Love Cool James, and also stands as an important Critical Reasoning lesson. When you think about it, LL Cool J’s name is kind of absurd. It’s three initials and a full word, and it forms a complete sentence (L = subject; L = verb; Cool = adjective; J = object). And he somewhat arbitrarily chose to spell out what, in the sentence, is the least important part. Ladies love James is the operative part of the sentence. “Cool” doesn’t add a ton of real value. So why did LL Cool J (James Smith, for those keeping score at home) choose to spell out what seems like the least critical word in the sentence?
Because LL Cool J is a marketing and Critical Reasoning genius.
Think about it: Ladies LCJ would be a terrible name. So would L Love CJ. And LLC James has a nice businessman ring to it, but lacks the kind of street cred it took to rise through the early NYC rap ranks in the 1980s. It had to be LL Cool J.
And what’s more: LL Cool J is telling you how you can master Critical Reasoning by calling attention to the modifier or adjective that adds specificity. Consider these two arguments.
1) The Something Like a Phenomenon award each year goes to the highest selling album of the year. This year, the highest selling album of the year was LL Cool J’s Mr. Smith, so Mr. Smith will be awarded the Something Like a Phenomenon award.
2) The Something Like a Phenomenon award each year goes to the highest selling rap album of the year. This year, the highest selling album of the year was LL Cool J’s Mr. Smith, so Mr. Smith will be awarded the Something Like a Phenomenon award.
What’s the difference? Like the name LL Cool J itself, it really comes down to an adjective. In #2, the first premise has an adjective qualifier – to win the award, it has to be a rap album. And since the second premise doesn’t tell us that Mr. Smith was a rap album, that argument is now vulnerable to criticism…that one adjective that made things a little bit more specific left a hole in the argument for us to attack.
And noticing that hole is everything on Strengthen and Weaken CR questions. If it’s a Strengthen question and we’ve noticed that gap, you should be looking for something that demonstrates that Mr. Smith was a rap album. And if it were a Weaken question, you’re looking for a reason to believe that it wasn’t. But either way, by finding that hole in the argument you now know what you’re looking for.
So the lesson? Train yourself – like the man who could have just been called LLJ did – to spot those extra adjectives or modifiers that make the conclusion or major premise of an argument that much more specific. Look for things like:
“Def Jam Records must find a way to reduce its costs.” vs. “Def Jam records must find a way to reduce its distribution costs.”
“Mr. Smith was host of the Grammys.” vs. “Mr. Smith was host of the 2015 Grammys.”
“LL Cool J proclaimed himself the greatest.” vs. “LL Cool J proclaimed himself the greatest of all time.”
and
“I need a girl.” vs. “I need an around-the-way girl.”
So step one is to notice those (cool) qualifiers that lend themselves to gaps in arguments, in which a more-generic premise just can’t connect to that more-specific statement. Once you’ve identified that potential for a gap in logic, check the other statements to see if they match the specificity. If they don’t – as they generally won’t in GMAT Critical Reasoning – well, then get critical. As mama says…knock it out.
Are you studying for the GM Admissions T? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: The Dress is White and Gold and Your GMAT Score Can Become Golden
The Dress” is white and gold, as all reasonable people can certainly agree. But a sizable, misguided percentage of the internet vehemently disagrees with that fact, proving two major points:
1) You can’t trust what people say on the internet.
2) Your five major senses can deceive you, so you can’t rely on them when approaching GMAT Sentence Correction problems.
If you want to avoid leaving the GMAT test center black-and-blue, beaten up by tricky Sentence Correction problems, make sure you do better than trusting your ear. Much like the powers that launched The Dress on us, the GMAT testmakers know that our senses don’t always hold true to logic and reason, and so they mine Sentence Correction problems with opportunities to be misled by your ear. Consider the example:
While Jackie Robinson was a Brooklyn Dodger, his courage in the face of physical threats and verbal attacks was not unlike that of Rosa Parks, who refused to move to the back of a bus in Montgomery, Alabama.
(A) not unlike that of Rosa Parks, who refused
(B) not unlike Rosa Parks, who refused
(C) like Rosa Parks and her refusal
(D) like that of Rosa Parks for refusing
(E) as that of Rosa Parks, who refused
For many, the phrase “not unlike” is a red (or black-and-blue) flag right away. Your ear may very well abhor that language, and if so you’ll quickly eliminate the white-and-gold answer A and answer B right away. But A is actually correct, as this sentence requires:
-“that of” (to compare Jackie Robinson’s courage with Rosa Parks’s courage)
-“who refused” (to make it clear that Rosa Parks was the one who refused to the back of the bus; with “for refusing” in D it’s unclear who that last portion of the sentence belongs to)
And only choice A includes both, so it has to be right. What makes this problem tricky? The GMAT testmakers know that:
1) You read left to right and top to bottom
2) Your ear likely won’t take kindly to “not unlike” even though it’s not wrong. “Not unlike” is saying “it’s not totally different from, even though it’s not the same thing,” whereas “like” indicates a much closer relationship. There’s a continuum there, and the phrase “not unlike” has a valid meaning on that continuum of similarity.
And so what do the testmakers do? They:
1) Make “not unlike” vs. “like” the first difference between answer choices, daring you to use your ear before you use your Sentence Correction strategy (look for modifiers, verbs, pronouns, and comparisons first)
2) Put the answer you won’t like (but should pick) first at answer choice A, making it easy for you to eliminate the right answer right away before you start considering the core skills listed in the parentheses above
And the lesson?
Don’t trust your ear as your primary deciding factor on Sentence Correction problems. Your senses – as The Dress shows – are prone to deceiving you, and what’s more the testmakers know that and will use it against you! They want to reward critical thinking, the use of logic and reason, the adherence to proven systems and processes. So they give you the opportunity to use your not-always-reliable senses, and reward you for learning the lesson of The Dress. Your senses can fool you, so on important decisions like The Dress and Sentence Correction, don’t simply rely on your senses: they may just leave you black and blue.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Data Sufficiency and The Imitation Game
With Oscar weekend upon us, it’s only fitting that this week’s GMAT Tip comes courtesy of Alan Turing. Of course the brilliant math mind featured in Best Picture nominee The Imitation Game would crush GMAT Data Sufficiency. But the mere title of the film provides a GMAT tip that can help bring Data Sufficiency success to even us mere mortals who can’t quite use math to save Britain from peril. How can you use The Imitation Game to succeed on Data Sufficiency?
When you’re asked a Yes/No Data Sufficiency question that asks whether an algebraic relationship is true, play The Imitation Game. Which means: if you can get one of the statements to directly imitate the question, you can definitively get the answer “yes” and prove that it’s sufficient.
Consider a few examples of questions that make for great Imitation Game candidates:
Is x – y > a – b?
(1) x + b > a + y
Here you can try to imitate the question with the statement. You want the statement to look more like the question, where x and y are paired together on the left and a and b are paired together on the right. so subtract y from both sides (to get it from the right to the left) and subtract b from both sides (to move it to the right), and the statement becomes:
x – y > a – b
Which directly answers the question “yes” – the question asks if the relationship is true, and by using the statement to imitate the question you can get the statement to directly answer it.
If the product abc does not equal 0, does a/b = c?
(1) bc = a
Here you can again use the statement to imitate the question, dividing both sides by b to get c on its own (which you’re allowed to do since no values are 0), and you have your answer:
c = a/b
Sometimes you’ll be able to imitate the question to get a definite “no” answer, which is still sufficient:
Is x – y > a – b?
(1) a > x and y > b
Here you can combine the inequalities to get them all in to one inequality. By adding the inequalities together (which you can do since the signs point in the same direction), you have:
a + y > x + b
And then you want to imitate the question, which has a and b on one side and x and y on the other. So subtract y and b from both sides to get:
a – b > x – y
Which is the opposite of the question, and therefore says “no, x – y is not greater than a – b” providing you with sufficient information.
The real lesson here? When you’re being asked a yes/no question with lots of algebra, it pays to play The Imitation Game. See if you can get the statement to imitate the question, and you’ll often find that it directly answers the question.
But be careful! As the second example showed you, you need to be careful when diving into algebra that you don’t:
*Divide by a variable that could be 0
*Multiply or divide by a variable in an inequality if you don’t know the sign
Keep those two caveats in mind and you can imitate math legend Alan Turing while you play the Data Sufficiency Imitation Game. And the winner is…you.
# GMAT Tip of the Week: The Corrupt Mechanic Explains Sentence Correction
Your parallelism knowledge is paramount. You’re a pro when it comes to pronouns. You relax when you see that the problem involves verb tense. You can’t find a modifier error that’s even moderately challenging anymore. You should be a Sentence Correction sensei. So why are Sentence Correction problems still such a problem?
You’re being taken for a ride by a corrupt mechanic.
Let’s explain. The GMAT testmakers are committed to testing the same concepts over and over again: Modifiers, Verbs, Pronouns, Parallel Structure, Logical Meaning… And at a certain point it’s difficult to make those concepts any harder; they are what they are. So the testmakers resort to a time-honored tradition among corrupt mechanics; when oil changes and tire rotations and front-end-alignments aren’t bringing in enough profit, what do corrupt mechanics do?
They fix things that don’t need to be fixed.
The corrupt mechanic never simply fixes, flushes, or replaces the part you came in asking about; he always “strongly recommends” that you add on another service. If you’re not careful, your \$30 oil change becomes a several-hundred dollar outing and your car comes back with shiny new parts that replaced perfectly-functional components, all with a nice labor surcharge on top. As Seinfeld’s George Costanza put it:
Well of course they’re trying to screw you! What do you think? That’s what they do. They can make up anything; nobody knows! “Why, well you need a new Johnson rod in here.” Oh, a Johnson rod. Yeah, well better put one of those on!
Now, in defense of the GMAT testmakers, they’re not trying to steal your money for unnecessary services. But in their quest to reward the kinds of business skills that are associated with avoiding unnecessary expenses and wasted time on ineffective initiatives, the GMAT testmaker does act like a Corrupt Mechanic on Sentence Correction problems. By fixing problems that don’t need fixing, the testmaker steals your attention, not your money. And in doing so, the testmaker baits the unwitting into bad decisions, while also rewarding those who prioritize their Decision Points properly. Consider this example:
Immanuel Kant’s writings, while praised by many philosophers for their brilliance and consistency, are characterized by sentences so dense and convoluted as to pose a significant hurdle for many readers who study his works.
(A) so dense and convoluted as to pose
(B) so dense and convoluted they posed
(C) so dense and convoluted that they posed
(D) dense and convoluted enough that they posed
(E) dense and convoluted enough as they pose
To those who know their role in the GMAT, the verb difference along the right hand side of the answer choices should loom large. “Pose” (present) vs. “Posed” (past) is a very actionable decision and a very common decision on the test. Like an oil change or the replacement of brake pads, verb tense decisions are something you should do regularly! So what does the Corrupt Mechanic do? He takes something uncomfortable – the structure “so dense and convoluted as to…” – but that doesn’t need fixing, and it fixes it. And since that choice comes along the left-hand side, many of us go right along with that and eliminate A with a preference for the more-familiar structures in B and C, without ever realizing that we’ve been “Johnson rodded” into ignoring the ever-important verb tense decision at the ends of the choices.
That’s how the testmaker’s Corrupt Mechanic works in Sentence Correction. He changes things that didn’t need changing and dares you to accept those “repairs” as necessary. So how can you avoid these traps? Be a savvy customer. Know what you want before you start listening to the Corrupt Mechanic’s menu of possible changes; you want to make Verb, Modifier, Pronoun, and Parallelism decisions before you even listen to anything else. Make the common repairs first, and then with the choices that are left you can start to get creative with add-ons.
The GMAT testmakers act like Corrupt Mechanics when they write Sentence Correction problems, so beware that not every change is actually a necessary repair. It’s on you to determine which fixes truly need to be made, so stick to the recommended SC maintenance schedule – the errors most commonly tested – and you’ll avoid falling victim to the Corrupt Mechanic.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# Should You Purchase The New Enhanced Score Report?
For anyone who has ever underperformed their goals on the GMAT, the first question they’ve asked is usually “where did it all go wrong?”. And for those who have asked that question since October 1, 2013, or will ask it soon, there may be an answer waiting for you.
The GMAT Enhanced Score Report is here.
This new add-on report, which costs \$24.95 USD, will provide you with diagnostic feedback from your official GMAT exam, including such information as:
-Performance (percentile ranking) by question type, with question types including Data Sufficiency vs. Problem Solving; Arithmetic vs. Algebra/Geometry; Critical Reasoning vs. Sentence Correction vs. Reading Comprehension
-Time management by question type, broken down by the same categories above
-Time management by correct vs. incorrect answers for Integrated Reasoning
-Percent of Integrated Reasoning questions answered correctly
So those are the features, but the question remains…is this worth \$25? And the answer is a little less concrete than you might like: it depends. Why?
*The report won’t give you question-by-question feedback, so you’ll never know if you got that crazy coordinate geometry problem at #17 right or wrong, and you won’t know which individual problems you spent way too much time on. You’ll get much more aggregate data, which may or may not help.
*If your performance was pretty similar to that of your practice tests – which ought to be the case for most examinees who have taken several practice tests – the report should likely match your expectations. If you’ve prepared well for the test, there shouldn’t be many surprises in that report.
*However, some users will see some VERY enlightening information. Say, for example, you were quite strong on Critical Reasoning and Reading Comprehension (~80th percentile each) but significantly less adept on Sentence Correction (
So who will benefit from the report? Those who have some outliers or anomalies in their performance. If you were 60th percentile on quant, and a combination of 55th, 62nd, 63rd, and 59th on DS, PS, Arith, Alg/Geo, you’re not going to learn very much from that report. But if one area is significantly higher or significantly lower than the others, you’ll learn something.
-If you’re going to retake the exam, the Enhanced Score Report is essentially a 10% increase on your next registration fee, and has the potential to be pretty enlightening. Especially if you’re likely to spend \$25 over the next month on Starbucks or Amazon impulse purchases or anything else extraneous, it’s a good idea to put that \$25 toward the score report. You might not learn anything, but the chance that you’ll learn something is substantial enough that you should leave no stone unturned.
-But if you have \$25 left in your GMAT budget and the choice is between the Enhanced Score Report or a tool like the GMAT Question Pack or one of the Official Guide supplements, choose the extra practice. If you’ve prepared thoroughly there shouldn’t be too many surprises on that report, and whatever you’d learn you’d have to improve by practicing anyway.
So in sum, GMAT retakers should probably pony up the \$25 because the more you know about your performance, the higher the likelihood that you can improve it. Almost all Veritas Prep instructors agree – we want to see those reports from our students! But don’t be surprised if the report only confirms what you suspected. The Enhanced Score Report is a tool to guide your hard work, not a substitute for the effort required to improve.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: All About That Base
It’s Grammy Weekend here in Los Angeles. All local sports teams have cleared out of the LA Live / Staples Center / Nokia Theater area and local citizens are humming along to the song of the year nominees. How can you (Taylor) Swiftly make your GMAT Quant score (Ariana) Grande, even without the help of an expensive GMAT (Meghan) Trainor? The process isn’t So Fancy, so take that stress and Shake It Off. When you see exponent-based questions, the #1 thing you can do:
What does that mean? Nearly every exponent rule you’ll learn requires common bases. For example:
So when you’re presented with an exponent problem, one question you should always ask yourself is “Can I get all these terms to have the same base?”. That step allows you to use the exponent rules you’ve memorized to solve complicated problems. Consider an example:
For integers a and b, 16*a = 32^b. Which of the following correctly expresses a in terms of b?
(A) = 2^b
(B) a = 4^b
(C) a = 2^5b-4
(D) a = 4^5b-4
(E) a = 2^5b
Here there’s only one exponential term, 32 to the b power. But if you recognize that both 32 and 16 are powers of 2, you can quickly transform the problem, coming up with:
2^4 * a = (2^5)^b
And that allows you to dive right into exponent rules. First deal with the parentheses on the right, using the third exponent rule in the list above so that that term becomes 2^5b. That means you have:
2^4 * a = 2^5b
Then to isolate a, you divide both side by 2^4, getting to:
a = 2^5b / 2^4
And now since your bases are the same, you can use the second exponent rule in the list above to subtract the exponents and get to:
a = 2^(5b – 4), matching answer choice C.
More important than this problem is the lesson: when problems deal with exponents, and particularly with non-prime bases (like 16 and 32), one of your first mantras should be “All About That Base (no treble).” See if you can get multiple terms to have the same base, and you can simplify the expression using common exponent rules. Then, with your monster GMAT quant score, Harvard can take its Blank Space and write your name…
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: The Super Bowl Provides Super GMAT Lessons
It’s Super Bowl weekend, one of the busiest gambling weekends of the year. Maybe you’ll play a squares pool and end up with the dreaded 6:5 combination, maybe you’ll parlay three prop bets and lose on the third, and maybe you’ll bet on your team to win and lose both the game and your cash. How can you turn your gambling losses into investments?
Well, if you’re a GMAT student, you can think about what the odds mean in terms of probability and you can watch the announcers miss Critical Reasoning lesson after Critical Reasoning lesson. For example:
Probability
Before the last piece of confetti hits the turf on Sunday, oddsmakers will have posted their odds on next year’s winner. For example, New England and Seattle might open at 4:1, Green Bay might come in at 7:1, etc. And while you might look at those odds and think “if I bet \$100 on the Packers I’ll win \$700!” you should also think about what those mean. 7:1 for Green Bay is really a ratio: 7 parts of the money says that Green Bay will not win, and 1 part says that it will. So that’s a good bet if you think that Green Bay has a better than 1 out of 8 chance (so better than 12.5%) to win next year’s Super Bowl. And if those are, indeed, the odds (4:1 for two teams and 7:1 for another), Vegas is essentially saying that there’s a less than likely chance (1/5 + 1/5 + 1/8 = 52.5% chance that one of those two teams wins) that someone other than Green Bay, New England, or Seattle will win next year.
So consider what the probability of those bets means before you make them. Individually odds might look tempting, but when you consider what that means on a fraction or percent basis you might have a different opinion.
Probability #2
As you watch the Super Bowl, there’s a high likelihood that at some point the screen will start showing a line indicating the season-long field goal for either Steven Hauschka or Stephen Gostkowski (the Seattle and New England kickers…there’s a huge probability that someone named Steve will be incredibly important in this game!). And the announcers will use that line to say that it’s likely field goal range for that team to win or tie the game.
Where’s the flawed logic? If that’s the longest field goal he’s made all year, is it really likely that he’ll make another one from a similar spot with all that pressure? Or, in the case of a low-scoring game like many predict between these two elite defenses, how likely is either kicker to make two consecutive field goals from a relatively far distance?
Sports fans are pretty bad with that probability. Say that a kicker has been 70% accurate from over 50 yards. Is it likely that he’ll make two straight 50-yard field goals on Sunday (assuming he gets those attempts)? Check the math: that’s 7/10 * 7/10 or 49/100 – it’s less than likely that he makes both! Even a kicker with 80% accuracy is only 8/10 * 8/10 = 64% likely to make two in a row…meaning that fail to perform that feat 1 out of every 3 times he had the chance! Think of the probability while announcers talk about field goals as a near certainty on Sunday.
Critical Reasoning
The announcers on Sunday will try to use all kinds of data to predict the outcome, and in doing so they’ll give you plenty of opportunities to think critically in a Critical Reasoning fashion. For example:
“For the last 40 Super Bowls, the team with the most rushing yards has won (some massive percent) of them; it’s important for New England to get LeGarrette Blount rolling early.”
This is a classic causation/correlation argument. Do the rushing yards really win the game? It could very well be true (Weaken answer!) that teams that build a big lead and therefore want to run out the clock run the ball a lot in the second half (incomplete passes stop the clock; runs keep it going). Winning might cause the rushing yards, not the other way around.
Similarly, the announcers will almost certainly make mention at halftime of a stat like:
“Team X has won (some huge percentage) of games they were leading at halftime, so that field goal to put them up 13-10 looms large.”
Here the announcer isn’t factoring in a couple big factors in that stat:
-A 3-point lead isn’t the same as a 20-point lead; how many of those halftime leads were significantly bigger?
-You’d expect teams leading at halftime to win a lot more frequently; based on 30 minutes they may have shown to be a better team plus they now have a head start for the last 30 minutes. Over time those factors should bear out, but in this one game is a potentially-flukey 3-point lead significant enough?
Regardless of how you watch the game, it can provide you with plenty of opportunities to outsmart friends and announcers and sharpen your GMAT critical thinking skills. So while Tom Brady or Russell Wilson runs off the field yelling “I’m going to Disneyland!”, if you’ve paid attention to logical flaws and probability opportunities during the game, you can celebrate by yelling “I’m going to business school!”
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Learn from DeflateGate and Don’t Get Caught Unintentionally Cheating
It’s Super Bowl week, and instead of Seattle’s miracle comeback over Green Bay or a fantastically-intriguing matchup between the longstanding dynasty in New England and the up-and-coming dynasty in Seattle, all anyone wants to talk about is DeflateGate. Did the Patriots knowingly underinflate or consciously deflate footballs? Did doing so provide a competitive advantage? Will/should they be punished?
Some will say it’s a heinous act committed by serial cheaters. Others will say it’s a minor violation and that “everybody does it.” And still others will say it’s an inadvertent mistake that happened to run afoul of a technicality. What does it mean for you, a GMAT aspirant?
Be careful about honest mistakes that could be construed as cheating!
While the NFL isn’t going to kick the Patriots out of the Super Bowl, the Graduate Management Admission Council won’t hesitate to cancel your score if you’re found to be in violation of its test administration rules. So beware these rules that honest examinees have accidentally violated:
1. You cannot bring “testing aids” into the test center.
Don’t bring an Official Guide, a test prep book, or study notes into the test center with you. You may want to have notes while you’re waiting to check in, but if you’re caught with “study material” in your hands during one of your 8-minute breaks – which has happened to students who were rearranging items in their lockers to grab an apple or a granola bar – you’ll be in violation of the rule, and GMAC has cancelled scores for this in the past. Don’t take that risk! Leave watches, cell phones, and study aids in your car or at home so that there’s no chance you violate this rule simply by having a forbidden item in your hand during a break.
You’ll be at the test center with other people, and someone’s break might coincide with yours. Holding a restroom door or crossing paths near a drinking fountain, you might be tempted to socially ask “how is your test going?” or sympathetically mention “man these tests are hard.” But since those innocent phrases could be seen as “talking about the test” you would technically be in violation of the rule, and GMAC has cancelled scores for this in the past. Your 8-minute break isn’t the time to make new friends – don’t take the risk of being caught talking about the test.
You know that you’re not a cheater, but as most New Englanders feel today it’s very possible to be considered a cheater if you end up on the wrong side of a rule, however accidentally. Learn from the lessons of test-takers before you: avoid these common mistakes and ensure that the score you earn is the score you’ll keep.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Stop Trying to Re-Write the Verbal Section of the Test
Which ineffective habit do nearly all GMAT aspirants have when it comes to studying for the verbal section?
Thou doth protest too much. Meaning:
We all think we can write verbal questions better than the authors of the test.
When it comes to GMAT verbal questions, we critique but don’t solve Critical Reasoning problems, we correct rather than solve Sentence Correction problems, and we try to write but don’t thoroughly read Reading Comprehension questions. And this hubris can be the death of your GMAT verbal score, even if it comes from a good place and a good knowledge base.
Wander into a GMAT class or scan a GMAT forum and you’ll see and hear tons of comments like:
“I feel like the question should say people and not individuals.”
“I would never use the word imply like that.”
“I don’t think that’s the right idiom.”
“I would have gotten it right if it said X…I think it should have said X.”
Or you’ll hear questions like:
“But what if answer choice D said and and not or?”
“If that word were different would my answer choice be right? And if so would it be more right than B?”
And while these questions often come from a genuine desire to learn, they more often come from a place of frustration, and they’re the type of hypothetical thinking that doesn’t lend itself to progress on this test. Even if it’s not always perfect, the GMAT chooses its words very carefully. When the word in the Reading Comprehension correct answer choice isn’t the word you were hoping it would be (but it’s close), they picked that word for a reason – it makes the problem more difficult. When none of the Sentence correction answer choices match the way you or your classmates would have phrased it, that’s not a mistake – that’s an intentional device to make you eliminate four flawed answers and keep the strange-but-correct one. The GMAT can’t always match your expectations, not just because doing so would make it too easy but also because it’s trying to test other critical-thinking skills. It has to test your ability to see less-clear relationships, to make logical decisions amidst uncertainty, to find the least of five evils, and it has to punish you for jumping to unwarranted conclusions.
GMAT verbal is constructed carefully, and as you study it you have to learn how to answer questions more effectively, not to write better questions. The only thing you get to write on test day is the AWA essay; everything else you must answer on the GMAT’s terms, not on your own, so as you study you have to resist the urge to protest the problem and instead learn to see the value in it.
So as you study, remember your mission. Your job isn’t to find a flaw with the logic of the question, but rather with the logic of the four incorrect answers. When you get mad at a wrong answer, use that energy to attack the next problem with the lessons you learned from that frustrating mistake. Take the GMAT as it is and don’t try to justify your mistakes or fight the test.
Save your writing energy for the AWA essay; on the verbal section, you only get to answer the problem in front of you. When you accept that the test is what it is and commit yourself to learning how to attack it through critical thinking and not just general angst, you’ll have a competitive advantage over most frustrated examinees. Think like the testmaker, but don’t try to be the testmaker.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: It’s Always Darkest Before Sunrise
With the winter solstice behind us here in the Northern Hemisphere, you’re probably noticing that the daylight is starting to return; this week we begin the steady climb toward summertime and you’ll see a few extra minutes of daylight after work each week from here until June. For many GMAT applicants, the darkest days of the year in December and early January match with the darkest days of their admissions journey, hustling to post a competitive GMAT while also scrambling on essays for Round 2. But this, too, shall pass.
If your New Year’s Resolution is to make 2015 the year that you ace the GMAT, you can take a lesson from this time of year. The darkest points always give way to enlightenment, and that secret will get you through some very difficult GMAT problems. There are two very common structures for challenging GMAT quant problems:
1) It looks easy, but the last step or two are tricky.
2) It looks impossible, but once you’ve found the right foothold it gets easy quickly.
This post is all about #2, those problems where it looks incredibly dark right up until that moment that you reach enlightenment. Veritas Prep’s own Jason Sun recounts the first quant question en route to his official 780 score: “I stared at a nasty sequence problem for probably 45 seconds with my jaw open thinking ‘there’s no way to solve this’. Then I remembered the strategy of starting with small numbers and finding a pattern, and 10 seconds later the answer was obvious.”
That’s common on the GMAT, and step one for you is to realize that problems are designed to look like that. When things look darkest, have faith that they’ll clear up. Here are a few ways that that occurs on the GMAT.
Calculations look awful, but work themselves out before you get to the answer.
Consider this problem:
If the product of the integers a, b, c, and d is 1,155 and if a > b > c > d > 1, then
what is the value of a – d?
(A) 2
(B) 8
(C) 10
(D) 11
(E) 14
Upon first glance, 1155 and four variables might look really messy. But take the first step – you know it’s divisible b y 11 and that you have to factor it. 1100 is 11*100 and 55 is 11*5, so you have 11*105. And 105 is much easier to divide out since it ends in a 5. That’s 21*5, which is 7*3*5. Once you’ve factored it down, it’s 11*7*5*3, which are all prime, so when 1 has to be less than any of these, that’s exactly a, b, c, and d. You need the biggest minus the smallest, and 11-3 is 8. What may have looked like a big, intimidating number was actually not so bad once you took the first step. It’s always darkest before the light goes on.
The problem is abstract, but comes into focus when you test small numbers.
What is the units digit of 2^40?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0
2^40 is an insanely large number. You’ll never be able to calculate it. But if you take the first few steps with small numbers, you’ll see a pattern:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
And since you only care about the units digits, you should see a pretty firm pattern emerging. 2, 4, 8, 6, 2, 4, 8, 6. If you repeat through this pattern, you’ll see that every 4th number is a 6, and since 2^40 will be the finish of the tenth run of that every-fourth-number cycle, the answer has to be 6. The GMAT loves to give you problems with big or abstract numbers that seem unfathomable, but if you test properties with small numbers you can often find a pattern or some other way to determine what you have.
It’s always the last place you look.
Another common theme is specific to geometry problems – the GMAT often constructs them so that a seemingly irrelevant piece of information (like the measure of a far, far away angle, or the area of a figure when you’re only solving for the length of one line) is crucial to the answer…it’s just that you don’t even consider filling in that piece of information that seems so far away from what you’re really trying to solve for. So FILL IN EVERYTHING! Even if it seems irrelevant, fill in every piece of information you can solve for and you’ll give yourself a better shot of finding that unlikely relationship that cracks the code.
You’re not supposed to be able to solve for it, but you can estimate or use answer choices.
Plenty of GMAT questions beg you to do some horrifying math, but if you look at the answer choices ahead of time you can see that they’re either spread incredibly far apart and ready to be estimated or they have easy-to-plug-in properties that allow you to just test them. It’s crucial to remember that the GMAT isn’t a test of pure math, but of problem solving using math. Heed this advice: if you think the calculations are too detailed to do in two minutes, you’re probably right. That’s when you should look to estimate or backsolve.
So if your GMAT study sessions are growing longer as the daylight does, keep this wisdom in mind. It always looks darkest before sunrise, and the same is true of many tough GMAT quant problems. As you struggle through practice problems, pay attention to all those times that the solution wasn’t nearly as bad as it seemed it would have to be upon first glance.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Serial and Sufficiency
Like most offices in the United States today, Veritas Prep’s headquarters had its fair share of water cooler and coffemaker discussions about yesterday’s final episode of the Serial podcast. Did Adnan do it? Did Jay set him up? Why does Don get a free pass based on a LensCrafters time-card punch? Does Best Buy have pay phones? The one answer we can give you is “we used MailChimp” so there’s that at least.
The other answer we can give you? Sarah Koenig would do pretty darned well on Data Sufficiency questions, where often it’s just as important to determine what you don’t know as it is to determine what you do. While the internet buzzed with theories certain that Adnan did it, that Jay did it, that a recently-released serial killer did it, Koenig was often ridiculed for being so noncommittal in her assessment of whether Adnan is guilty or not. But that’s an important mentality on Data Sufficiency questions, as one of the common ways that the GMAT will bait you is giving you information that seems overwhelmingly sufficient (The Nisha call! The phone was in Leakin Park!) but that leaves just enough doubt (Why did Jay’s story change so much?) that you can’t prove a definitive answer. And like the jury in the Serial case, we all have that tendency to jump to conclusions (“well if he didn’t kill her, who did?”) and filter out information that we don’t like (Christina Gutierrez’s performance…). This Serial-themed Data Sufficiency problem should exemplify (forgive the lack of subscript formatting, but a sequence problem in a Serial blog post seemed fitting):
The infinite (serial) sequence a1, a2, …, an, … is such that a1 = x, a2 = y, a3 = z,a4 = 3 and an = a(n-4) for n > 4. What is the sum of the first 98 terms of the sequence?
(1) x = 5
(2) y + z = 2
As people unpack the mystery in this problem, they start to see what’s going on. If an = a(n-4), then each term equals the term that came four prior. So the sequence really goes:
x, y, z, 3, x, y, z, 3, x, y, z, 3…
So although it looks like a pretty massive mystery, really you’re trying to figure out x, y, and z because 3 is just 3. And here’s a common way of thinking:
Statement 1 is not sufficient, but it gets you one of the terms. And Statement 2 is not sufficient but it gets you two more. So when you put them together, you know that the sum of one trip through the 4-term sequence is 5 + 2 + 3 = 10, so you should be able to extrapolate that to the whole thing, right? Just figure out how many trips through will get you to term 98 and you have it; like the Syed jury, you have the motive and the timeline and the cell phone records and Jay’s testimony, so the answer has to be C. Right?
But let’s interview Sarah Koenig here:
Sarah: The pieces all seem to fit but I’m just not so sure. Statement 2 looks really bad for him. If we can connect those dots for y and z, and we already have x, we should have all variables converted to numbers. Literally it all adds up. But I feel like I’m missing something. I can definitely get the sum of the first 4 terms and of the first 8 terms and of the first 12 terms; those are 10, and 20, and 30. But what about the number 98?
And that’s where Sarah Koenig’s trademark thoughtfulness-over-opinionatedry comes in. There is a giant hole in “Answer choice C’s case” against this problem. You can get the sequence in blocks of 4, but 98 is two past the last multiple of 4 (which is 96). The 97th term is easy: that’s x = 5. But the 98th term is tricky: it’s y, and we don’t know y unless we have z with it ( we just have the sum of the two). So we can’t solve for the 98th term. The answer has to be E – we just don’t know.
Now if you’ve heard yesterday’s episode, think about Dana’s “think of all the things that would have to have gone wrong, all the bad luck” rundown. “He lent his car and his phone to the guy who pointed the finger at him. That sucks for him. On the day that his girlfriend went missing. That’s awful luck…” And in real life she may be right – that’s a lot of probability to overcome. But on the GMAT they hand pick the questions. On this problem you can solve for the 97th term (up to 96 there are just blocks of 4 terms, and you know that each block sums to 10, and the 97th term is known as 5) or the 99th term (same thing, but add the sum of the 98th and 99th terms which you know is 2). But the GMAT hand-selected the tricky question just like Koenig hand-selected the Adnan Syed case for its mystery. GMAT Data Sufficiency questions are like Serial…it pays to be skeptical as you examine the evidence. It pays to think like Sarah Koenig. Unlike Jay, the statements will always be true and they’ll always be consistent, but like Serial in general you’ll sometimes find that you just don’t have enough information to definitively answer the question on everyone’s lips. So do your journalistic due diligence and look for alternative explanations (Don did it!). Next thing you know you’ll be “Stepping Out!!!” of the test center with a high GMAT score.
This message presented by Mail Chimp. Send us a high five.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# 4 Questions To Ask Yourself On Min/Max GMAT Problems
Min/Max problems can be among the most frustrating on the GMAT’s quantitative section. Why? Because they seldom involve an equation or definite value. They’re the ones that ask things like “did the fisherman who caught the third-most fish catch at least 12 fish?” or “what is the maximum number of fish that any one fisherman caught?”. And the reason the GMAT loves them? It’s precisely because they’re so much more strategic than they are “calculational.” They make you think, not just plug and chug.
However…
There are three knee-jerk questions that you should plug (if not chug) into your brain to ask yourself every time you see a Min/Max problem before you ask that fourth question “What’s my strategy?”:
1. Do the numbers have to be integers?
2. Is zero a possible value?
3. Are repeat numbers possible?
In the Veritas Prep Word Problems lesson we refer to these problems as “scenario-driven” Min/Max problems precisely because of the above questions. The scenario created by the problem drives the whole thing, related mainly to those three above questions. Consider these four prompts and ask yourself which ones can definitively be answered:
#1: “Four friends go fishing and catch a total of 10 fish. How many fish did the friend with the highest total catch?”
#2: “Four friends go fishing and catch a total of 10 fish. If no two friends caught the same number of fish, how many fish did the friend with the highest total catch?”
#3: “Four friends go fishing and catch a total of 10 fish. If each friend caught at least one fish but no two friends caught the same number, how many fish did the friend with the highest total catch?”
#4: “Four friends go fishing and catch a total of 10 pounds of fish. If each friend caught at least one fish but no two friends caught the same number, how many pounds of fish did the friend with the highest total catch?”
Hopefully you can see the progression as this set builds. In the first problem, there’s clearly no way to tell. Did one friend catch all ten? Did everyone catch at least two and two friends tied with 3? You just don’t know. But then it gets interesting, based on the questions you need to ask yourself on all of these.
With #2, two big restrictions are in play. Fish must be integers, so you’re only dealing with the 11 integers 0 through 10. And if no two friends caught the same number there’s a limited number of unique values that can add up to 10. But the catch on this one should be evident after you’ve read #3. Zero *IS* possible in this case, so while the totals could be 1, 2, 3, and 4 (guaranteeing the answer of 4), if the lowest person could have caught 0 (that’s where “min/max” comes in – to maximize the top value you want to minimize the other values) there’s also the possibility for 0, 1, 2, and 7. Because the zero possibility was still lurking out there, there’s not quite enough information to solve this one. And that’s why you always have to ask yourself “is 0 possible?”.
#3 should showcase that. If 0 is no longer a possibility *AND* the numbers have to be integers *AND* the numbers can’t repeat, then the only option is 1 (the new min value since 0 is gone), 2 (because you can’t match 1), 3, and 4. The highest total is 4.
And #4 shows why the seemingly-irrelevant backstory of “friends going fishing” is so important. Pounds of fish can be nonintegers, but fish themselves have to be integers. So even though this prompt looks very similar to #3, because we’re no longer limited to integers it’s very easy for the values to not repeat and still give wildly different max values (1, 2, 3, and 4 or 1.5, 2, 3, and 3.5 for example).
As you can see, the scenario really drives the answer, although the fourth question “What is my strategy?” will almost always require some real work. Let’s take a look at a couple questions from the Veritas Prep Question Bank to illustrate.
Question 1:
Four workers from an international charity were selling shirts at a local event yesterday. Did one of the workers sell at least three shirts yesterday at the event?
(1) Together they sold 8 shirts yesterday at the event.
(2) No two workers sold the same number of shirts.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked
(C) Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient
(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed
Before you begin strategizing, ask yourself the three major questions:
1) Do the values have to be integers? YES – that’s why the problem chose shirts.
2) Is zero possible? YES – it’s not prohibited, so that means you have to consider zero as a min value.
3) Can the numbers repeat? That’s why statement 2 is there. With the given information and with statement 1, numbers can repeat. That allows you to come up with the setup 2, 2, 2, and 2 for statement 1 (giving the answer “NO”) or 1, 2, 2, and 3 (giving the answer “YES” and proving this insufficient).
But when statement 2 says on its own that, NO, the numbers cannot repeat, that’s a much more impactful statement than most test-takers realize. Taking statement 2 alone, you have four integers that cannot repeat (and cannot be negative), so the smallest setup you can find is 0, 1, 2, and 3 – and with that someone definitely sold at least three shirts. Statement 2 is sufficient with really no calculations whatsoever, but with careful attention to the ever-important questions.
Question 2:
Last year, Company X paid out a total of \$1,050,000 in salaries to its 21 employees. If no employee earned a salary that is more than 20% greater than any other employee, what is the lowest possible salary that any one employee earned?
(A) \$40,000
(B) \$41,667
(C) \$42,000
(D) \$50,000
(E) \$60,000
Here ask yourself the same questions:
1) The numbers do not have to be integers.
2) Zero is theoretically possible (but probably constrained by the 20% difference restriction)
3) Numbers absolutely can repeat (which will be very important)
4) What’s your strategy? If you want the LOWEST possible single salary, then use your answer to #3 (they can repeat) and give the other 20 salaries the maximum. That way your calculation looks like:
x + 20(1.2x) = 1,050,000
Which breaks out to 25x = 1,050,000, and x = 42000. And notice how important the answer to #3 was – by knowing that numbers could repeat, you were able to quickly put together a smart strategy to minimize one single value.
The larger lesson is crucial here, though – these problems are often (but not always) fairly basic mathematically, but derive their difficulty from a situation that limits some options or allows for more than you’d think via integer restrictions, the possibility of zero, and the possibility of repeat values. Ask yourself these four questions, and your answer to the first three especially will maximize your efficiency on the strategic portion of the problem.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Today’s Date in Geometry History
Today is December 5, or in date form it’s 12/5. And if you hope to score 700+ on the GMAT, you should see those two numbers, 5 and 12, and immediately also think “13”!
Why?
There are certain combinations of numbers that just have to be top of mind when you take the GMAT. The quantitative section goes quickly for almost everyone, and so if you know the following combinations you can save extremely valuable time.
Based on Pythagorean Theorem, a^2 + b^2 = c^2, these four ratios come up frequently with right triangles:
a_______b________c
3_______4________5
5______12_______13
x_______x______x*(sqrt 2)___(in an isosceles right triangle)
x____x*(sqrt3)___2x________(in a 30-60-90 triangle)
These four ratios come up frequently when right triangles are present, so they’re about as high as you can get on the “should I memorize this?” scale. But just as important is using these ratios wisely and appropriately, so make sure that when you see the opportunity for them you keep in mind these two important considerations:
1) These “Pythagorean Triplets” are RATIOS, not just exact numbers.
So a 3-4-5 right triangle could also be a 6-8-10 or 15-20-25, and an isosceles right triangle could very well have dimensions a = 4(sqrt 2), b = 4(sqrt 2), and c = 8 (which would be one of the short sides 4(sqrt 2) multiplied by (sqrt 2) ). An average level question might pair 5 and 12 with you and reward you for quickly seeing 13, while a harder question could make the ratio 15, 36, 39 to reward you for seeing the ratio and not just the exact numbers you memorized.
Similarly, people often memorize the 45-45-90 and 30-60-90 triangles so specifically that the test can completely destroy them by making the “wrong” side carry the radical. If the short sides are 4 and 4, you’ll naturally see the hypotenuse as 4(sqrt 2). But if they were to ask you for the length of the hypotenuse and tell you that the area of the triangle is 4 (so 1/2 * a * b = 4, and with a equal to b you’d have 1/2 a^2 = 4, so a^2 = 8 and the short side then measures 2(sqrt 2)), it’s difficult for many to recognize that the hypotenuse could be an integer. So be careful and know that the above chart gives you *RATIOS* and not fixed numbers or fixed placements for the radical sign that denotes square root.
2) In order to apply these ratios, you MUST know which side is the hypotenuse.
In a classic GMAT trap, they could easily ask you:
What is the perimeter of triangle ABC?
(1) Side AB measures 5 meters.
(2) Side AC measures 12 meters.
And it’s common (in fact a similar problem shows that about 55% of people make this exact mistake) to think “oh well this is a 5-12-13, so both statements together prove that side BC is 13 and I can calculate that the perimeter is 30 meters.” But wait – 5 and 12 only lead to a third side of 13 when you know that 5 and 12 are the short sides. If you don’t know that, the triangle could fit the Pythagorean Theorem with 12 as the hypotenuse, meaning that you’re solving for side b:
5^2 + b^2 = 12^2, so 25 + b^2 = 144, and b then equals the square root of 119.
So while it’s critical that you memorize these four right triangle ratios, it’s just as important that you don’t fall so in love with them that you use them even when they don’t apply.
Important caveats aside, knowing these ratios is crucial for your ability to work quickly on the quant section. For example, a problem that says something like:
In triangle XYZ, side XY, which runs perpendicular to side YZ, measures 24 inches in length. If the longest side of the the triangle is 26 inches, what is the area, in square inches, of triangle XYZ?
(A) 100
(B) 120
(C) 140
(D) 150
(E) 165
Those employing Pythagorean Theorem are in for a fight, calculating a^2 + 24^2 = 26^2, then finding the length of a and calculating the area. But those who know the trusty 5-12-13 triplet can quickly see that if 24 = 12*2 and 26 = 13*2, then the other short side is 5*2 which is 10, and the area then is 1/2 * 10 * 24, which is 120. Knowing these ratios, this is a 30 second problem; without them it could be a slog of over 2 minutes, easily, with a higher degree of difficulty due to the extensive calculations. So on today of all days, Friday, the 5th day of the 12th month, keep that 13th in there as a lucky charm.
On the GMAT, these ratios will get you out of lots of trouble.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
# GMAT Tip of the Week: Sentence Correction in Real Life
Totes McGotes. FML. Sorry for partying. I know, right? Of the common phrases that have permeated pop culture and everyday conversation, easily one of the most common is, wait for it…
Wait for it.
And that one phrase can totes make your GMAT score supes high. Like, for real.
How?
Perhaps the best example comes from an all-staff email sent at Veritas Prep headquarters this week regarding the holiday vacation schedule. It began “With pumpkin spice season nearing its apex, it’s…” Seeing that introduction, multiple Veritas Prep staffers commented later that “it’s” after the comma made them nervous, as the possessive of “season” is its, not it’s (which grammatically means “it is”).
Now later in that sentence it became clear that the intention was “it is” (…”it’s time to start making holiday vacation plans.”), but the fact that so many Sentence Correction experts were on the edge of their seats just seeing that contraction “it’s” next to a possessive should demonstrate for you how to become great at Sentence Correction. To be efficient and effective with Sentence Correction, it’s helpful to anticipate what types of errors you might see, rather than simply sit back and wait for them to appear. Those who are most successful at Sentence Correction read sentences looking for signs of potential danger; they’re proactive as they search for likely Decision Points. For example, if you were to read the introduction:
Particularly for a leadership or management role, it is important that a candidate be both…
your senses should be heightened for parallel structure with “both X and Y,” number one, and secondly you should be acutely aware that the word “be” precedes the word “both,” so there is a very high likelihood that there will be an extraneous “be” after the word “and” to follow. In other words, when you see “both,” wait for it…where’s the “and,” and is the portion directly after it parallel to the first portion?
Correct:
(A) qualified to perform the duties of most subordinates and able to inspire subordinates to perform those duties at a higher level.
Incorrect:
(B) qualified to perform the duties of most subordinates and be able to inspire subordinates to perform those duties at a higher level.
While the grammar of this problem is crucial, true expertise comes from knowing where to focus your attention and expend your mental energy. Analyzing every word of every answer choice is exhausting, so the experts train themselves to see clues and “…wait for it” focusing back in on the parts of the sentence most highly correlated with errors. Clues can be:
Signals of parallel structure: both, either, neither, not only
Signals of verb tense: since, from, until
Signals of pronoun or subject/verb agreement: it, they, its, their
To train yourself to spot those clues that tell you to “wait for it…”, pay attention not only (wait for it…) to the grammatical reasons that an answer choice is right or wrong in your homework, but also (here it is…is it parallel?) to the signals outside the underline that required the application of that grammar. Sentence Correction is to an extent about “what do you know” but to really excel it also has to be about “what do you do” – the clues and signals that tell you what to look for and where to spend your time and energy.
Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
By Brian Galvin
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offers hundreds of practice questions and video explanations. Go there now.
# GRE Math
The GRE Quantitative Reasoning measure is NOT a test of advanced math. In fact, remembering all the geometry proofs, calculus, and trigonometry that you learned back in high school won’t help you at all on the GRE Math section. The difficulty of GRE Quantitative Reasoning comes from the need to logically reason your way through each problem. Once you figure out what the question is asking (often easier said than done), the math involved in solving the problem is actually fairly basic. Let’s take a look at how to study for GRE math and the tools you’ll need to do for an efficient GRE math review.
## GRE Math Review
Technically, GRE QR assesses your ability to deal with number properties and standard geometric figures. But what the GRE Quantitative Reasoning measure really tests is the logic you use to approach problem solving. Just as GRE Verbal tests your ability to analyze or “reason with” written English, GRE Math tests your ability to reason using numbers. The key to finding the answer to GRE Math problems lies in finding a way to unwrap the problem using logic.
And remember: The GRE doesn’t reward you for process. Unlike your math teacher who gave you partial credit for showing your work, the GRE only cares that you select the correct answer choice.
## Four GRE Quant Concept Categories
Most of GRE Quant appears in word problem format. The rest appears in purely mathematical form. In both cases, the mathematical concepts and abilities tested fall into four main categories:
• Arithmetic
• Algebra
• Geometry
• Data Analysis
For a full list of the topics that fall under these categories, be sure to check out the Introduction to the Quantitative Reasoning section article on the Educational Testing Service (ETS) website. Luckily, these concepts don’t get much harder than the Algebra II level math classes that most of us took in high school.
ETS also offers a Math Review PDF that promises to help you understand the concepts needed to solve problems as well as how to reason quantitatively. Both of these ETS links are great places to start your GRE math review online.
## GRE Quantitative Reasoning Question Types
There are four types of questions on the GRE Math test:
### 1) Quantitative Comparison Questions
You’ll be given two values: Column A and Column B. Your goal is to determine the relationship between the two.
Example:
Choose the correct statement.
When Car S covered a distance of D on a track, it covered 25% more distance than Car T had covered on the same track.
Column AColumn B
The distance covered by Car T0.80D
The quantity in Column A is greater
The quantity in Column B is greater
The two quantities are equal
The relationship cannot be determined from the information given
### 2) Multiple-Choice Questions—Select One Answer Choice
Example:
The price of a pair of sneakers was \$80 for the last six months of last year. On January first, the price increased 20%. After the price increase, an employee bought these sneakers with a 10% employee discount. What price did the employee pay?
\$70.40
\$82.00
\$83.33
\$86.40
\$88.00
### 3) Multiple-Choice Questions—Select One or More Answer Choices
Exactly how it sounds…multiple-choice on steroids.
Example:
Both P and Q are positive numbers, and S is a negative number. Which of the following fractions could be undefined?
P/(Q + S)
Q/(P + S)
S/(P + Q)
Q/(S − P)
S/(P − Q)
### 4) Numeric Entry Questions
Example:
If what is the value of n?
## GRE Math Practice
Want even more practice? For a high-quality GRE math review of all different question types, check out the following Clemmonsdogpark quant resources:
## GRE Math Prep Resources
Struggling with a particular question type? As you prepare to take GRE quant, here are some other GRE math prep resources you can use to keep your studying on the right track!
## Other Important Info
### GRE Assumptions
Math on the GRE follows the basic number conventions that you learned in high school:
• The positive direction of a number line is to the right and the negative direction is to the left.
• Distances are nonnegative.
• Prime numbers are greater than 1.
The same ETS article from earlier also states additional assumptions in its instructions. Hint: It’s best to memorize these instructions before taking the GRE exam so you don’t have to spend your time reading and processing basic instructions.
### Data Interpretation Sets
Some of the GRE Quant questions appear on their own, independent of other information. Others appear as a set of questions called a Data Interpretation set. Data Interpretation questions are all based on the same set of data presented in tables, graphs, or some other sort of display.
Here’s an example of how data interpretation can show up on GRE math.
### GRE Calculator
Let’s keep this short and sweet: Can you use a calculator on the GRE? Yes, but it looks something like this:
You also get scratch paper. 🙂
## How to Study for GRE Math
So you’ve bought a few of the major GRE prep books, and you’re ready to rip into the quantitative part. You’ll read through each book, page by page, and by the end, GRE math mastery will be yours. If only!
Studying for GRE quant is actually much more complicated than the above. Indeed many become quickly stymied by such an approach, feeling that after hundreds of pages and tens of hours they’ve learned very little, and asking themselves … “But, how can I ever learn GRE math?!”
To avoid such a thing befalling you, keep in mind the following important points on how to study for the GRE quantitative section.
But first, watch this video for the top 3 GRE math study tips:
### Using Formulas on GRE Math
How can formulas be bad, you may ask? Aren’t they the lifeblood of the GRE math? Actually, formulas are only so helpful. And they definitely aren’t the lifeblood of the quant section. That would be problem-solving skills.
Many students feel that all they have to do is use the formulas and they can solve a question. The reality is you must first decipher what the question is asking. Only at the very end, once you know how the different parts come together, can you “set up” the question.
All too often, many students let the formulas do the thinking. By that I mean they see a word problem, say a distance/rate question, and instead of deconstructing the problem, they instantly come up with d(distance) = r(rate) x t(time) and start plugging in parts of the question. In other words, they expect the question to fall neatly into the formula.
If you find yourself stuck in a problem with only a formula or two in hand, remember that the essence of problem solving is just that: solving the problem using logic, so you can use the formula when appropriate.
### Start Slowly With GRE Math—Then Build
Many students learn some basic concepts/formulae and feel that they have the hang of it. As soon as they are thrown into a random fray of questions, they become discombobulated, uncertain of exactly what problem type they are dealing with.
Basic problems, such as those you find in the Manhattan GRE math books, are an excellent way to begin studying. You get to build off the basic concepts in a chapter and solve problems of easy to medium difficulty. This phase, however, represents the “training wheels.”
Actually riding a bike, much like successfully answering a potpourri of questions, hinges on doing GRE math practice sessions that take you out of your comfort zone. In other words, you should try a few GRE math practice questions chosen at random in your GRE math review. Opening up the Official Guide to the GRE and doing the first math questions you see is a good start. Even if you haven’t seen the concept, just so you can get a feel for working through a question will limited information.
Oftentimes students balk at this advice, saying, “but I haven’t learned how to X, Y, or Z yet.” The reality is that students can actually solve many problems based on what they already know. However, because the GRE “cloaks” its questions, many familiar concepts are disguised in a welter of verbiage or other such obfuscation.
### Study All Concept Areas
Some students become obsessed with a certain question type, at the expense of ignoring equally important concepts. For instance, some students begin to focus only on algebra, forgetting geometry, rates, counting and many of the other important concepts.
This “tunnel vision” is dangerous; much as the “training wheels” phase lulls you into a false sense of complacency, only doing a certain problem type atrophies the part of your math brain responsible for being able to identify the type of question and the steps necessary to solve it.
### Focus on the Most Commonly Tested Areas
This is a subset of “tunnel vision.” Really speaking, it is a more acute case. To illustrate, some students will spend an inordinate amount of time learning permutations and combinations problems. The time they could have spent on more important areas, such as number properties and geometry, is squandered on a question type that, at most, shows up twice on the GRE. Make sure your GRE math review focuses on this!
### Quantitative Comparison
Answer: C. The two quantities are equal
Watch the video explanation here!
### Multiple Choice: One Answer Choice
Watch the video explanation here!
### Multiple Choice: Multiple Answer Choices
A. P/(Q + S)
B. Q/(P + S)
E. S/(P − Q)
Watch the video explanation here!
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# chemistry
You are a workstudy for the chemistry department. Your supervisor has just asked you to prepare 500mL of 3M HCl for tomorrow's undergraduate experiment. In the stockroom explorer, you will find a cabinet called "Stock Solutions". Open this cabinet to find a 2.0L bottle labeled "11.6M HCl". The concentration of the HCl is 11.6M. Please prepare a flask containing 500 ml of a 3 M (+/- 0.005M) solution and relabel it with its precise molarity. Note that you must use realistic transfer mode, a buret, and a volumetric flask for this problem.
As a reminder, to calculate the volume needed to make a solution of a given molarity, you may use the following formula:
C1V1 = C2V2
C[mol/l], v[mL]
C[mol/l] X v/1000 [L] = CV/1000 [mol]
11.6M HCI c2=3M V2= 500 mL
C1=11.6 M V1(mL)
the answer is Volume of HCl used 500ml
1. 👍 0
2. 👎 0
3. 👁 1,302
1. 129.31 ml
1. 👍 2
2. 👎 0
2. 22.5 ml?
1. 👍 0
2. 👎 0
3. 500mL of 3M HCl contains 1.5 moles of HCl
so 129.31 mL of the 11.6 M acid is needed
acid is added to water (NOT vice versa)
using a volumetric cylinder, add the necessary acid to 350 ml of water
... then add more water to make the 500 mL of solution required
1. 👍 1
2. 👎 0
4. 2.25
1. 👍 0
2. 👎 0
5. use the following formula:
C1V1 = C2V2
11.6M HCI
c2=3M
V2= 500 mL
v1=129.31
C1=11.6 M V1(mL)
1. 👍 0
2. 👎 0
6. the answer is 2.25 ml
1. 👍 0
2. 👎 0
7. what is the percentage error % error = (experimental concentration – 3.0 M) / 3.0 M * 100, in absolute value.
one of the people answered and said it is 129.31 ml is that the correct answer
1. 👍 0
2. 👎 0
## Similar Questions
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3. How does the study of science benefit society? A. Non scientists in society often use the same experiments conducted by scientists in the laboratory. B. The study of science helps slow down a fast paced society. C. The study of
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Problem Description: You are a workstudy for the chemistry department. Your supervisor has just asked you to prepare 500mL of 3M HCl for tomorrow's undergraduate experiment. In the stockroom explorer, you will find a cabinet
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describe how you would prepare 250.0 mL of a .500 M solution of NaCl by using apparatus found in a chemistry lab.
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Describe how you would prepare 250.0 mL of a 0.500 M solution of NaCl by using apparatus found in a chemistry lab.
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In this paragraph, any errors on capitalization, commas or semicolons? Also, should "they are all to report directly to me be part of the next sentence instead of the first sentence? Thanks again! Howard Clark will be the
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Please read to the bottom - This help was wonderful and absolutely correct, I'm asking a bit more regarding the same question, please read below. Imagine that you have been asked to prepare 500ml of a 0.25 mol L solution of Sodium
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http://www.gurufocus.com/term/Accts+Rec./ONNN/Accounts%2BReceivable/ON%2BSemiconductor%2BCorporation
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Switch to:
ON Semiconductor Corp (NAS:ONNN)
Accounts Receivable
\$418 Mil (As of Dec. 2014)
Accounts Receivable are created when a customer has received a product but has not yet paid for that product. ON Semiconductor Corp's accounts receivables for the quarter that ended in Dec. 2014 was \$418 Mil.
Accounts receivable can be measured by Days Sales Outstanding. ON Semiconductor Corp's Days Sales Outstanding for the quarter that ended in Dec. 2014 was 44.08.
In Ben Grahams calculation of liquidation value, accounts receivable are only considered to be worth 75% of book value. ON Semiconductor Corp's Liquidation Value for the quarter that ended in Dec. 2014 was \$-1,001 Mil.
Definition
Accounts Receivable is money owed to a business by customers and shown on its Balance Sheet as an asset.
Explanation
1. Accounts Receivable are created when a customer has received a product but has not yet paid for that product. Days sales outstanding measures of the average number of days that a company takes to collect revenue after a sale has been made. It is a financial ratio that illustrates how well a company's accounts receivables are being managed.
ON Semiconductor Corp's Days Sales Outstanding for the quarter that ended in Dec. 2014 is calculated as:
Days Sales Outstanding = Account Receivable / Revenue * Days in Period = 417.5 / 864.2 * 91 = 44.08
2. In Ben Grahams calculation of liquidation value, ON Semiconductor Corp's accounts receivable are only considered to be worth 75% of book value:
ON Semiconductor Corp's liquidation value for the quarter that ended in Dec. 2014 is calculated as:
Liquidation value = Cash and Cash Equivalents - Total Liabilities + (0.75 * Account Receivable) + (0.5 * Inventory) = 517.8 - 2196.5 + 0.75 * 417.5 + 0.5 * 729.9 = -1,001
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Be Aware
Net receivables tells us a great deal about the different competitors in the same industry. In competitive industries, some attempt to gain advantage by offering better credit terms, causing increase in sales and receivables.
If company consistently shows lower % Net receivables to gross sales than competitors, then it usually has some kind of competitive advantage which requires further digging.
Average Days Sales Outstanding is a good indicator for measuring a companys sales channel and customers. A company may book great revenue and earnings growth but never receive payment from their customers. This may force a write-off in the future and depress future earnings.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
ON Semiconductor Corp Annual Data
Dec05 Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Accts Rec. 160 178 175 199 261 295 457 358 383 418
ON Semiconductor Corp Quarterly Data
Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Accts Rec. 415 358 367 408 419 383 417 437 489 418
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# 無窮公理
## 形式陳述
${\displaystyle \exists \mathbf {N} :\varnothing \in \mathbf {N} \land (\forall x:x\in \mathbf {N} \implies x\cup \{x\}\in \mathbf {N} )}$
## 解釋
1 = 0 ∪{0} = {} ∪ {0}= {0}
2 = 1 ∪ {1} = {0} ∪ {1} = {0,1}
${\displaystyle \exists \mathbf {N} \forall n(n\in \mathbf {N} \iff ([\forall k\in n(\bot )\lor \exists k\in n(\forall j\in k(j\in n)\land \forall j\in n(j=k\lor j\in k))]\land }$
${\displaystyle \forall m\in n[\forall k\in m(\bot )\lor \exists k\in n(k\in m\land \forall j\in k(j\in m)\land \forall j\in m(j=k\lor j\in k))]))}$
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Questions tagged [entropy]
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# Index-Velocity Rating Development (Calibration) for H-ADCP Real-Time Discharge Monitoring in Open Channels
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2 Figure 2 After calibration, a ChannelMaster H-ADCP will turn into a flow meter that can directly output Q, V, H data in a text format through a serial port. This allows an H-ADCP being easily integrated with a telemetry system. The Index-Velocity method is especially useful at a site where a stage-discharge rating does not exist. An advantage of the Index-velocity method is that it can be used for a channel with its width much greater than the H-ADCP profiling range. This technical note describes the Index-velocity method and rating development or calibration of the H-ADCP built-in Index-velocity rating model. The Appendix describes the site selection considerations. 2.0 Index-Velocity Method and the H-ADCP Built-in Index-Velocity Rating Model The principle of the Index-velocity method is to establish a relation between the channel mean velocity and the Index-velocity. The Index-velocity defined here is a time and range averaged velocity measured by a ChannelMaster H-ADCP. The channel discharge Q is the product of the wetted cross-section area and the channel mean velocity: Q AV (1) where A = wetted area of the standard cross-section, and V = mean velocity of the channel. It should be pointed out that the standard cross-section used in the Index-velocity method does not need to be the same cross-section where an H-ADCP is mounted at. The wetted cross-section area depends on the cross-section geometry and water level. For a specific site, the wetted cross-section area is a function of water level only: A f (H) (2) 2
3 where H = water surface level (stage) above the local datum. Equation (2) is a stage-area rating that can be developed from the depth survey data. This rating is usually available at a hydrology station. The channel mean velocity V may be a function of the Index-velocity and the water level H: V f ( V, H) (3) I where V I = Index-velocity. The Index-velocity rating model that is built in the ChannelMaster H-ADCP firmware is: V b b b H ) (4) 1 ( 2 3 V I where b B1B bb2b b B3B bb4b are coefficients to be determined from field calibration tests. However, the use of the two parameter rating model, Eq. (4), beyond the parameter calibration range must be with caution. In addition, the model requires multiple regression analysis. In most cases, the channel mean velocity may be a function of the Index-velocity only. Thus, Eq. (4) is reduced to: V b1 b2 V I (5) In some cases, the coefficient b 1 is small and may be neglected. Thus, Eq. (5) is reduced to a simple coefficient rating: V b2 V I (6) The user needs to make a judgment for forcing b B1=0 (i.e., forcing the rating through zero) or not during a regression analysis. The key in using the Index-velocity method is to calibrate the rating model at a site because the model coefficients are site specific. The calibration, also called rating development, involves two steps. The first step is to conduct field calibration test. The test involves collecting H-ADCP velocity data from which the Index-velocity is obtained, and concurrently conducting discharge measurement from which the channel mean velocity is obtained. The second step is to create a relation between the channel mean velocity and the Index-velocity by regression analysis of the data obtained from the field calibration test. It should be note that other rating models such as power law and second-order polynomial may be used. However, at the present time, only the above rating models, Eqs. (4), (5), and (6), are built in the ChannelMaster H-ADCP firmware. Therefore, other rating models are not described here. 3.0 Field Calibration Test While an H-ADCP is sampling velocities, the calibration discharge measurements are conducted concurrently using a moving boat ADCP or a conventional current meter. It is 3
4 important to note that the calibration test must cover a flow or water level range from low to high to obtain a set of data for channel mean velocity, water level (stage), and Index-velocity. This is because, like a stage-discharge rating, an index-velocity rating will have uncertainties outside the range of the calibration measurements particularly if channel conditions (flow in the floodplain) change with stage/flow. Some sites may require a year to complete the calibration to cover the whole stage/flow range. 3.1 H-ADCP installation and configuration Referring to Figure 3, an H-ADCP may be mounted on a bank at an elevation of Z adcp above the local datum. A mounting structure must be designed and constructed based on site conditions. The H-ADCP must be mounted with its two horizontal beams in the horizontal plane. Tilt, either pitch or roll, must be zero or close to zero. The mount must be stable without changing over time. Z 0 H-ADCP Cell 1 Water Surface Cell j Y H Canal Bottom Canal Bank Acoustic Beams Canal Bank 0 Y H-ADCP X Mean Flow Direction Figure 3 A sketch of H-ADCP setup The H-ADCP s orientation should be perpendicular to the mean flow direction. That is, its instrument coordinate X should be parallel to the mean flow direction and Y is to the crosssection direction. The up-direction is defined as Z coordinate. Z = 0 is the local datum used or defined by the user or a local authority. For a channel with a flat bottom, the bottom may be set as Z = 0. 4
8 Note again that the wetted cross-section area A is taken at the so-called Standard Crosssection that may not be the same as the cross-section where the H-ADCP is mounted at. 4.5 Example of organized data Table 1 shows an example of the organized data from a field calibration test at the irrigation canal in California. A 600 khz ChannelMaster H-ADCP was used for Index-velocity measurements. A StreamPro ADCP was used for the calibration discharge measurements. Table 1 Example of organized StreamPro ADCP and ChannelMaster H-ADCP data from the field calibration test at the California irrigation canal StreamPro ADCP Measurement ChannelMaster H-ADCP Measurement Canal Wetted Transect Start Time Measured Discharge (Q measured ) [m 3 /s] Mean Velocity (V) [m/s] Sample Start Time Water Level (H) [m] Index- Velocity (V I ) [m/s] Cross- Section Area (A) [m 2 ] 12:44: :44: :49: :49: :57: :57: :01: :01: :11: :11: :14: :14: :21: :21: :24: :24: :36: :36: :40: :40: :53: :50: :58: :58: :21: :20: :24: :24: :43: :43: :47: :47: :51: :51: :54: :54: :57: :57: :00: :00: :09: :08: :15: :15: :19: :18: :23: :23: :27: :27: :31: :31: :36: :36: :42: :42: :06: :06: :11: :11: :15: :15:
9 Vm (m/s) Note that the calibration discharge measurement time, i.e., the transect start time, and Index velocity measurement time, i.e., the sample start time, must match as close as possible if the flow changes rapidly with time as it was at this site. The next step is to plot the data for channel mean velocity V vs. Index-velocity V I. Figure 5 shows the plot of the data shown in Table Vm = Vi Vi (m/s) Figure 5 Plot of data for V and V I shown in Table Regression analysis to determine rating model coefficients First, select a rating model based on the observation of the V vs. V I plot. Apparently, the plot shown in Figure 3 indicates a leaner relation. Second, conduct regression analysis for the mean velocity and index velocity using Excel or a rating creation software such as IVR-Creator. Below is the rating result obtained from the regression analysis for the data shown in Table 1: V (10) mean V I with a correlation coefficient of Note that in this case, a judgment of forcing b B1=0 was made during the regression analysis. Although the Excel spread sheet may be used for regression analysis to determine the rating model coefficients, its use requires knowledge of Excel and it is time consuming. IVR- Creator, the software that is specially designed for Index-velocity rating development is recommended. IVR-Creator does linear and nonlinear regression analysis using the leastsquare method. It accepts field data for channel cross-section geometry, discharge, water level (stage), and Index-velocity. Figure 6 shows a screen shot of the IVR-Creator software. IVR-Creator is very easy to use and saves a lot of time. It is a useful tool for Index-velocity rating development. 9
10 4.7 Rating quality evaluation Figure 6 Screenshot of the IVR-Creator software The quality of a calibrated rating model needs to be evaluated. Below are the parameters for rating quality evaluation. Correlation coefficient The correlation coefficient (r 2 ) indicates what proportion of variation in the dependent variable (V) can be explained by the independent variable (V I ). An r 2 of 0.97 indicates that 97% of the variation in V is explained by variation in V I. Lower values of r 2 could indicate that other parameters (variables) may be significant and missing in the rating. Residual plot A residual plot shows the departures of observed samples from the regression line (i.e., the calibrated rating model). The residuals should have a random pattern, be distributed equally above and below zero line, and have reasonable magnitudes. Attention should be given for the residual plots that reveal non-linearity and outliers. Figure 7 shows three example residual plots. 10
11 Residual (Vm - Vr) (m/s) Figure 7 Examples of residual plots It can be seen from Figure 7 that, Plot A looks good, Plot B has an outlier, and Plot C has a pattern or non-linearity. The outlier may be due to an error in the calibration discharge measurement. The pattern or non-linearity may be an indication that another independent variable (e.g., the water level) may be missing in the rating. Figure 8 shows the residual plot for the California irrigation canal rating example. The red lines are the 95% confidence intervals. Note that the residuals are random and relative uniformly distributed between the 95% confidence intervals Vm = Vi Vrated (Vr) (m/s) Figure 8 Residual plot for the California irrigation canal rating example Residual distribution plot The residuals from a good rating should have a nearly Gaussian distribution. Figure 9 shows the residual distribution plot for the California irrigation canal rating example. 11
12 Distribution (%) Vm = Vi^ Residual Figure 9 Residual distribution plot for the California irrigation canal rating example Standard error Standard error of estimate (S e ) is a measure of how nearly values of the dependant variable (V) predicted from the regression line agree with observed values. The smaller it is, the better the rating. The standard error is particular useful when comparing several rating models. Percent difference between the rated Q and the measured Q Regardless of the rating model used, you need to check the calibration measurement data points against your rating. That is, first calculate the rated channel mean velocity using your rating model. Secondly, calculate the rated Q = VA, where the wetted area A is calculated from the water level data and the cross-section geometry data. Then calculate the residual Q and the percent difference between the rated Q and the measured Q: Q residual Q Q (11) measured rated Percent Q difference measured Q Q measured rated 100 (%) (12) Table 2 shows the percent difference between the rated Q and the measured Q for the irrigation canal rating example. Note that some percent differences are greater than 14% during low flow conditions. But most of the differences are less than 5%. Table 2 Percent difference between the rated Q and the measured Q for the California irrigation canal rating example No. Measured Q (m 3 /s) Rated Q (m 3 /s) Residual Q (m 3 /s) Percent difference (%)
13 In theory, at least 30 data sets are required to obtain a statistically sound calibration. However, field calibration tests are time consuming and costly. Therefore, less data sets, say, 10 may be fine, particularly for initial calibration. The calibration may be refined when more calibration data are available. 4.8 Rating verification It is important to note that the created rating (i.e., the calibrated rating model) for a site may change over time in the following conditions: Cross-section change due to sediments deposition or bottom erosion H-ADCP sampling volume change due to mounting settlement, tilt, or high solids concentration Therefore, periodically filed discharge measurements may be required to verify the rating. Recalibration may be needed if the rating has a significant change. 13
15 Figure A-1 Wake flow generated by a pier. If an H-ADCP has to be mounted on a pier, avoid the H-ADCP sampling volume in the wake flow. Figure A-2 Bi-directional flow. It may be difficult or impossible to gage bi-directional flow with horizontal systems such as an H-ADCP 15
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# algebra
posted by .
completetly factors the followin expression 4a^2-12ab+9b^2
• algebra -
4a^2-12ab+9b^2 = (2a-3b)^2
## Similar Questions
1. ### Algebra
I'm having trouble working out these problems. Could someone please assist?
2. ### math, algebra,correction & help
directions are: write each fraction in simplest form: (-15x^3y^3)/(-20x^1y^4) My answeris: (3x^2)/(4y) THESE TWO NEXT ONES ARE THE ONES THAT I DON'T UNDERSTAND: pROBLEM #1 DIRECTIONS: Write each expression in simplest form. (4x-28)/(5x-35) …
3. ### math,algebra
Can someone explain this to me plese: to factor a number you.... and then can you explain to me To factor an algebraic expression into prime factors means..... thanks This is how I would complete these two sentences. To factor a number, …
4. ### algebra
a^3 - 8b^3/6b^2 - ba- a^2 / 3a^3b + 6a^2b^2 + 12ab^3/ a^2 - 9b^2 perform indicated operation and reduce to lowest terms
5. ### Algebra
Find the quotients and remainder when: 4a^2 + 9b^2 + 25c^2 - 12ab + 20ac - 30bc divided by 2a - 3b + 5c
4a^2+9b^2+25c^2-12ab+20ac-30bc divided by 2a-3b+5c.
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## Sunday, December 23, 2012
### Work Problem - Farming
Category: Algebra
"Published in Newark, California, USA"
Joaquin can plow his farm in 4 days. He and his eldest son can finish plowing the farm in 2 days. With his youngest son helping, the three can finish the work in 1 ½ days. Alone, how long it will take the youngest son to plow the farm?
Solution:
The word problem is about a work problem. If there are at least two or more people to work in a certain job, then they will finish their work in lesser time. If there are lesser people to work in a certain job, then they will finish their work in longer time. Because of these statements, the working equation for this type of problem is rational linear equation. Let's start to analyze the word problem as follows
Joaquin can finish his work alone = 4 days
Joaquin and his eldest son = 2 days
Joaquin, his eldest son, and youngest son = 1 ½ days
His eldest son can finish his work alone = x days
His youngest son can finish his work alone = y days
Next, write the working equation for Joaquin and his eldest son in order to get the number of days that his eldest son can plow the farm alone as follows
Multiply the both sides of the equation by their LCD, which is 4x as follows
His eldest son can plow the farm alone in 4 days also. Finally, write the working equation for Joaquin, his eldest son, and his youngest son in order to get the number of days that his youngest son can plow the farm alone as follows
Multiply the both sides of the equation by their LCD, which is 6y as follows
Therefore, his youngest son can plow the farm alone in 6 days
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COGS 501, Fall 2011 Schedule
[Note: some of these topics (and many of these links) will change, reflecting updates, improvements, and adaptation to the interests of this year's students.]
Week
Date
Description
Assignments
1. 9/12++ Overview & organization
Mathematical background
Basic: Eero Simoncelli, "Geometric Review of Linear Algebra" Michael Jordan, "An Introduction to Linear Algebra..." Intermediate: Connected Curriculum Project linear algebra materials Advanced: Desmond Fearnley-Sander "Hermann Grassman and the Creation of Linear Algebra" Stephen Gull et al. "Imaginary Numbers are not Real" (Note that these describe types of linear algebra significantly different from those that will be used in this course. They are offered as enrichment for any students to whom standard linear algebra is old news.)
Placement test, to be completed in class.
If you miss the first class, please fill this out and return to Laurel Sweeney at IRCS.
Instructions: there are 20 questions whose answers should be obvious if you know and remember the material, and impossible otherwise. Don't spend more than 10 minutes on it. If you don't know the answer to a question, leave it blank rather than guessing.
We don't expect you to know all this materials, or even a majority of it. The point is to give us a sense of who needs (or doesn't need) help with what.
2a Lab session Ed Neuman's Matlab tutorials #1, #2, #3
Problem set #1
Note on r2 -- how to calculate it and what it means
Hamming's Rule
2b Lecture: Linear algebra review
3a Lecture: Variance and covariance
Classifying multivariate data
3b Lab session Problem set #2
HW2 hints (.m file to start you off)
The same hints for Octave rather than Matlab
(explanation of the differences)
Solution (matlab version)
4a Lecture: Introduction to Subspace Methods
4b Lab Session Problem set #3
5a Lecture: Linear shift-invariant systems
Impulse Response and Convolution
(Deconvolving the hemodynamic response...)
5b Lab session Problem set #4
6a Lecture: The discrete fourier transform
6b Lab session Problem set #5
7a Lecture FIR & IIR Filters
Poles, zeros and the z-transform
Problem set #6
8 Reading James Hillenbrand, Laura A. Getty, Michael J. Clark, and Kimberlee Wheeler, "Acoustic Characteristics of American English Vowels", JASA 1995
Problem set #7
Original data from Hillenbrand et al. 1995
[Discussion and code]
Simple models of speech community dynamics
Hariharan Narayanan and Partha Niyogi, "Language Evolution, Coalescent Processes, and the Consensus Problem on a Social Network"
Adding phonetics to speech community dynamics
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# Derivation of Gravitational Potential
by AbsoluteZer0
Tags: derivation, gravitational, potential
P: 126 Hi, The derivation of the Gravitational Potential formula, as I understand, is: $W = Fd$ (1) $W = G \frac{M_1m_2}{r^2}d$ (2) Substituting the Gravitational Force formula $W = - \int_R^∞G \frac{M_1m_2}{r^2} \, dr$ (3) Integrating within the boundaries of the initial distance (R) and Infinity Which allows us to arrive at: $E_p = - \frac{GM_2m_1}{R}$ (4) However, what I don't understand is how we are able to proceed from step 3 to step 4. What method must be used in order to proceed as such? My proficiency with Calculus is still in the works. Thanks,
Mentor P: 11,231 Do you know how to do integrals like this one? $$\int_a^b {x^n dx}$$ If so, here's a hint: ##\frac{1}{r^2} = r^{-2}##. If no, then you'd best develop your calculus up to that point.
P: 542 $$\lim_{t \to \infty} \int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr$$ Do you know how to solve that?
P: 126
## Derivation of Gravitational Potential
Do you know how to do integrals like this one?
I can use integrals like these, to an extent.
Do you know how to solve that?
Unfortunately not.
P: 542 $$\lim_{t \to \infty} -\int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr$$ If you're taking an integral with respect to r and G, m1, and m2 are all constants, then what happens to the integral: $$\lim_{t \to \infty} -Gm_{1}m_{2}\int_{R}^{t} \frac{1}{r^{2}}dr$$ Now take the fact that 1/r^2 = r^-2 So the integral is then solvable: $$\lim_{t \to \infty} (-Gm_{1}m_{2} \frac{-1}{r})|_{t}^{R}$$ So then this becomes: $$\lim_{t \to \infty} (\frac{-Gm_{1}m_{2}}{R}-\frac{-Gm_{1}m_{2}}{t})$$ Finally take the limit, anything over infinity tends to 0. So you end up with: $$\frac{-Gm_{1}m_{2}}{R}$$ Pretty sure the math is correct, someone might be able to fix any physics errors I have.
P: 783 Also, note that potential at infinity is conventionally defined to be 0. The convention, being what it is, isn't derivable mathematically, so you need to use it as a given in solving the problem when evaluating potential at infinity in your integral. BiP
P: 542 Quick question (for my knowledge), why are the limits of intergration from R to infinity?
P: 783
Quote by iRaid Quick question (for my knowledge), why are the limits of intergration from R to infinity?
The gravitationalpotential is defined as the work done by gravity to bring an object from infinity to a distance R from an object along a straight line, hence the limits of integration. A theorem from vector calculus shows that it does not matter what path the object travels, so the definition can be adjusted to an "arbitrary path from infinity to a distance R" from an object. But that is an offshoot of vector calculus.
BiP
P: 126
I think I figured it out, please correct me if I'm wrong.
$W = Fd$
$W = G\frac{M_1m_2}{r^2}d$
$W =- \int_R^∞ G\frac{M_1m_2}{r^2}\,dr$
$W = - G M_1m_2 \int_R^∞ r^{-2}\,dr$ (Initially I was uncertain about pulling $GM_1m_2$ out)
$W = - G M_1m_2 [\frac{1}{r}]^R_∞$
anything over infinity tends to 0.
Am I right in assuming that this is the reason why $-G\frac{M_1m_2}{∞}$ produces zero?
$W = [-G \frac{M_1m_2}{r} - -G \frac{-GM_1m_2}{∞}]$
Which leads to
$E_p = -G\frac{M_1m_2}{r}$
Thanks for the help
Related Discussions Special & General Relativity 33 Special & General Relativity 3 General Physics 8 Introductory Physics Homework 7 Classical Physics 3
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### Find the final temperature of the system
Assignment Help Physics
##### Reference no: EM1312416
The specific heat of a certain 50 g metallic block is 900 J/kg K. If the block is at an initial temperature of 80 degrees Celsius, & dropped into a 50 g bucket of water at 1 degree Celsius, Find the final temperature of the system?
1. 30 degreees Kelvin
2. 39 degrees Celsius
3. 50 degrees
4. None of the above. (show your work)
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# How to Calculate Stack Back Curtains
#### Things You'll Need
• Metal tape measure
• Calculator
A longer rod for the stack back helps make the room look larger.
Image Credit: Jupiterimages/Photos.com/Getty Images
When you want your pair of curtains to open fully to display the window and the view, a simple measurement and formula will help you calculate the stack back in front of the wall on either side of the window. Although a longer rod will cost more and take up more wall space, positioning an extended rod above the window trim helps make the room look larger. This formula will be your approximate guide to help you achieve an unobstructed view through your windows.
## Step 1
Measure with a tape measure the width of the window opening. Record this number.
Video of the Day
## Step 2
Divide this number by 6 to calculate the length of the rod extensions or the stack back on either side of the window. Work with a calculator, if necessary. Round up to the nearest inch. For example, a 48-inch-wide window opening divided by 6 equals 8 inches for the rod extensions on each side of the window. If your curtain is a single panel that pulls back to one side or the other, divide the window width by 3 to calculate the stack back.
## Step 3
Add the window width plus the rod extensions for each side of the window in inches to calculate the rod length. In this example, 48 inches plus 8 inches for each side equals 64 inches. The face of the rod should measure approximately 64 inches to draw the pair of curtains off the window.
#### Tip
Factors such as the weight, lining and thickness of the fabric can affect how much of the curtain will actually stack in front of the wall. For example, a thinner sheer curtain will take up less space than a thicker damask curtain.
#### Warning
Avoid headers with tab tops or grommets, because these styles work best as stationary curtains.
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# How do you solve x/5=y+4 for y?
##### 1 Answer
May 12, 2016
color(blue)(y = (x - 20 ) / 5
#### Explanation:
$\frac{x}{5} = y + 4$
$\frac{x}{5} = \textcolor{b l u e}{y} + 4$
$\frac{x}{5} - 4 = \textcolor{b l u e}{y}$
$\textcolor{b l u e}{y} = \frac{x}{5} - 4$
$\textcolor{b l u e}{y} = \frac{x}{5} - \frac{20}{5}$
color(blue)(y = (x - 20 ) / 5
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Thread: easy way to check ordering of numbers?
1. easy way to check ordering of numbers?
I want to check the ordering of numbers in an if statement, for example, in pseudo-code:
Code:
```if a < b < c < ... < z
then do something magical
else quit```
Is there an easy way to do this? Obviously doing all the boolean comparisons is not possible. Maybe sorting the numbers and then doing the respective comparisons is the best way?
2. Yes, a sort() will accomplish a sort. But what do you mean by "order"? Like as in a polynomial?
3. Oh sorry, by 'order' I meant just numerical order. For example
Code:
`3 < 4 < 5 < 6`
should be true and
Code:
` 3 < 5 < 4 < 6`
should be false
4. I would imagine the easiest way is to place these numbers in a list or vector, and traverse it checking if prev < curr.
If you make it to the end of the list, the order is correct.
5. sorry I must have had a brain fart ...
Code:
`a < b && b < c && c < d`
etc...
my post probably wasn't clear, I was just looking for that
.... I should probably take a break
6. Originally Posted by Marksman
sorry I must have had a brain fart ...
Code:
`a < b && b < c && c < d`
etc...
Well sure that will work for just four numbers, but what if you have 50 numbers? Will you write out all those comparison tests?
7. Yeah that's true. In my actual program I only had 4 variables, so it was manageable.
Otherwise the vector/list approach would definitely be better.
8. How are the numbers stored? In a list/array/vector of some kind? You can potentially keep track if the values are ordered as they are entered and simply have a flag (initialized to true for the empty container). Whenever you add a new number, simply check if it is more than the previous value - if not then set the flag to false meaning the list is not ordered. When you do your if test, then just check this flag (an O(1) operation).
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# Simplify (5-2a)(7+3a)
(5-2a)(7+3a)
Expand (5-2a)(7+3a) using the FOIL Method.
Apply the distributive property.
5(7+3a)-2a(7+3a)
Apply the distributive property.
5⋅7+5(3a)-2a(7+3a)
Apply the distributive property.
5⋅7+5(3a)-2a⋅7-2a(3a)
5⋅7+5(3a)-2a⋅7-2a(3a)
Simplify and combine like terms.
Simplify each term.
Multiply 5 by 7.
35+5(3a)-2a⋅7-2a(3a)
Multiply 3 by 5.
35+15a-2a⋅7-2a(3a)
Multiply 7 by -2.
35+15a-14a-2a(3a)
Multiply a by a.
35+15a-14a-2⋅3a2
Multiply -2 by 3.
35+15a-14a-6a2
35+15a-14a-6a2
Subtract 14a from 15a.
35+a-6a2
35+a-6a2
Simplify (5-2a)(7+3a)
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# Mechanics 2 MEI June 06 stuck
Watch
Announcements
#1
Currently stuck on this last part of the question on June 2006 Mei M2 where the answer is apparently 15N whereas I keep getting 35N as I think that force at C is acting upwards and is acting anticlockwise and B and Q are both acting clockwise, can anyone tell me any rules I'm missing out here or mistakes I am blindly making?
0
#2
Image of question is here
https://onedrive.live.com/redir?resi...nt=photo%2cjpg
756
0
5 years ago
#3
Currently stuck on this last part of the question on June 2006 Mei M2 where the answer is apparently 15N whereas I keep getting 35N as I think that force at C is acting upwards and is acting anticlockwise and B and Q are both acting clockwise, can anyone tell me any rules I'm missing out here or mistakes I am blindly making?
0
#4
829
(Original post by lizard54142)
https://onedrive.live.com/redir?resi...nt=photo%2cjpg
0
5 years ago
#5
I agree that the reaction at B is 60N. You then need to take moments about another point, but this time in the rod BC. Note that the reaction at B is still acting in BC.
0
5 years ago
#6
Currently stuck on this last part of the question on June 2006 Mei M2 where the answer is apparently 15N whereas I keep getting 35N as I think that force at C is acting upwards and is acting anticlockwise and B and Q are both acting clockwise, can anyone tell me any rules I'm missing out here or mistakes I am blindly making?
Do moments about A to get the force at C, and do moments about C to get the force at A. Remember to include every force along the poles
0
5 years ago
#7
Have you got horizontal and vertical components of the contact force at A. did you take moments about A?
0
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889. Unlike a hurricane, which can be observed from within, a tornado is so small that such a study has not been practical.
(A) that such a study has not been practical
(B) that studying it that way has not been impractical
(C) for such studies as this to have been impractical
(D) as to not make such a study practical
(E) as to be impractical of study
between A and D which one is best. are there any rules for usage of "so that" and "so as to ".
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# A High frequency circuits
#### mertcan
Hi, initially I have seen that when circuits are exposed to high frequency kirchoff laws become invalid. According to my search I can not find any derived circuit equations for that case, instead computational electromagnetism method is suggested using maxwell. Is it the only way to make calculation while high frequencies exist? Or any other methods exist as we have in low frequencies ?
Related Classical Physics News on Phys.org
#### Cryo
Gold Member
There are plenty of laws, all derived from Maxwell's laws under certain approximations :-), but the simple fact is that when the size of your device/apparatus/probe is comparable to wavelength, the physics becomes more complex (and more interesting)
What is the frequency range? What is application?
Have a look at theory of guided waves, transmission line theory, waveguides etc. IMHO. There are a lot of approximations and useful techniques developed for this field.
#### mertcan
There are plenty of laws, all derived from Maxwell's laws under certain approximations :-), but the simple fact is that when the size of your device/apparatus/probe is comparable to wavelength, the physics becomes more complex (and more interesting)
What is the frequency range? What is application?
Have a look at theory of guided waves, transmission line theory, waveguides etc. IMHO. There are a lot of approximations and useful techniques developed for this field.
Thanks for return, when I endeavor to dig information out of Internet I can not find valuable things could you share me nice sources book files or videos?
#### mertcan
There are plenty of laws, all derived from Maxwell's laws under certain approximations :-), but the simple fact is that when the size of your device/apparatus/probe is comparable to wavelength, the physics becomes more complex (and more interesting)
What is the frequency range? What is application?
Have a look at theory of guided waves, transmission line theory, waveguides etc. IMHO. There are a lot of approximations and useful techniques developed for this field.
Besides, I have seen that transmission line theory applies kirchoff laws altough at high frequencies kirchoff laws can not be applied. Could you explain this contradiction??
#### Cryo
Gold Member
Besides, I have seen that transmission line theory applies kirchoff laws altough at high frequencies kirchoff laws can not be applied. Could you explain this contradiction??
Not quite sure what you mean here. Do you mean that the common way to derive the telegraf equation is to introduce distributed inductance/capacitatne/resistance and then talk about a small section of the transmission line as if we were dealing with low-frequency circuits? Well this bit makes perfect sense. They start with a transmission line which is small in one direction (thickness), but large in the other (length-wise). The size here is relative to wavelength of the radiation. Then they isolate a small section of the transmission line, which is now small in both directions, so the low-freqency Kirchhoff's laws apply. Finally they link these small bits together and take a continuum limit. If you want to discuss this point (telegraf equation) you need to be more specific, i.e. give the derivation and ask about specific parts you don't like.
#### Cryo
Gold Member
Thanks for return, when I endeavor to dig information out of Internet I can not find valuable things could you share me nice sources book files or videos?
I don't work with guided waves myself. Most of my work is with propagating free-space waves, but I can suggest Pozar's "Microwave Engineering". Maybe experts around here will give better suggestions.
#### mertcan
@Cryo hi again I just would like to express I can not establish any connection between size of device being comparable to wavelength and kirchoff law conservation. Could you help me understand better?
#### tech99
Gold Member
@Cryo hi again I just would like to express I can not establish any connection between size of device being comparable to wavelength and kirchoff law conservation. Could you help me understand better?
If you imagine a network consisting of just one resistor at the end of long wires, at high frequencies the distributed L and C along the wires causes the wires to act as transmission lines. The resistance is then transformed to a totally different impedance at the opposite end of the wires. For this reason, complex networks which are physically large are a problem at high frequencies. I have not had to tackle a network of this sort before, but I think an RF bridge would be an example where Kirchoff might be necessary. For the most part, networks are ladder or tee types, where Kirchoff is not necessary.
#### vanhees71
Gold Member
The intuitive answer is that Kirchhoff's Laws of AC circuit theory are derived using the quasi-static approximations (approximations, because two different ones are involved), i.e., in conductors (Ohmic resistances and coils) the "displacement current" is neglected (magneto-quasi-static approximation), approximating the Ampere-Maxwell Law of the full theory by the magnetostatic Ampere Law. In the space between capacitor plates the displacement current cannot be neglected but $\dot{\vec{B}}$ in Faraday's Law of induction (electro-quasi-static approximation). The upshot is that the quasi-static approximations used are good as long as the spatial extension of the circuit is small compared with the wavelength of the electromagnetic waves with the frequency of the AC, i.e., if $\lambda \ll L$, where $$\lambda=2 \pi/k=2 \pi c/\omega=c/f.$$
#### mertcan
The intuitive answer is that Kirchhoff's Laws of AC circuit theory are derived using the quasi-static approximations (approximations, because two different ones are involved), i.e., in conductors (Ohmic resistances and coils) the "displacement current" is neglected (magneto-quasi-static approximation), approximating the Ampere-Maxwell Law of the full theory by the magnetostatic Ampere Law. In the space between capacitor plates the displacement current cannot be neglected but $\dot{\vec{B}}$ in Faraday's Law of induction (electro-quasi-static approximation). The upshot is that the quasi-static approximations used are good as long as the spatial extension of the circuit is small compared with the wavelength of the electromagnetic waves with the frequency of the AC, i.e., if $\lambda \ll L$, where $$\lambda=2 \pi/k=2 \pi c/\omega=c/f.$$
If you imagine a network consisting of just one resistor at the end of long wires, at high frequencies the distributed L and C along the wires causes the wires to act as transmission lines. The resistance is then transformed to a totally different impedance at the opposite end of the wires. For this reason, complex networks which are physically large are a problem at high frequencies. I have not had to tackle a network of this sort before, but I think an RF bridge would be an example where Kirchoff might be necessary. For the most part, networks are ladder or tee types, where Kirchoff is not necessary.
Thanks for your your return, My first question is : is there a proof when size of device on circuit gets smaller than wavelength, then kirchoff law is much more true??
also I would like to say that I can not find nice examples and solution pertain to high frequency circuits on internet. Lots of stupid videos exist on youtube . Would you mind suggesting a book or pdf files or nice lecture notes or videos related to that topic in order to learn more effectively?
#### vanhees71
Gold Member
The standard textbook is of course Jackson, Classical electrodynamics.
#### mertcan
The standard textbook is of course Jackson, Classical electrodynamics.
Thanks @vanhees71 by the way is there mathematical proof when size of device on circuit gets smaller than wavelength, then kirchoff law is much more true? ?
#### vanhees71
Gold Member
Sure, you can estimate the error made when doing the approximations leading from the full Maxwell equations to AC circuit theory. It's nicely discussed in Jackson, if I remember right.
#### tech99
Gold Member
Sure, you can estimate the error made when doing the approximations leading from the full Maxwell equations to AC circuit theory. It's nicely discussed in Jackson, if I remember right.
Does it mean that I am slightly in error if I calculate the input impedance of a line, which is in free space, using ordinary transmission line theory? For example, using Zo and Zload?
#### mertcan
Sure, you can estimate the error made when doing the approximations leading from the full Maxwell equations to AC circuit theory. It's nicely discussed in Jackson, if I remember right.
@vanhees71 I have already the book but I can not find the proof that when size of device on circuit gets smaller than wavelength, then kirchoff law is much more true??
#### Cryo
Gold Member
Which Kirchoff law?
The current law simply states that current is incompressible (what goes into the node, must come out). In electrodynamics you would instead say that the integral of the current density over any closed surface must be zero (in simple terms current density*area = current )
Then, using Gauss' law on current density $\boldsymbol{J}$:
$\oint d^2r \boldsymbol{J}.\boldsymbol{\hat{n}}=\int d^3 r \boldsymbol{\nabla.J}$
So the Kirchhoff's Current Law requires the divergence of current density to vanish. Take the divergence of the fourth Maxwell's law:
$\boldsymbol{\nabla}.\boldsymbol{\nabla}\times\boldsymbol{B}=0=\mu_0\boldsymbol{\nabla}.\boldsymbol{J}+\frac{n^2}{c^2}\boldsymbol{\nabla}.\boldsymbol{\dot{E}}$
So to get the Kirchhoff' Current Law you need $\boldsymbol{\nabla}.\boldsymbol{\dot{E}} \to 0$
The Kirchoff's voltage law states, that voltages around a closed loop add up to zero. Voltage between points $a$ and $b$ is given by: $V=\int^{b}_{a} \boldsymbol{E}.d\boldsymbol{r}$. So KVL is basically, $\oint \vec{E}.d\boldsymbol{r}=0$. From Maxwell's third law $\boldsymbol{\nabla}\times\boldsymbol{E}=-\boldsymbol{\dot{B}}$, so to get KVL you need
$\boldsymbol{\dot{B}}\to 0$.
At this point you have to start hand-waveing. For example, you could say that both conditions are satisfied if the time-scale of your oscillations is long (time-derivatives vanish). But what does long mean? Wave-equations couple space and time, so long time-scales are equivalent to short distances. The conversion factor is the speed of light. Have a look at microwave engineering books for more details. Did you check Pozar?
Homework Helper
Gold Member
2018 Award
Dr. Walter Lewin makes an addition to the Kirchhoff Voltage Law (KVL)=see e.g. https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/ so that the requirement that $\frac{dB}{dt}=0$ is not necessary. $\\$ However, the assumption in KVL that the current is the same everywhere in the circuit requires that wavelengths involved are large compared to the size of the circuit. Kirchhoff's voltage law does not allow for a current of the form $i(x,t)=i_o \cos(kx-\omega t)$ that is different for different locations $x$ in the circuit. With Kirchhoff, we can have $i=i_o \cos(\omega t)$ , but that current is the same everywhere in the circuit, i.e. we can have a branch, etc, but what goes into the junction comes out of the junction, etc.$\\$ See also post 52 of https://www.physicsforums.com/threads/why-does-a-voltmeter-measure-a-voltage-across-inductor.880100/page-3 for Dr. Walter Lewin's video. IMO, Dr. Walter Lewin makes an important addition to the KVL theory with the paradox that he presents for the case where $\frac{dB}{dt} \neq 0$. (It would be helpful to watch this video before reading the more in-depth discussion of it in the "link" of the first paragraph). $\\$ Edit: Additional item: With KVL, there is no provision for a circuit to be able to radiate power. When power is radiated, KVL will not compute the power correctly. If there is significant power radiated, the results that KVL gets must thereby be an approximation. @vanhees71 Please confirm these last statements, but I believe they are correct.
Last edited:
#### mertcan
Which Kirchoff law?
The current law simply states that current is incompressible (what goes into the node, must come out). In electrodynamics you would instead say that the integral of the current density over any closed surface must be zero (in simple terms current density*area = current )
Then, using Gauss' law on current density $\boldsymbol{J}$:
$\oint d^2r \boldsymbol{J}.\boldsymbol{\hat{n}}=\int d^3 r \boldsymbol{\nabla.J}$
So the Kirchhoff's Current Law requires the divergence of current density to vanish. Take the divergence of the fourth Maxwell's law:
$\boldsymbol{\nabla}.\boldsymbol{\nabla}\times\boldsymbol{B}=0=\mu_0\boldsymbol{\nabla}.\boldsymbol{J}+\frac{n^2}{c^2}\boldsymbol{\nabla}.\boldsymbol{\dot{E}}$
So to get the Kirchhoff' Current Law you need $\boldsymbol{\nabla}.\boldsymbol{\dot{E}} \to 0$
The Kirchoff's voltage law states, that voltages around a closed loop add up to zero. Voltage between points $a$ and $b$ is given by: $V=\int^{b}_{a} \boldsymbol{E}.d\boldsymbol{r}$. So KVL is basically, $\oint \vec{E}.d\boldsymbol{r}=0$. From Maxwell's third law $\boldsymbol{\nabla}\times\boldsymbol{E}=-\boldsymbol{\dot{B}}$, so to get KVL you need
$\boldsymbol{\dot{B}}\to 0$.
At this point you have to start hand-waveing. For example, you could say that both conditions are satisfied if the time-scale of your oscillations is long (time-derivatives vanish). But what does long mean? Wave-equations couple space and time, so long time-scales are equivalent to short distances. The conversion factor is the speed of light. Have a look at microwave engineering books for more details. Did you check Pozar?
Yes I have checked Pozar but could not see any clue or derivation of the relation between wavelength size and kirchoff laws? Could you help me a little bit more to understand it better?
Last edited:
#### mertcan
Dr. Walter Lewin makes an addition to the Kirchhoff Voltage Law (KVL)=see e.g. https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/ so that the requirement that $\frac{dB}{dt}=0$ is not necessary. $\\$ However, the assumption in KVL that the current is the same everywhere in the circuit requires that wavelengths involved are large compared to the size of the circuit. Kirchhoff's voltage law does not allow for a current of the form $i(x,t)=i_o \cos(kx-\omega t)$ that is different for different locations $x$ in the circuit. With Kirchhoff, we can have $i=i_o \cos(\omega t)$ , but that current is the same everywhere in the circuit, i.e. we can have a branch, etc, but what goes into the junction comes out of the junction, etc.$\\$ See also post 52 of https://www.physicsforums.com/threads/why-does-a-voltmeter-measure-a-voltage-across-inductor.880100/page-3 for Dr. Walter Lewin's video. IMO, Dr. Walter Lewin makes an important addition to the KVL theory with the paradox that he presents for the case where $\frac{dB}{dt} \neq 0$. (It would be helpful to watch this video before reading the more in-depth discussion of it in the "link" of the first paragraph). $\\$ Edit: Additional item: With KVL, there is no provision for a circuit to be able to radiate power. When power is radiated, KVL will not compute the power correctly. If there is significant power radiated, the results that KVL gets must thereby be an approximation. @vanhees71 Please confirm these last statements, but I believe they are correct.
As far as I see Walter Lewin does not link wavelength up with kirchoff conservation laws. It is always written that wavelength size affects kirchoff laws conservation. When it comes to find some mathematical proof if it, I can not find valuable things. I do not want to memorize those things just verbally, also I would like to examine the mathematical derivation of the relation between wavelength size and kirchoff conservation laws. Could you help me in that regard?
Gold Member
#### mertcan
Unfortunately, I have written this up only in German in my lecture notes about E&M. Maybe you can nevertheless follow it since the equation density is quite high ;-)):
https://th.physik.uni-frankfurt.de/~hees/publ/theo2-l3.pdf
Thanks for sharing actually I have a little bit german but English would be so nice so could you tell me on which page I should focus for my question?
#### jasonRF
Gold Member
My first question is : is there a proof when size of device on circuit gets smaller than wavelength, then kirchoff law is much more true??
First, i hope you have looked at transmission line theory enough so that you have seen the equation after the title "input impedance of lossless transmission line" at
https://en.m.wikipedia.org/wiki/Transmission_line
This of course aasumes $\exp(j \omega t )$ time dependence.
Consider the simple case where you have a transmission line of length $\ell$ and characteristic impedance $Z_0$ that is shorted on one end. Of course the low frequency approximation (Kirchoff law) tells us that the impedance looking into the other end will be zero. However, transmission line theory tells us the impedance looking into the other end will be
$$Z_{in}=j Z_0 \tan(2 \pi \ell / \lambda)$$
where $\lambda$ is the wavelength of a signal on the line. $Z_{in}$ is clearly not zero. However, when $2 \pi \ell << \lambda$ then we have
$$Z_{in}\approx j Z_0 2 \pi \ell / \lambda$$
This clearly approaches the Kirchoff law approximation of zero as $\ell/\lambda \rightarrow 0$.
While this is not a completely general proof it shows you the basic idea. Hope that helps.
Jason
#### vanhees71
Gold Member
Thanks for sharing actually I have a little bit german but English would be so nice so could you tell me on which page I should focus for my question?
It's the beginning of Chpt. 4 (p. 103).
"High frequency circuits"
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# help exponents
0
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How many different numbers between 1/1000 and 1000 can be written either as a power of 2 or as a power of \(5\), where the exponent is an integer?
May 6, 2022
### 1+0 Answers
#1
0
1 - Powers of 2: 2^-9, 2^-8, 2^-7, 2^-6, 2^-5, 2^-4, 2^-3, 2^-2, 2^-1, 2^0, 2^1, 2^2, 2^3, 2^4, 2^5 2^6, 2^7, 2^8 and 2^9 ==Total 19 powers of 2.
2 - Powers of 5: 5^-4, 5^-3, 5^-2, 5^-1, 5^0, 5^1, 5^2, 5^3, and 5^4==Total 9 powers of 5
3 - Total powers of 2 + powers of 5 ==19 + 9==28
May 6, 2022
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# In 2007, the Seattle Mariners signed their all-star outfielder Ichiro Suzuki to a five-year contract extension that will pay him \$90 million over its...
I do not understand the questions and how to solve it. explanation need :(
In 2007, the Seattle Mariners signed their all-star outfielder Ichiro Suzuki to a five-year contract extension that will pay him \$90 million over its lifespan. Several news outlets reported this as a “five year contract worth \$90 million”. But being an astute finance student familiar with the concept of present value, you know that the actual worth of the contact is quite a bit less. The contract is structured with a \$5M signing bonus, paid immediately; and \$17M in annual salary from 2008-2012. However, each year \$5M of that \$17M is deferred and will grow at 5.5% interest for as long as it is deferred. Furthermore, the deferred money does not have to be paid in full until 2032. For all questions below, assume that the signing bonus is paid immediately and the annual salaries are paid once yearly, starting exactly one year from the signing date. a) First assume that all of the deferred money and accumulated interest is paid out along with the final salary payment at the end of 2012. Use 5.5% as the discount rate. What is the present value (as of the signing date) of Ichiro’s contract under these circumstances? (Don’t forget the singing bonus.) b) How much (measured in present value) in total have the Mariners saved by deferring the \$5M every year instead of paying the \$17M in full every year? c) Now assume a 15% discount rate, but keep the interest rate for deferred money at 5.5%. How much (measured in present value) in total have the Mariners saved by deferring the payments?
d) Finally suppose that instead of paying out a lump sum of deferred money and accumulated interest at the end of 2012, the Mariners decide to pay Ichiro a fixed payment every year from 2013 until 2032. Still using a 15% discount rate and 5.5% interest rate on the deferred payments, calculate how much in sum the Mariners are saving over a contract that would pay Ichiro a straight \$17M yearly payment.
In 2007, the Seattle Mariners signed their all-star outfielder Ichiro Suzuki to a five-year contract extension that will pay him \$90 million over its lifespan. Several news outlets reported this as...
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shunfurh 于 2017.08.27 16:58 提问
Master Spark
Master Spark is a powerful bomb that can deal the highest damage with good stability. Kirisame Marisa (霧雨魔理沙) uses her Hakkero (八卦炉) to power the Master Spark. The outline of Master Spark is a parabola with the vertex at (0, 0), where Marisa is located. The second-order coefficient of the parabola is a specific real number a.
Knowing that there are n fairies in the 2D-plane, Marisa wants to damage as many of them as possible by powering a single Master Spark. She can choose any direction to power as she likes.
Input
There are multiple cases. Each case is made up of three lines, the first line contains an integer n (0 ≤ n ≤ 30000) -- the number of fairies in the plane and an real number a (0.1 ≤ a ≤ 10) -- the second-order coefficient of the Master Spark. Both the next two lines contains n real numbers xi and yi -- the coordinates of the i-th fairy.
Output
For each test case, output the the maximum number of fairies Marisa can damage, in format "%d daze\n".
Sample Input
10 0.5
-10 -6 20 1 0 -8 5 -5 30 40
-10 0 0 -5 20 33 13 13 40 30
Sample Output
4 daze
1个回答
caozhy 2017.09.10 23:53
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## Square root of decimal number :[Procedure ,examples and practice problem}
In this post, we’re going to learn to “find the square root of decimal numbers using the long division method”.So, till now we have learned to find square root using factorization,one tens method, and long division method. If you didn’t read these stuff, check now Today we’ll learn the complete concept of finding the square … Read more
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Algebra and Trigonometry 10th Edition
This conic is an ellipse. Notice from the graph that the center of the ellipse is shifted 2 units right and 1 unit down. This can be verified from the equation: $\frac{(x-2)^2}{9}-\frac{(y+1)^2}{4}=1$ The $(x-2)$ means that the center is shifted two units right (across the x-axis). The $(y+1)$ means that the center is shifted one unit down (the y-axis).
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# solve.QP in R doesn't give a solution satisfying the constraint
I've dealing with a quadratic programming problem as following, and my objective is to get a 49*1 vector.
$$\widehat{\boldsymbol{w}}^{0}(\tau):=\underset{\boldsymbol{w}(\tau) \in \mathbb{W}}{\arg \min }\left(\hat{\boldsymbol{\psi}}^{*}(\tau)-\hat{\boldsymbol{\Psi}}^{*}(\tau) \boldsymbol{w}(\tau)\right)^{\top}\left(\hat{\boldsymbol{\psi}}^{*}(\tau)-\hat{\Psi}^{*}(\tau) \boldsymbol{w}(\tau)\right)$$
where $$\mathbb{W}:=\left\{\boldsymbol{w} \mid \boldsymbol{w} \in[0,1]^{N-1} \text { and } \boldsymbol{\iota}_{N-1}^{\top} \boldsymbol{w}=1\right\}$$
Thus when I use solve.QP function in R, I define $$psi1 = \hat{\boldsymbol{\psi}}^{*}(\tau)$$ $$psi2 = \hat{\boldsymbol{\Psi}}^{*}(\tau)$$ I suppose my constraint $$A^{\top}\boldsymbol{w}\geq b_{0}$$ are $$A=\begin{bmatrix} 1&1&1&\ldots&1&1\\ 1&0&0&\ldots&0&0\\ 0&1&0&\ldots&0&0\\ 0&0&1&\ldots&0&0\\ \vdots & \vdots & \vdots& \ddots & \vdots &\vdots \\ 0&0&0&\ldots&1&0\\ 0&0&0&\ldots&0&1\\ -1&0&0&\ldots&0&0\\ 0&-1&0&\ldots&0&0\\ 0&0&-1&\ldots&0&0\\ \vdots & \vdots & \vdots& \ddots & \vdots &\vdots \\ 0&0&0&\ldots&-1&0\\ 0&0&0&\ldots&0&-1\\ \end{bmatrix}$$ $$b_{0}=(1,0,0,0,\cdots,0,0,-1,-1,-1,\cdots,-1,-1)^{\top}$$
Dmat = (t(psi2)%*%psi2)/2
dvec = t(2*(t(psi1)%*%psi2))
Amat = t(rbind(rep(1,(unit-1)),diag(1,49,49),diag(-1,49,49))
bvec = c(1,rep(0,(unit-1)),rep(-1,(unit-1)))
w_hat=solve.QP(Dmat,dvec,Amat,bvec,meq=1)\$solution
However, I get a solution like this:
Obviously there are some negative numbers in this solution and it doesn't satisfy my constraint, and I'm wondering why this happens. Thanks a lot.
• Please show all your code. – Bob Jansen Apr 3 at 16:49
• I have edited my question and updated my code, thanks a lot. – Toxique Apr 3 at 17:21
• Can you elaborate further on what quadratic programming problem you're trying to solve (maybe formulate it mathematically)? – Pleb Apr 3 at 18:35
• I have edited my question with all codes and the objective function, thanks a lot. – Toxique Apr 4 at 1:30
• Please stop significantly changing the question. My answer was valid until the problem turned out to be something else entirely. – Bob Jansen Apr 4 at 5:55
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# 1.8: Activity 1G - Measuring Angles
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
## You will need the following equipment for these lab activities:
1. Using your protractor, measure the angle of the arrow below.
2. Using your protractor, measure the angle of the arrow below.
3. Using your protractor, measure the angle of the arrow below.
4. Using your protractor, measure the angle of the arrow below.
5. Below is a compass rose, which indicates the cardinal directions. North (N) is always represented as 0°/360°. On the compass below, label:
a. The remaining cardinal directions (S, W, E) in black.
b. The intermediate points (NE, SE, NW, SW) in brown.
c. The intermediate points of the intermediate points (NNE, ENE, ESE, SSE, SSW, WSW, NNW, WNW) in blue.
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GMATCLUB TEST Help M01
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Author Message
Intern
Affiliations: Omicron Delta Kappa
Joined: 16 May 2011
Posts: 8
Kudos [?]: [0], given: 0
Schools: Union Graduate College, Howard, Quinnipiac University
GMATCLUB TEST Help M01 [#permalink]
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23 Jun 2011, 11:27
00:00
Difficulty:
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Question Stats:
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Today John is 3 times older than Mark. If in 13 years, John will be one year younger than twice the age of Mark on that day, how old is John today?
A. 13
B. 33
C. 36
D. 38
E. 39
I just cant understand how this was solved.
[Reveal] Spoiler: OA
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Current Student
Joined: 26 May 2005
Posts: 554
Kudos [?]: 244 [0], given: 13
Re: GMATCLUB TEST Help M01 [#permalink]
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23 Jun 2011, 11:35
adelabu44 wrote:
Today John is 3 times older than Mark. If in 13 years, John will be one year younger than twice the age of Mark on that day, how old is John today?
A. 13
B. 33
C. 36
D. 38
E. 39
I just cant understand how this was solved.
Its C
today john is 3 times mark , this means J = 3M...........................1
in 13 years
j = J + 13
M = M+13
John will be one year younger than twice the age of Mark on that day
J+13= 2(M+13)-1
j+13= 2m+25
from 1;
3m+13= 2m+25
m = 12
therefore j = 36
Kudos [?]: 244 [0], given: 13
Intern
Affiliations: Omicron Delta Kappa
Joined: 16 May 2011
Posts: 8
Kudos [?]: [0], given: 0
Schools: Union Graduate College, Howard, Quinnipiac University
Re: GMATCLUB TEST Help M01 [#permalink]
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23 Jun 2011, 11:39
! I see my mistake. I grouped (2M+13)-1 instead of 2(M+13)-1
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Director
Joined: 01 Feb 2011
Posts: 725
Kudos [?]: 147 [0], given: 42
Re: GMATCLUB TEST Help M01 [#permalink]
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23 Jun 2011, 20:32
today +13yrs
J =3M 3M+13
M M+13
Given after 13yrs john will be 1 year younger than twice mark's age
i.e 3M+13 = 2(M+13)-1
M=12
=>John's current age = 3M = 36
Answer is C.
Kudos [?]: 147 [0], given: 42
Re: GMATCLUB TEST Help M01 [#permalink] 23 Jun 2011, 20:32
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Left Videos
# Logical Reasoning
## Questions and answers on Alphabet Reasoning
### Alphabet reasoning
1. Which letter will be 4th to the left of 6th to the right of 13th letter from the left end of the alphabet?
A. M B. N C. O D. P
2. Which letter will be 5th to the right of 7th to the left of 15th letter from the right end of the alphabet?
A. K B. L C. I D. J
3. Which letter will be 6th to the right of 16th letter from the right end of the alphabet?
A. U B. V C. I D. J
4. How many pairs of letters are their in the word MASTER in the english alphabet?
A. 3 B. 2 C. 1 D. 4
5. NOSTRIL in the word how many pairs in english alphabet?
A. 3 B. 5 C. 1 D. 4
6. The word MOTHERLND in this word how many pairs in english alphabet?
A. 1 B. 3 C. 2 D. 4
7. How many english meaningful words can be made with the word IDET using each letter only once in each word?
A. 4 B. 3 C. 1 D. 2
8.
A) ROBBER B) RANDOM C) RESTRICT D) RESTAURANT E) ROCKET
Find the middle word after arranging the given words in alphabetical order?
A. A B. C C. D D. E
9.
A) PARASITE B) PASTE C) PRAWNS D) PETAL E)PARTY
Arrange the given words in alphabetical order and find the middle word?
A. B B. A C. E D. C
10.
A) BRISH B) BICYCLE C) BISHOP D) BITTER E) BIFOCAL
Which is the middle word after arranging the given words in alphabetical order?
A. D B. C C. B D. A
11.
A) NOZZLE B) NAUSEA C) NOMENCLATURE D) NOSTRIL E) NORMAL
What is the middle word after arranging the given words in alphabetical order?
A. B B. D C. C D. E
12.
A) WRONG B) ZYGOTE C) WROUGHT D) YIELD E)YEARN
Arrange the given words in alphabetical order and find the center word?
A. C B. E C. B D. D
13.
1) Late 2) Load 3) Letter 4) Liver 5) Long
Which is the correct alphabetical sequence?
A. 1, 3, 4, 2, 5 B. 1, 2, 4, 3, 5 C. 1, 3, 5, 2, 4 D. 1, 3, 4, 5, 2
14.
1) Pest 2) Pestle 3) Pessimist 4) Pestilence 5) Pester
What is the correct alphabetical sequence?
A. 1, 3, 5, 4, 2 B. 3, 1, 5, 4, 2 C. 3, 1, 2, 4, 5 D. 3, 5, 2, 4, 1
15.
1) Writhe 2) Wrinkle 3) Wretch 4) Wrath 5) Wriggle
Choose the correct alphabetical sequence?
A. 4, 2, 5, 3, 1 B. 5, 3, 1, 2, 4 C. 4, 3, 5, 2, 1 D. 4, 3, 2, 5, 1
16.
In the word ' ACCURATE ' find the number of pairs of letters which consists as many letters between them in the word like in the alphabet ?
A. Two B. Three C. One D. Six
17.
How many pairs of letters are there in the word 'TIGER' which have as many letters between them in the word as in the alphabet?
A. Three B. Two C. four D. None
18.
How many independent words can 'WILLIAMSON' be divided into without changing the order of the letters and using each letter only once?
A. 4 B. 3 C. 1 D. None
19.
Which letter is exactly in middle between H and T in the english alphabets?
A. N B. M C. P D. S
20.
In the english alphabet, find the letter which is 4th to the right of D?
A. E B. G C. H D. None
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HOW TO CALCULATE VOLUME OF CEMENT BAG
HOW TO CALCULATE VOLUME OF CEMENT BAG
In order to calculate the volume of any cement bag,
We need to know the density of cement bag.
The unit weight of the bag is 50 kgs
Density of cement in loose condition is 1440 kgs/m3
Density = mass/volume (ρ=m/V)
So, V=m/ρ
50 kg ÷ 1440 kg/m3 = 0.0347 m3 = 1.23 CFT
A step by step explanation in this video
Continue Reading about the "HOW TO CALCULATE VOLUME OF CEMENT BAG" on the next page below
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You are Here: Home >< Physics
# Rounding - which way?
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1. Hi all,
I'm doing OCR AS Physics.
If I do a calculation and get a decimal answer of 14.5485, I'm a bit confused how to round this to, say, 3 SF.
My maths teacher says I should round it down to 14.5, but my Physics teacher rounds from the right, which would make it 14.6 (because the 8 takes the 4 up to 5, which rounds up to 14.6)
Any ideas? Which one's right?
Thanks!
Ryan
2. I've always been told 14.5 and it's the only one that makes sense.
3. (Original post by mf2004)
I've always been told 14.5 and it's the only one that makes sense.
This.
Basic rule of mathematics.
0.0485 < 0.0500
Must be over 5 to round up, less than 5 you do not.
4. that's what I thought.
Great, thanks.
Ryan
5. (Original post by dazed&confused)
that's what I thought.
Great, thanks.
Ryan
You're welcome Good luck in your exams!
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http://www.lmfdb.org/Character/Dirichlet/6025/339
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# Properties
Label 6025.339 Modulus $6025$ Conductor $6025$ Order $10$ Real no Primitive yes Minimal yes Parity even
# Related objects
Show commands for: Pari/GP / SageMath
sage: from sage.modular.dirichlet import DirichletCharacter
sage: H = DirichletGroup(6025)
sage: M = H._module
sage: chi = DirichletCharacter(H, M([3,8]))
pari: [g,chi] = znchar(Mod(339,6025))
## Basic properties
Modulus: $$6025$$ Conductor: $$6025$$ sage: chi.conductor() pari: znconreyconductor(g,chi) Order: $$10$$ sage: chi.multiplicative_order() pari: charorder(g,chi) Real: no Primitive: yes sage: chi.is_primitive() pari: #znconreyconductor(g,chi)==1 Minimal: yes Parity: even sage: chi.is_odd() pari: zncharisodd(g,chi)
## Galois orbit 6025.bo
sage: chi.galois_orbit()
pari: order = charorder(g,chi)
pari: [ charpow(g,chi, k % order) | k <-[1..order-1], gcd(k,order)==1 ]
## Values on generators
$$(2652,2176)$$ → $$(e\left(\frac{3}{10}\right),e\left(\frac{4}{5}\right))$$
## Values
$$-1$$ $$1$$ $$2$$ $$3$$ $$4$$ $$6$$ $$7$$ $$8$$ $$9$$ $$11$$ $$12$$ $$13$$ $$1$$ $$1$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$1$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{3}{10}\right)$$
value at e.g. 2
## Related number fields
Field of values: $$\Q(\zeta_{5})$$ Fixed field: Number field defined by a degree 10 polynomial
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https://www.coursehero.com/file/6147295/vj1334-NASAccedil%C5%A1eacutepound%C5%B8ccedilcopyegraveregiexclaring%CB%86/
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vj1334 NASA的食物计划
# Vj1334 NASA的食物计划
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Unformatted text preview: const maxn=50; maxv=400; var n,m,vinteger; a,b,carray[0..maxn] of integer; farray[0..maxv,0..maxv] of longint; function max(x,ylongint)longint; begin if x=y then exit(x) else exit(y); end; procedure init; var iinteger; begin readln(v,m); readln(n); for i=1 to n do readln(a[i],b[i],c[i]); end; procedure main; var i,j,kinteger; begin fillchar(f,sizeof(f),0); for i=1 to n do for j=v downto a[i] do for k=m downto b[i] do f[j,k]=max(f[j,k],f[j-a[i],k-b[i]]+c[i]); end; procedure print; begin writeln(f[v,m]); end; begin init; main; print; end. ...
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## This note was uploaded on 02/21/2011 for the course CS 001 taught by Professor 001 during the Spring '11 term at Shandong University.
Ask a homework question - tutors are online
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# Darcy number
In fluid dynamics through porous media, the Darcy number (Da) represents the relative effect of the permeability of the medium versus its cross-sectional area—commonly the diameter squared. The number is named after Henry Darcy and is found from nondimensionalizing the differential form of Darcy's Law. This number should not be confused with the Darcy friction factor which applies to pressure drop in a pipe. It is defined as
$\mathrm{Da} = \frac{K}{d^2}$
where
• K is the permeability of the medium;
• d is the diameter of the particle.[1]
Alternate forms of this number do exist depending on the approach that Darcy's Law is made dimensionless and the geometry of the system.[2] The Darcy number is commonly used in heat transfer through porous media.[3]
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https://www.teacherspayteachers.com/Product/Common-Core-Word-Problem-of-the-Day-1345327
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# Common Core Word Problem of the Day
Subject
Resource Type
Common Core Standards
Product Rating
File Type
PDF (Acrobat) Document File
232 KB|16 pages
Share
Product Description
This printer friendly black and white packet includes a Common Core math word problem of the day for 30 weeks (120 total word problems). These word problems can be used for many activities but are designed as a Problem of the Day (POTD).
How does it work?
Students will cut out the word problems on Monday and glue them on a page in a math journal. Each day, they will have to solve one problem and write a paragraph explaining what strategies they used and the steps they took to reach a solution.
How is this beneficial?
Research shows that students often do not make sense of word problems. Instead, to come up with an answer, they just apply the most recent algorithm taught, chose operations based on the types of numbers involved, or apply other approaches that do not require sense making. Having students analyze the problems and explain their thinking requires them to think more carefully about what the problem is asking and how they can come up with a solution.
Requiring students to write during math is a win-win for both teacher and student. Although it may be difficult to introduce this practice, it is well worth the effort. I have noticed tremendous growth in my students’ problem-solving and writing skills through this simple POTD (Problem of the Day) practice.
What types of problems are included?
Place Value (4 weeks, 16 word problems)
Example: Nicole has a big bag of buttons. When she counts, she has 4 stacks of ten buttons, 3 stacks of one hundred buttons, and 2 single buttons left over. How many buttons does she have?
Addition/Subtraction without Regrouping (10 weeks, 40 problems)
Example: Abril counted 194 ants on the ant hill. Some of the ants left to get crumbs and now there are 121 ants on the ant hill. How many ants left to get crumbs?
Addition/Subtraction with Regrouping (8 weeks, 32 problems)
Example: Demiya saw 50 butterflies at the park. Then 16 of them flew away. A few minutes later 8 came back. How many butterflies are in the park now?
Counting Money (6 weeks, 24 problems)
Example: Esther has 56 cents in her piggy bank. What are 4 different combinations of coins she could have in her piggy bank?
Measurement (2 weeks, 8 problems)
Example: Kimberly’s dog is 30 centimeters long. Jessica’s dog is 12 inches long. Whose dog is longer?
Total Pages
16 pages
N/A
Teaching Duration
1 Year
Report this Resource
\$6.00
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http://gitweb.dragonflybsd.org/~tuxillo/dragonfly.git/blob_plain/4ae2eec871242422d6d867846150bfe859dc937a:/secure/lib/libcrypto/man/BIO_f_null.3
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# 3 Digit Math Coloring Worksheet For 2nd Grade
In Free Printable Worksheets183 views
4.25 / 5 ( 138votes )
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## 3 Digit Math Worksheets Printable Worksheets
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Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
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Keeping Score
# Knicks Not a Pretty Picture When Painted by Numbers
An old Johnny Mercer song said you should “accentuate the positive” and “eliminate the negative.” For the Knicks, at the season’s midpoint, that is proving to be a difficult task.
Determining what the Knicks are bad at is fairly easy. They are bad on the road (8-12); they are bad at home (7-14, the second-fewest home victories in the N.B.A.). They are bad against teams with winning records (7-17); they are bad against teams with losing records (8-9). They are bad in games decided by 3 points or less (1-5); they are bad in games decided by more than 10 points (6-13).
They are especially bad while wearing orange (0-6 before abandoning the alternate uniforms) or when playing matinees at Madison Square Garden (0-5, including four blowouts). Over all, they are 15-26, the opposite of where they were at this point last season.
Then again, it’s not completely negative. While the Knicks have a losing record against five of the N.B.A.’s six divisions, they have a winning record against the Southeast, including an upset victory over the Miami Heat earlier this month. Going 6-5 against a single division may not seem like much — in fact, it’s pretty trivial — but other than a 2-1 record in overtime games, it is the only way to sort the standings and have the Knicks come out with a positive result.
Along the road to losing in so many different ways, the Knicks produced a player who was so marginal that he didn’t leave a fingerprint on the two games in which he made cameo appearances before being released.
That player, Chris Smith, J. R.’s younger brother, saw less than two minutes of action over all, hardly the time needed to put up some numbers. But Smith, seemingly on the Knicks’ roster as a favor to his brother, had zeros across the board, failing to register a single statistic that appears in a box score. When a player does that, it is known as a trillion because of all of the zeros that follow minutes played. In addition, Smith registered a 0.0 in player-efficiency rating, an advanced statistic that measures a player’s overall effectiveness and uses 15 as the predetermined league average.
Perhaps Smith, 26, will get another shot at the N.B.A. some day and get rid of some of those zeros.
Beyond Smith it does not get that much better. Just four Knicks have reached the league average in player-efficiency, and one of those players, Jeremy Tyler, has played all of 28 minutes.
Meanwhile, lost in the losing, the bad dramatics (J. R. Smith easily warrants his own article in that regard) and the ongoing speculation (Coach Mike Woodson may not have much time left to turn things around) is a nice season by Carmelo Anthony.
Accused of being a shot-happy ball-hog in the past, Anthony has sacrificed some of his own offense this season while trying to do more to help the team win. His 9.1 rebounds a game are a career high, and he has increased his per-game averages in assists, steals and blocks while reducing his turnovers and personal fouls. He is playing more minutes than ever and shooting a career-high 39.5 percent from 3-point range.
Despite Anthony’s positive work, the indelible image of him this season will be of him shaking his head following a J. R. Smith blunder, as has been captured in a variety of shots.
Currently on a four-game losing streak, the Knicks have, amusingly, not lost much ground in the playoff race. A .366 winning percentage puts the Knicks on pace for 30 wins this season, yet leaves them just two games behind Charlotte for the last playoff spot in the Eastern Conference.
If the positive is being accentuated, Knicks fans can at least console themselves with the knowledge that, in terms of postseason contention, their team picked a fantastic year to play poorly.
It stands to reason, however, that the last few spots in the noncompetitive East will go to the teams that come into their own in the second half. There is little reason to believe the Knicks have that in them, since every time they appear to improve, they immediately short circuit. But stranger things have happened.
And if the negative is being eliminated, the good news is that there are plenty of games remaining against the Southeast, including three against the Heat. Miami may be the two-time defending N.B.A. champions, but this season they are 0-1 against the Knicks.
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You are Here: Home >< Physics
# how exactly does changing magnetic flux result in current induced watch
1. yeah I get the whole explanation of moving a magnet in a coil but then most sources just say "current is then induced" how? or am I just supposed to get it that way?
2. (Original post by sarah99630)
yeah I get the whole explanation of moving a magnet in a coil but then most sources just say "current is then induced" how? or am I just supposed to get it that way?
As a magnet is moving toward a coil, there is a change in the magnetic flux linkage with the coil. The change in magnetic flux linkage would “give rise” to an induced e.m.f. according to Faraday’s law. This induced e.m.f. will result in having an induced current in the coil.
Recall that a battery would provide the emf in a circuit to cause the current to flow in the circuit.
3. (Original post by Eimmanuel)
As a magnet is moving toward a coil, there is a change in the magnetic flux linkage with the coil. The change in magnetic flux linkage would “give rise” to an induced e.m.f. according to Faraday’s law. This induced e.m.f. will result in having an induced current in the coil.
Recall that a battery would provide the emf in a circuit to cause the current to flow in the circuit.
ahaa i get it thanks!! so for an emf to be induced it does not need a complete circuit, but for a current there has to be a battery right?
4. (Original post by sarah99630)
ahaa i get it thanks!! so for an emf to be induced it does not need a complete circuit
Correct. For an open circuit, the changing flux forces electrons in the conductor to bunch up at one end of the conductor. There will be a momentary flow of current as electrons still move within the limited length of the conductor, but they cannot move further than the end. So at one end there will be a deficit of electrons (net +ve charge build up) and at the other end there will be a surplus of electrons (net -ve charge build up). The difference of charge measured across the ends of the conductor is the e.m.f.
That e.m.f build up, is the pressure that provides the potential energy (hence voltage potential) to push electrons around a closed circuit.
,but for a current there has to be a battery right?
Noooooo. A battery is one way of providing an emf.
However, as I explained above, it's not the only way. If the conductor experiencing a changing magnetic flux is part of a closed loop, then current will flow around the loop.
Perform a thought experiment:
Think of a bicycle where the chain is a conductor full of electrons. If the cycle is secured to a post, then trying to pedal away will not result in motion, the pedal force (changing magnetic flux) simply applies pressure (emf) to the chain but it cannot move.
Only when the cycle is not secured, will the chain move freely in a loop and work can be performed.
5. (Original post by sarah99630)
ahaa i get it thanks!! so for an emf to be induced it does not need a complete circuit, but for a current there has to be a battery right?
We don't need to have a complete circuit to have induced emf but complete circuit is required for induced current like closed loop coil.
For example, when you drop a magnet through a coil of wire, there will be induced emf and induced current in the coil.
Watch the following video to see that the LED light up for a short while when the magnet falls through the coil.
But if the magnet is dropped through a loop that has an opening (as shown in the picture below on the right), there is will be induced emf but no induced current because of the open circuit.
We don’t need a battery source to have induced current.
You can think of the flow of “concepts” as follow:
Changing magnetic flux linkage --> Induced emf --> Induced current provided it is a closed circuit.
6. (Original post by uberteknik)
Noooooo. A battery is one way of providing an emf.
However, as I explained above, it's not the only way. If the conductor experiencing a changing magnetic flux is part of a closed loop, then current will flow around the loop.
Perform a thought experiment:
Think of a bicycle where the chain is a conductor full of electrons. If the cycle is secured to a post, then trying to pedal away will not result in motion, the pedal force (changing magnetic flux) simply applies pressure (emf) to the chain but it cannot move.
Only when the cycle is not secured, will the chain move freely in a loop and work can be performed.
wow! Great explanation. thanks a LOT. you've helped me understand it. God bless you.
7. (Original post by Eimmanuel)
We don't need to have a complete circuit to have induced emf but complete circuit is required for induced current like closed loop coil.
For example, when you drop a magnet through a coil of wire, there will be induced emf and induced current in the coil.
Watch the following video to see that the LED light up for a short while when the magnet falls through the coil.
But if the magnet is dropped through a loop that has an opening (as shown in the picture below on the right), there is will be induced emf but no induced current because of the open circuit.
We don’t need a battery source to have induced current.
You can think of the flow of “concepts” as follow:
Changing magnetic flux linkage --> Induced emf --> Induced current provided it is a closed circuit.
God, I thought ill never understand this. Cant thank you enough!! great explanation. God bless you.
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You are Here: Home >< Maths
# inflections Watch
1. can someone tell me how to know if there is going to be an inflection on a curve just by looking at an equation. there was this question that asked me to sketch the curve, y=x^3+3 and i knew that it would be like wavy like --> /\/ but i didnt know that there was going to be an inflection.
the whole idea of sketching curves scares me because substituting values is very time consuming in exams.
2. In this case, it's just a vertical translation of the graph . Another way of telling is that when you differentiate you're left with a square, which means the only stationary point is at , and differentiating again tells you that at this stationary point, . This doesn't necessarily mean it's a point of inflection in general, but in the case of a cubic with just one stationary point, it does.
For cubic graphs, there's no real way of noticing straight away whether it's going to have a point of inflection or not, unless you're good at differentiating and factorising quadratics in your head.
3. (Original post by nuodai)
In this case, it's just a vertical translation of the graph . Another way of telling is that when you differentiate you're left with a square, which means the only stationary point is at , and differentiating again tells you that at this stationary point, . This doesn't necessarily mean it's a point of inflection in general, but in the case of a cubic with just one stationary point, it does.
For cubic graphs, there's no real way of noticing straight away whether it's going to have a point of inflection or not, unless you're good at differentiating and factorising quadratics in your head.
so basically you're saying that if you differentiate it and you are left with, (something)^2, it means that its an inflection for cubic graphs?
btw i wont need to know anything too advanced as i am only doing C1
4. (Original post by cooldudeman)
so basically you're saying that if you differentiate it and you are left with, (something)^2, it means that its an inflection for cubic graphs?
btw i wont need to know anything too advanced as i am only doing C1
Yes, or indeed (something)×(something else)^2 + (something else again)
Essentially, it has a point of inflection if it's a stretch/translation of the graph . From your graph transformations stuff you should know that these are things like and and so on; combining all the results together, the general rule is that anything of the form has a point of inflection. The only downside is that to be able to see whether not it has a point of inflection you have to check if it can be put in this form, which is a lot harder than completing the square!
5. (Original post by nuodai)
Yes, or indeed (something)×(something else)^2 + (something else again)
Essentially, it has a point of inflection if it's a stretch/translation of the graph . From your graph transformations stuff you should know that these are things like and and so on; combining all the results together, the general rule is that anything of the form has a point of inflection. The only downside is that to be able to see whether not it has a point of inflection you have to check if it can be put in this form, which is a lot harder than completing the square!
oh okok thanks for your help!
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0
# 1 degree Fahrenheit equal what degree Celsius?
Wiki User
2010-02-28 04:28:49
One degree Fahrenheit is equal to -17.22222222222222...repeating degrees Celsius.
Wiki User
2010-02-02 00:39:59
Study guides
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## The ideal gas law measures pressure in
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Wiki User
2010-02-28 04:28:49
1 degree Fahrenheit = -17.2 degrees Celsius.
Wiki User
2009-10-13 00:14:41
0, celsius = 32, fahrenheit
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# What is friction Class 11 and its types?
Friction is the force that opposes the motion of a solid object over another. There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction.
## In which chapter is friction in Class 11?
NCERT Solutions for Class 11 Science Physics Chapter 6 – Friction.
## What is friction with Example Class 11?
It is exerted by one surface to another when two surfaces rub against each other. The magnitude of the force will depend on the coefficient of kinetic friction between the types of materials of the matter. For Example: When you rub your hands a kind of heat produces because of kinetic friction.
## What is a friction in physics?
friction, force that resists the sliding or rolling of one solid object over another. Frictional forces, such as the traction needed to walk without slipping, may be beneficial, but they also present a great measure of opposition to motion.
## What are laws of friction?
Solution : Following are the laws of friction:
i) Friction always opposes the motion.
ii) Friction acts tangentially along the surface of contact.
iii) Force of friction is independent of area of contact of two bodies.
iv) Force of friction depends on the nature of the surfaces in contact.
## What is the SI unit for force?
The SI unit of force is the newton, symbol N. The base units relevant to force are: The metre, unit of length — symbol m. The kilogram, unit of mass — symbol kg. The second, unit of time — symbol s.
## What is friction Class 11 and formula?
Friction is the hindering force that comes into play when two items are interacting with each other. Friction formula is typically given as. Ff = μFn. The numeric’s are, the magnitude of friction is Ff , the coefficient of friction is μ and the magnitude of the normal force is Fn.
## What are uses of friction?
Friction is a resistance force that slows down or prevents motion, it is necessary for many applications where you might want to hold items or do things and prevent slipping or sliding. Friction is used in car brakes, when we walk or climb a hill, making a fire, skiing down a hill, and more.
## How is friction caused?
Surface roughness: When two rough surfaces come into contact with each other, they generate frictional force or an opposing force, which can sometimes be converted into heat. Body deformation: During motion, deformations in the body or the surface of the body in contact may cause friction.
## What are three types of friction?
The reason we are able to control cars at all is because of friction between the car’s tires and the road: more accurately, because there are three kinds of friction: rolling friction, starting friction, and sliding friction.
## What are 5 examples of friction?
• Driving of a a vehicle on a surface.
• Applying brakes to stop a moving vehicle.
• Skating.
• Writing on notebook/ blackboard.
• Flying of aeroplanes.
• Drilling a nail into wall.
• Sliding on a garden slide.
## What are the two main types of friction?
There are two main types of friction: static and kinetic, according to the journal The Physics Teacher (opens in new tab). Static friction operates between two surfaces that aren’t moving relative to each other, while kinetic friction acts between objects in motion.
## What type of force is friction?
Friction is a type of contact force. It exists between the surfaces which are in contact.
## What is called frictional force?
Frictional force refers to the force generated by two surfaces that contact and slide against each other. A few factors affecting the frictional force: These forces are mainly affected by the surface texture and the amount of force impelling them together.
## What is friction simple answer?
Friction is a force between two surfaces that are sliding, or trying to slide, across each other. For example, when you try to push a book along the floor, friction makes this difficult. Friction always works in the direction opposite to the direction in which the object is moving, or trying to move.
## Who is the father of friction?
John Theophilus Desaguliers (1734) first recognized the role of adhesion in friction. Microscopic forces cause surfaces to stick together; he proposed that friction was the force necessary to tear the adhering surfaces apart. The understanding of friction was further developed by Charles-Augustin de Coulomb (1785).
## What is angle of friction?
The angle between the resultant of frictional force and the normal reaction makes with the normal force is called the angle of friction.
## What are the effects of friction?
Friction reduces the speed of the moving objects and it even stops the motion of the object. Friction between the objects produces heat. This results in the wastage of energy in the machines. There will be wear and tear of the machine parts due to friction.
## Why is force a vector?
Answer: Force has both direction and magnitude and obeys the vector law of addition. Hence, it is a vector.
## What is Newton’s first law?
Newton’s First Law of Motion (Inertia) An object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line unless acted on by an unbalanced force.
## What is CGS full form?
The full form of CGS system is Centimeter Gram Second system. In the CGS system, fundamental units are Centimeter, Gram, and second. The full form of FPS system is Foot Pound Second system.
## What is the general formula of friction?
coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.
## What causes friction Class 11?
Friction is a resistive force, which comes into play when there is a relative motion between two bodies in contact. The frictional force between two bodies depends mainly on three factors: (i) the adhesion between body surfaces (ii) roughness of the surface (iii) deformation of bodies.
## How do I calculate friction?
The coefficient of friction (fr) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together. It is represented by the equation: fr = Fr/N.
## What are advantages of friction?
• Friction enables us to walk freely.
• It helps to support ladder against wall.
• It becomes possible to transfer one form of energy to another.
• Objects can be piled up without slipping.
• Breaks of vehicles work due to friction.
• It always resists the motion, so extra energy is required to overcome it.
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# Potential of a Charged Conducting Sphere
## Homework Statement
A conducting sphere with radius R is charged to voltage V0 (relative to a point an infinite distance from the sphere where the potential is zero). What is the surface charge density σ? Express your answer in terms of the given quantities and ϵ0.
## Homework Equations
Electric field on a sphere, E=(1/4πϵ0)*(Q/r^2)
Surface charge density: σ=Q/A, where Q is the charge and A is the area of the surface.
A for a sphere: A=4πr^2
## The Attempt at a Solution
I need to find σ and I have Q in two of the equations I replaced A with the area of a sphere, so I get
σ=Q/4πr^2
Then I solved for Q from this, so I have Q=σ4πr^2, now I substitute this into the first equation and solve for σ, so now I have, E=(1/4πϵ0r^2)*(σ4πr^2/r^2), this gives me E*ϵ0, but this is not the correct answers. Does E=V0?
Related Introductory Physics Homework Help News on Phys.org
What do you mean by 0 here? Are you referring to the electric permittivity of free space, ##ε_0## ?
ehild
Homework Helper
Hi Estefania, welcome to PF!
## Homework Statement
A conducting sphere with radius R is charged to voltage V0 (relative to a point an infinite distance from the sphere where the potential is zero). What is the surface charge density σ? Express your answer in terms of the given quantities and ϵ0.
## Homework Equations
Electric field on a sphere, E=(1/4πϵ0)*(Q/r^2)
Surface charge density: σ=Q/A, where Q is the charge and A is the area of the surface.
A for a sphere: A=4πr^2
## The Attempt at a Solution
I need to find σ and I have Q in two of the equations I replaced A with the area of a sphere, so I get
σ=Q/4πr^2
Then I solved for Q from this, so I have Q=σ4πr^2, now I substitute this into the first equation and solve for σ, so now I have, E=(1/4πϵ0r^2)*(σ4πr^2/r^2), this gives me E*ϵ0, but this is not the correct answers. Does E=V0?
You need to give σ in terms of V0 , the potential of the sphere. What is the formula for the potential of a charged sphere ?
What do you mean by 0 here? Are you referring to the electric permittivity of free space, ##ε_0## ?
Yes, that's what I meant.
Hi Estefania, welcome to PF!
You need to give σ in terms of V0 , the potential of the sphere. What is the formula for the potential of a charged sphere ?
The potential is given by V=Q/4pi*epsilon naught*r, so I would use this equation and set V0 equal to this?
ehild
Homework Helper
The potential is given by V=Q/4pi*epsilon naught*r, so I would use this equation and set V0 equal to this?
Yes, ##V_0=\frac{Q}{4\pi\epsilon_0 r}## . (Use parentheses!)
Find Q from this.
Estefania_8
Yes, that's what I meant.
Then you have all the variables you'll need. Follow ehild's procedure and eliminate ##Q## from the equations. This should be easy as you've already stated yourself that ##σ=\frac Q{4πr^2}##.
Estefania_8
Yes, ##V_0=\frac{Q}{4\pi\epsilon_0 r}## . (Use parentheses!)
Find Q from this.
Yes, ##V_0=\frac{Q}{4\pi\epsilon_0 r}## . (Use parentheses!)
Find Q from this.
OK, so now I have V0*4πε0*r=Q, so now I would just use σ=Q/A, which would translate into σ=Q/4πr^2, like PWiz said. Then I would use σ4πr^2=Q, which would give me V0*4πε0*r=σ4πr^2, so solving for σ I get Voε0/r=σ and that is the answer. Thank you to you both!
ehild
Homework Helper
You are welcome
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# I Directional Derivatives and Derivations - Tangent Spaces
#### Math Amateur
Gold Member
I am reading John M. Lee's book: Introduction to Smooth Manifolds ...
I am focused on Chapter 3: Tangent Vectors ...
I need some help in fully understanding Lee's conversation on directional derivatives and derivations ... ... (see Lee's conversation/discussion posted below ... ... )
Lee defines a directional derivative and notes that taking a directional derivative of a function $f$ in $\mathbb{R}^n$ at a tangent vector $v_a$ is a linear operation ... ... and follows a product rule ... ... that is, for two functions at $v_a$ we have
$D_{v|_a} (fg) = f(a) D_{v|_a} g + g(a) D_{v|_a} f$
Lee then defines a derivation ... that seems to generalise the directional derivative to any linear map that satisfies the product rule (see definition and (3.2) below ...
... ... BUT ... ... why does Lee need a 'derivation' ... why not stay with the the directional derivative ... especially as Proposition 3.2 (see below) establishes an isomorphism between $\mathbb{R}^n_a$ by using a map:
$v_a \mapsto D_{v|_a}$
... that is a map that is onto the directional derivative ...
Can someone please explain why Lee is introducing the derivation ... ? ... and not just staying with the directional derivative ... ?
The relevant discussion in Lee, referred to above, is as follows:
#### Attachments
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Related Differential Geometry News on Phys.org
#### fresh_42
Mentor
2018 Award
Without knowing the textbook I assume it is about Lie groups, too. They play a major role in physics and one way to study them is to consider the representations of their tangent space, their Lie algebras. Derivations play a major role there, e.g. the left multiplication in a Lie algebra is one. And if you study the multiplication and automorphisms of Lie groups you will end up with derivations in their corresponding Lie algebra.
As you can see in what you uploaded a tangent vector is a function for which the product rule holds, a derivation.
For short: it is often easier to study the Euclidean tangent spaces of a manifold than the manifold itself and obtain important results about the latter.
#### JonnyG
I'm reading that same chapter in Lee's book too. From what I understand, the directional derivative works fine for a Euclidean manifold. But for an abstract manifold, you won't be be able to apply the same definition. So he generalizes the directional derivative to derivations. Then, given $p \in M$, he identifies $T_p(M)$ with $T_{\phi(p)}(\mathbb{R}^n)$, where this identification is independent of the coordinate chart. If you keep reading in the chapter and doing the mini-exercises along the way, you'll see how he does this. With this identification, we can think of the Euclidean tangent space instead, which is more intuitive.
EDIT: Look at the picture at the top of page 60. That should give you a picture of what he's doing.
#### Math Amateur
Gold Member
Thanks fresh_42 and JonnyG ... appreciate your help ...
Peter
#### stevendaryl
Staff Emeritus
I think that the author might be looking ahead to the case of more general manifolds than $R^n$.
If you have an arbitrary manifold $M$, and you have a parametrized path $\mathcal{P}(t)$ (that is a continuous, smooth function of type $R \rightarrow M$), you can implicitly define a kind of vector--the tangent vector to the path $\mathcal{P}(t)$--by just giving its directional derivative:
If $v = \frac{d \mathcal{P}}{dt}$, and $a = \mathcal{P(0)}$, then $D_{v|a}$ is that operator defined by:
$D_{v|a} f = \frac{d}{dt}|_{t=0} f(\mathcal{P}(t))$
This is a generalization of the original definition, which assumed a simple form of $\mathcal{P}(t)$: $\mathcal{P}(t) = a + vt$. Adding a vector to a point doesn't make sense for an arbitrary smooth manifold, but you can still make sense of a parametrized path.
Now, the problem with this generalization of the notion of a directional derivative is that it's not completely obvious that it forms a vector space. That is, if you have two directional derivatives: $D_{v|a}$ and $D_{u|a}$, can you always find a third directional derivative $D_{w|a} = D_{u|a} + D_{v|a}$? You can, but it's not obvious. In contrast, the proof that derivations form a vector space is pretty trivial.
"Directional Derivatives and Derivations - Tangent Spaces"
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The 2024 Developer Survey results are live! See the results
# Questions tagged [projection]
Projections are several different types of geometrical mapping functions. In 3D, a projection maps three-dimensional points to a two-dimensional plane.
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### How should I avoid the weird stretching that comes with a projection matrix?
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127 views
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285 views
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1 vote
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### How to make backface culling work correctly in both orthographic and perspective projection?
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389 views
### D3D12: how to enable perspective correction when interpolating UV coordinates?
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### Move player towards screen
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2k views
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### When moving the "camera" should one move the projection matrix or move the world?
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### Matrix math: How to specify camera distance so that an object with a given size is completely visible?
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1 vote
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### Custom directional shadow map from sun
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### Drawback of painters algorithm:
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### How to visualize angle of projection in Oblique Parallel Projection?
We know that in Oblique Parallel Projection Point (x,y,z) is projected to position (x_p,y_p) on the view plane.Projector (oblique) from (x,y,z) to (x_p,y_p) makes an angle alpha. with the line (L) on ...
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### 3D image projected to 2D image
We know real world images are 3D.when we projects image from real world this image projects into 2D image where Z component is constant or zero. But when capture the below image which seems to 3D(z ...
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### Painter's theorem inside and outside test [closed]
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### How to project 3D meshes onto a 2D plane producing a 2D polygon silhouette?
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### 3D projection, Similar to SceneCapture2D
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### Approximate doom/heretic-style freelook distortion with modern graphics APIs
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### Perspective of a Point Light
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4k views
### How to determine if a 2D image represents what player sees in 3D game world?
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199 views
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I am using an old Runescape model format, also used by Thief and Quake. In this format, instead of specifying UV coordinates for each vertex ABC, we specify a second trio of vertices PMN. Those ...
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I'm trying to find a formula that could convert my 3D global coordinates into local object space. Here are the inputs I have ...
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### How to project a 3D circle/ellipse to 2D?
The 3D ellipse is represented by its center and the vectors for its major and minor axes and the 2D ellipse by its center, the length of the major and minor axes and the angle between the major axis ...
1 vote
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### How to convert from frag position to UV coordinates when my viewport doesn't cover the screen?
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I found this algorithm in some old codebase, it takes two triangles from a mesh ABC and PMN, where ABC is the triangle that will be rendered and PMN is an extra triangle that is only used to generate ...
632 views
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I've found this article on Gamasutra on a custom camera projection to make 3D look more 2D and decided to implement it to make some gameplay tests. I had to make some changes to make it work on URP(on ...
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## N-Queens
### June 11, 2010
We present today a classic exercise that has been on my to-do list since the start of Programming Praxis.
The n-queens problem is to find all possible ways to place n queens on an n × n chess board in such a way that no two queens share a row, column, or diagonal. The diagram at right shows one way that can be done on a standard 8 × 8 chess board.
Your task is to write a program to find all such placements. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
### 13 Responses to “N-Queens”
1. Remco Niemeijer said
```import Data.List
queens :: Int -> [[Int]]
queens n = filter (safe . zip [1..]) \$ permutations [1..n]
safe :: [(Int, Int)] -> Bool
safe [] = True
safe (x:xs) = all (safe' x) xs && safe xs where
safe' (x1,y1) (x2,y2) = x1+y1 /= x2+y2 && x1-y1 /= x2-y2
```
2. […] Praxis – N-Queens By Remco Niemeijer In today’s Programming Praxis we have to solve the classic n-queens problem. The provided Scheme solution has […]
3. I was trying to make a version that printed and counted, and would fit in 4 lines so you could use it as a .signature file. Didn’t quite succeed, mostly because I need to include to get printf, but here’s where I got it.
```#include <stdio.h>
int t,i,j,b[8],r;c(){r=1;for(i=0;i<7;i++)for(j=i+1;j<8;j++)r&=(b[i]-b[j]&&abs(i
-j)!=abs(b[i]-b[j]));return r;}p(){for(i=0;i<8;i++){for(j=0;j<8;j++)putchar(
" Q"[j==b[i]]);puts("");}puts("");t++;}s(n){int i;if(n--){for(i=8;i;){b[n]=--i;
s(n);}}else if(c(b))p(b);}main(){s(8);printf("%d solutions.\n",t);}
```
4. N = 8
def valid? stack
q2 = stack.length – 1
(0..stack.length-2).each do |q1|
return false if stack[q1] == stack[q2] ||
(q1-q2).abs == (stack[q1]-stack[q2]).abs
end
end
def queens stack, n
if n == N
puts “[ #{stack.join(‘, ‘)} ]”
else
(1..N).each do |rank|
stack.push(rank)
queens(stack, n+1) if valid?(stack)
stack.pop
end
end
end
queens [], 0
5. slabounty said
Here’s a ruby version (commented at http://steamcode.blogspot.com/2010/06/n-queens-problem-in-ruby.html)
```class Board
def initialize(n)
@n = n
@board = Array.new(n)
@board.map! { Array.new(n, "b") }
end
def print_board
puts "Board:"
@board.each_index do |row|
@board.each_index do |col|
end
puts
end
end
def safe_row(suggested_row)
0.upto(@n-1) do |col|
return false if @board[suggested_row][col] == "Q"
end
return true
end
def safe_col(suggested_col)
0.upto(@n-1) do |row|
return false if @board[row][suggested_col] == "Q"
end
return true
end
def safe_diag(suggested_row, suggested_col, row_mod, col_mod)
row,col = suggested_row+row_mod, suggested_col+col_mod
while true do
break if (row >= @n) || (col >= @n) || (row < 0) || (col < 0)
return false if @board[row][col] == "Q"
row += row_mod
col += col_mod
end
return true
end
def safe(suggested_row, suggested_col)
return false if !safe_row(suggested_row)
return false if !safe_col(suggested_col)
return false if !safe_diag(suggested_row, suggested_col, 1, 1)
return false if !safe_diag(suggested_row, suggested_col, 1, -1)
return false if !safe_diag(suggested_row, suggested_col, -1, 1)
return false if !safe_diag(suggested_row, suggested_col, -1, -1)
return true
end
def solve
solve_1(0)
end
def solve_1(row)
0.upto(@n-1) do |col|
if safe(row, col)
@board[row][col] = "Q"
if row == (@n-1)
print_board
else
solve_1(row+1)
end
@board[row][col] = "b"
end
end
end
end
board = Board.new(8).solve
```
6. Jos Koot said
http://en.wikipedia.org/wiki/Eight_queens_puzzle section ‘constructing a solution’ shows an algorithm that is linear in the number of columns of the board when implemented with a vector. It allows you to go to boardsizes up to N=200000000 (I can’t go higher because of memory)
```(define (queens N)
(if (and (exact-nonnegative-integer? N) (> N 3))
(let* ((R (remainder N 12))
(n-even (floor (/ N 2)))
(n-odd (- N n-even))
(V (make-vector N))
(set-V! (lambda (f r) (vector-set! V f r)))
(sema (make-semaphore 0)))
(let-syntax
((do-loop (syntax-rules ()
((for (x range ...) body ...)
(for ((x (in-range range ...))) body ...)))))
(λ ()
(case R
((3 9)
(do-loop (F 0 (sub1 n-even)) (set-V! F (+ (* 2 F) 3)))
(set-V! (sub1 n-even) 1))
(else (do-loop (F 0 n-even) (set-V! F (add1 (* 2 F))))))
(semaphore-post sema))) ; Signal completion.
(λ ()
(case R
((2)
(do-loop (F 2 (- n-odd 1)) (set-V! (+ F n-even) (+ (* 2 F) 2)))
(set-V! n-even 2)
(set-V! (sub1 N) 4))
((3 9)
(do-loop (F 0 (- n-odd 2)) (set-V! (+ F n-even) (+ (* 2 F) 4)))
(set-V! (- N 2) 0)
(set-V! (sub1 N) 2))
((8)
(do-loop (i 0 (sub1 n-odd) 2)
(set-V! (+ i n-even) (* 2 (add1 i)))
(set-V! (+ i n-even 1) (* 2 i))))
(else
(do-loop (F 0 n-odd) (set-V! (+ F n-even) (* 2 F)))))
(semaphore-post sema))) ; Signal completion.
(semaphore-wait sema) ; Wait for both loops to complete.
(semaphore-wait sema)
V))
(raise-type-error 'queens "exact integer greater than 3" N)))
```
8. programmingpraxis said
Here’s a different, somewhat simpler, solution in Scheme:
```(define (attack? q1 q2) (or (= (car q1) (car q2)) (= (cadr q1) (cadr q2)) (= (abs (- (car q1) (car q2))) (abs (- (cadr q1) (cadr q2))))))```
```(define (safe? q qs) (cond ((null? qs) #t) ((attack? q (car qs)) #f) (else (safe? q (cdr qs)))))```
```(define (queens n) (let queen ((n n) (x 1) (y 1) (qs '()) (qss '())) (cond ((< n x) (cons (reverse qs) qss)) ((< n y) qss) ((safe? (list x y) qs) (queen n x (+ y 1) qs (queen n (+ x 1) 1 (cons (list x y) qs) qss))) (else (queen n x (+ y 1) qs qss)))))```
9. Ying Yin said
Java implementation:
Source
10. simonliu2008 said
can any one tell me how to convert n-queens problem to exact cover problem?
11. programmingpraxis said
Knuth discusses this in his dancing links paper.
12. […] problem with n queens on an n by n chessboard. After finding it again in older posts on both Programming Praxis and DataGenetics, I decided to go ahead and take a crack at it and I think the solution is pretty […]
13. Hi I am new to Ruby,I wrote a program in ruby which based on NQueen java program by Robert Sedgewick and Kevin Wayne http://introcs.cs.princeton.edu/java/23recursion/Queens.java.html,The problem is My program is not outputting the result,please help me here
my program source : http://repl.it/1jv
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https://physics.stackexchange.com/questions/590107/why-is-the-electrostatic-force-felt-in-straight-lines
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# Why is the electrostatic force felt in straight lines?
When two positive charges are kept close, they get repelled in the direction of a line joining both the charges. Why is it so?
Also, why is the repulsion in a straight path?
In both the cases, the potential energy of the charge which gets repelled decreases. What makes it repel in a straight line such that the line passes through both charges?
• You mention potential energy. Have you drawn the equipotential curves around one charge to see which way the gradient will act on the second (test) charge? Commented Oct 29, 2020 at 19:33
• I’m voting to close this question because Physics cannot tell us why the universe works the way it does. Commented Oct 29, 2020 at 19:34
• I'm no physicist, but you may want to see this: researchgate.net/publication/… . Unfortunately I believe it has a major flaw, namely that it fails to derive conservation of AM as it claims it does, for exactly the reason Alexander Issa says. Personally, I find this very question the reason to adopt conservation of angular momentum for rigid body mechanics, and also shows why it is necessary but often ignored. Commented Oct 30, 2020 at 0:31
• In truth, the answer to why conservation of angular momentum works is a little more complicated though. onlinelibrary.wiley.com/doi/pdf/10.1002/zamm.202000117 Commented Oct 30, 2020 at 0:31
• @BioPhysicist I don't think most people have studied Physics to such an extent to know what's a taken fact and what is provable. So how does one be certain that this cannot be proved and rather is an assumption of ours based on observations? Commented Oct 30, 2020 at 9:13
Just to elaborate on the symmetry argument - Let's suppose in your first diagram you are observing the two charges from the side and we assume, as you have done, that the repulsion direction is vertical and to the right. If we now observe the two charges from the top looking down, we are presented with the exact same situation as before and we would say the charge should now be repulsed horizontally and to the right. But that contradicts the direction we initially assumed - the repulsion cannot depend on how you look at the two charges. Continuing with this type of argument one can only conclude that the repulsion must be along the line connecting the two charges.
Consider an isolated system of two particles:
Since system is isolated, angular momentum (and linear) is conserved.
I.e. $$\vec\tau_{net}=\Sigma\space \vec r\times\vec F=0$$
But clearly from figure $$\vec\tau_{net}=\vec r\space\times\space\vec F_y\ne 0$$
Thus as R.W. Bird noted, the system violates the conservation of angular momentum.
• But this doesn't explain why it is this way. The angular momentum could be coming from some other thing (like a field, for example) and thus the total angular momentum would be constant. Commented Oct 29, 2020 at 19:20
• @BioPhysicist: Per Noether, that's because the universe is rotationnally symmetric. There's no preferred direction. You add "could, in the presence of a field (of some kind)", but then the premise is no longer true. A magnetic field will trivially affect the repelled electric charge. This answer correctly follows the assumption of the question. Commented Oct 30, 2020 at 16:37
• IMO angular momentum is conserved because the force is radial, and not the other way around. The deeper reason would just be symmetry of space. Commented Oct 31, 2020 at 3:46
If the forces were not co-linear, the system would violate conservation of angular momentum.
• Can you explain it more.. Commented Oct 28, 2020 at 14:52
• But electromagnetism does produce non-co-linear forces, so any argument along these lines is not sufficient. Commented Oct 29, 2020 at 14:27
• Segments of current carrying wire are generally subject to other forces. Two permanent magnets floating in space and interacting only with each other will conserve linear and angular momentum. Commented Oct 29, 2020 at 20:47
Maxwell's equations (along with the Lorentz force law) tell us that the force between two charges is radial (along the line connecting the charges).
You can also appeal to symmetry, which is related to R.W. Bird's point about conservation of angular momentum. Unless the charges have an inherent "orientation" associated with them (which might be in the form of angular momentum) or nature is "random" in the way it decides which way things move, there is no reason the charge on the right would move up or down.
• The force between two stationary charges is along the line connecting them. The force between two moving charges is not necessarily radial. See, for example, Section 8.2.1 of Griffiths. Commented Oct 29, 2020 at 14:39
Consider the case of two positively charged bodies, with gravitational forces similar to the electromagnetic repulsion, orbiting each other. The particles have to be travelling not on the same line to enter this system (eg opposite parallel paths), and we find the angular momentum for the orbiting situation comes from that, the ‘seesaw’ motion around their combined centre of mass.
Particles have electromagnetic properties which produce forces many orders of magnitude higher than gravitational, but all the forces momenta & potentials do always interact. There is no ‘pure’ electrostatic object. When you go beyond 2 bodies, and include magnetism, all kinds of strange things can happen. Have a look at pictures from cloud-chamber particle detectors.
Also, consider a delocalised electron - perhaps interacting with a hole in a semiconductor layer, to preserve having the same charges: It cannot be simplified like in your diagram, because the angular momentum of the electrons is 'smeared out’ through the conduction band.
This is about idealisations. The other idealisation, is nice neat spheres, like particles. What if they aren't: like say two negatively charged molecules, that have rotational energy. Changes in the rotation of the molecules, speeding up or slowing down, or changing the angle of tumbling, or adding momentum around another axis in the molecule, could give all different bounce directions. It's normal to deal with dynamics of spheres in particle mechanics, and gravitation. What that really means is asymmetries are small, compared to the size of the interaction being looked at. Quarks stop protons being points, Earth bulges in the middle and is lumpy. When something can be treated as a sphere, symmetry says it’s like all the forces act at the spheres centre, because that is where everything averages - but beware when that doesn’t hold.
Big picture, what you are asking about is symmetries, and it quickly gets you into our deepest ideas in physics. Why can particles only have curving interactions when they approach off-centre, and only linear interactions (where idealisations hold), is a deep question. Noether’s theorem relates conservation laws, like conservation of rotational momentum in this case (ie, if it starts at 0, both on same line, stays at zero), to continuous symmetries: these are generalisations of these 'little symmetries’, to the whole system, or to the universe itself. A lot of people think the total of important quantities like momentum and energy for the universe as a whole, will be zero. Occasional violations of symmetries/conserved quantities are key to the frontiers of physics, like charge-parity-time (CPT) violations, & CP violations that explain why there is more matter than antimatter.
Asking simple questions, and really digging until you get answers you are satisfied with, is key to really doing physics. Many important results came from people not satisfied with prevailing wisdom. If you really follow this question you will cover some of the deepest physics there is. Don’t let people fob you off! Keep digging. Keep asking simple questions, with tenacity.
Great question!
Perhaps you're familiar with the idea that the force on an object is the negative gradient of the potential energy created by that force*:
$$-\vec{\nabla} PE(x,y,z) = \vec{F}(x,y,z)$$
Now, imagine any slope. The gradient vector at any point on that slope points in the direction that you would step to increase you're altitude most quickly. The negative gradient vector does the opposite: it points in the direction that you would step to decrease you're altitude most quickly.
With this in mind, think about what this physics equation is really saying! The force on an object always points in the direction of travel that would most efficiently decrease it's potential energy.
So it's not JUST that the forces on the object want to get rid of it's potential energy, it's that (in this sense) they want to do it as efficiently as possible. This is why the positive charge wants to get away from the positive charge in a straight line: it wants to decrease it's potential energy as efficiently as possible.
*if the force is conservative, and here it is.
One answer is that forces are determined by the gradient of -U, where U is the potential energy. Of course, that just raises the question of why that is so. I believe that the Principle of Least Action prescribes that particles move along the gradient, but that similarly leads to the question of why particles are constrained to follow that. I believe that if you look at a Hamiltonian, the sum over histories of the paths other than following the gradient results in destructive interference, but I'm far from clear on that.
You could dig deeper into this and probably find a very complicated explanation, however there is a simple reason: Nobody knows why.
Yes, coloumb's law was empirical and it was observed how the charges behaved and then the law was derived. Most laws in classical physics are empirical and were observed at some point of time by really smart scientists and put into theory. Even the formula for the wavelength emitted for transition of an electron in an Hydrogen atom was discovered by trying to relate the numbers that were seen when the hydrogen spectrum was observed. This was accomplished by Balmer who was a teacher. Get used to this fact. If we could derive everything from scratch, why would there still be unanswered questions? If this answer was helpful, take some time off to accept it and if you require any further information, do not hesitate to comment!
Edit: Before you say that this isn't how stuff works and we can explain using this or that, remember that my point here is to tell that every theoretical concept requires some observational backing or assumption. Nothing can be derived from scratch. Nobel Prize Laureate for Physics Richard Peter Feynman has in his Feynman Lectures said that the universe is like a game, whose play we observe and based on our observation, we try to deduce the game's rules. This is the best explanation of the idea conveyed here. We did not make the game, we observe and deduce.
• We do know. It's en.m.wikipedia.org/wiki/Noether%27s_theorem What is your point about Coulomb’s law, that we didn’t used to understand it so give up trying to make sense of things? Or are you claiming it’s not understood? We have all the electron transitions nice & neat from first principles. Sure we start empirically, but that is no way to justify staying there Commented Oct 28, 2020 at 22:02
• @CrigICrigI I have said in my first line that you can find complicated explanations but none of the explanations can be something that starts off from scratch and derives something. You always have to assume that some particular law works which would lead to the result. And as regards coloumb's law, I spoke about it because a tag has been used and that is what anyone would think of first when two charges are given. The OP clearly asked why is the repulsion along the line joining the charges and not in any other direction. And as regards trying to make sense of things, as I said... Commented Oct 29, 2020 at 3:41
• ...nothing can be derived without a prior assumption that a certain law/principle is true. Moreover the OP who is maybe studying electrostatics as a beginner surely has no need to dive deep into noether theorem at this stage. At early stages of physics study we all have questions like why does this work and that does not. This does not mean we stray off to complicated physics at that moment. Commented Oct 29, 2020 at 3:48
• @CriglCragl ^^^ Commented Oct 29, 2020 at 4:38
• @CriglCragl also why is it that in Newton's laws the force varies as inverse square of distance and not the inverse cube, why is the Lorentz force what it is? Why is it that particles have charge? Why do electrons have magnetic moment?? They don't have current flowing in them. Moreover, if the magnetism is intrinsic well explain that. Could you solve equations and prove all the trajectories in our solar system are correct. Also derive the equations you use for that. These all questions are examples of when someone does not understand something. And that something is that.... Commented Oct 29, 2020 at 4:46
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# SOLUTION: Find the LCM of 6x^8 and 12x^5
Algebra -> -> SOLUTION: Find the LCM of 6x^8 and 12x^5 Log On
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Mathway solves algebra homework problems with step-by-step help!
Click here to see ALL problems on Exponents Question 81291: Find the LCM of 6x^8 and 12x^5Answer by stanbon(60774) (Show Source): You can put this solution on YOUR website!Find the LCM of 6x^8 and 12x^5 ----- The LCM is the product of all the factors of the component terms in their highest power. 6x^8 = 2*3*x^8 12x^5 = 2^2*3*x^5 ---------- So the LCM must have a 2^2; a 3; and x^8 So the LCM is 2^2*3*x^8= 12x^8 ============= cheers, Stan H.
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BUILDING MINIMUM SPANNING TREES BY LIMITED NUMBER OF NODES OVER TRIANGULATED SET OF INITIAL NODES
Authors
• Vadim Romanuke Vinnytsia Institute of Trade and Economics of State University of Trade and Economics, Ukraine
Keywords:
minimum spanning tree, triangulation, edge lengths, redundant nodes, root node
Abstract
Background. The common purpose of modelling and using minimum spanning trees is to ensure efficient coverage. In many tasks of designing efficient telecommunication networks, the number of network nodes is usually limited. In terms of rational allocation, there are more possible locations than factually active tools to be settled to those locations.
Objective. Given an initial set of planar nodes, the problem is to build a minimum spanning tree connecting a given number of the nodes, which can be less than the cardinality of the initial set. The root node is primarily assigned, but it can be changed if needed.
Methods. To obtain a set of edges, a Delaunay triangulation is performed over the initial set of nodes. Distances between every pair of the nodes in respective edges are calculated. These distances being the lengths of the respective edges are used as graph weights, and a minimum spanning tree is built over this graph.
Results. The problem always has a solution if the desired number of nodes (the number of available recipient nodes) is equal to the number of initially given nodes. If the desired number is lesser, the maximal edge length is found and the edges of the maximal length are excluded while the number of minimum spanning tree nodes is greater than the desired number of nodes.
Conclusions. To build a minimum spanning tree by a limited number of nodes it is suggested to use the Delaunay triangulation and an iterative procedure in order to meet the desired number of nodes. Planar nodes of an initial set are triangulated, whereupon the edge lengths are used as weights of a graph. The iterations to reduce nodes are done only if there are redundant nodes. When failed, the root node must be changed before the desired number of nodes is changed.
References
J. Nešetřil, E. Milková, and H. Nešetřilová, “Otakar Borůvka on minimum spanning tree problem: translation of both the 1926 papers, comments, history,” Discrete Mathematics, vol. 233, iss. 1 — 3, pp. 3 — 36, 2001.
R. L. Graham and P. Hell, “On the history of the minimum spanning tree problem,” Annals of the History of Computing, vol. 7, iss. 1, pp. 43 — 57, 1985. https://doi.org/10.1109/MAHC.1985.10011
Y. K. Dalal and R. M. Metcalfe, “Reverse path forwarding of broadcast packets,” Communications of the ACM, vol. 21, iss. 12, pp. 1040 — 1048, 1978. https://doi.org/10.1145/359657.359665
B. Stojanović, T. Rajić, and D. Šošić, “Distribution network reconfiguration and reactive power compensation using a hybrid Simulated Annealing — Minimum spanning tree algorithm,” International Journal of Electrical Power & Energy Systems, vol. 147, Article ID 108829, 2023.
H. Ahmadi and J. R. Martí, “Minimum-loss network reconfiguration: A minimum spanning tree problem,” Sustainable Energy, Grids and Networks, vol. 1, pp. 1 — 9, 2015.
S. Pettie and V. Ramachandran, “An optimal minimum spanning tree algorithm,” Journal of the Association for Computing Machinery, vol. 49, iss. 1, pp. 16 — 34, 2002. https://doi.org/10.1145/505241.505243
T. H. Cormen, C. E. Leiserson, R. L. Rivest, and C. Stein, “Chapter 23: Minimum Spanning Trees,” in Introduction to Algorithms, Second Edition. MIT Press and McGraw-Hill, 2001, pp. 561 — 579.
R. C. Prim, “Shortest connection networks and some generalizations,” Bell System Technical Journal, vol. 36, iss. 6, pp. 1389 — 1401, 1957.
E. W. Dijkstra, “A note on two problems in connexion with graphs,” Numerische Mathematik, vol. 1, iss. 1,
pp. 269 — 271, 1959.
T. Cormen, C. E. Leiserson, and R. L. Rivest, and C. Stein, Introduction to Algorithms, Third Edition. MIT Press, 2009.
H. Loberman and A. Weinberger, “Formal procedures for connecting terminals with a minimum total
wire length,” Journal of the ACM, vol. 4, iss. 4,
pp. 428 — 437, 1957.
R. E. Tarjan, “Chapter 6. Minimum spanning trees. 6.2. Three classical algorithms,” Data Structures and Network Algorithms, in CBMS-NSF Regional Conference Series in Applied Mathematics, vol. 44, Society for Industrial and Applied Mathematics, 1983, pp. 72 — 77.
S. Chakraborty, A. Mukherjee, V. Raman, and S. Rao Satti, “Frameworks for designing in-place graph algorithms,” Journal of Computer and System Sciences, vol. 123, pp. 1 — 19, 2022.
R. Jothi, S. K. Mohanty, and A. Ojha, “Fast approximate minimum spanning tree based clustering algorithm,” Neurocomputing, vol. 272, pp. 542 — 557, 2018. https://doi.org/10.1016/j.neucom.2017.07.038
H. Edelsbrunner, T. S. Tan, and R. Waupotitsch, “An time algorithm for the minmax angle triangulation,” SIAM Journal on Scientific and Statistical Computing, vol. 13, iss. 4, pp. 994 — 1008, 1992. https://doi.org/10.1137/0913058
J. A. De Loera, J. Rambau, and F. Santos, Triangulations, Structures for Algorithms and Applications, in Algorithms and Computation in Mathematics, vol. 25, Springer, 2010.
G. Xia, “The stretch factor of the Delaunay triangulation is less than 1.998,” SIAM Journal on Computing, vol. 42, iss. 4, pp. 1620 — 1659, 2013.
V. V. Romanuke, “Fast-and-smoother uplink power control algorithm based on distance ratios for wireless data transfer systems,” Studies in Informatics and Control, vol. 28, iss. 2, pp. 147 — 156, 2019.
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# 12sinx-5coax find the maximum and minimum belie of the question +2 1st year
Arun
25750 Points
6 years ago
We know that in
a sinx - b cosx
Maximum value = $\sqrt$(a2 + b2)
Minimum value = - $\sqrt$(a2 + b2)
hence in this equation
Maximum value = $\sqrt$(144+25) = 13
Minimum value = - $\sqrt$(144+25) = -13
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INDEX&MATCH – Excel Template
The combination of INDEX&MATCH is an interesting alternative to VLOOKUP. However, INDEX&MATCH has one major advantage compared to VLOOKUP – the lookup value doesn’t have to be in the leftmost column of the source table.
INDEX is a function, which returns the value located at a given intersection within an array. MATCH returns the relative position of an item within an array. The INDEX function requires the number of the row within a range as an input. And we can find that number with the MATCH function. In other words, MATCH indicates the position of the result, and the INDEX function extracts it.
Some other related topics you might be interested to explore are VLOOKUP, XLOOKUP, and HLOOKUP.
This is an open-access Excel template in XLSX format that will be useful for anyone who wants to work as a Financial Analyst, Business Analyst, and for everyone who works with a spreadsheet software.
3-Statement Model – Excel Template
The P&L, Balance sheet, and Cash flow statements are three interrelated parts. The P&L feeds net income on the
3-Statement Model – Excel Template
The P&L, Balance sheet, and Cash flow statements are three interrelated parts. The P&L feeds net income on the liabilities and equity side of the Balance sheet. At the same time, we obtain Cash (an asset) by summing the bottom-line result of the Cash flow statement with previous year...
Cash Flow – Excel Template
The cash flow statement shows how a company generated and spent cash throughout a given timeframe. An important truth
Cash Flow – Excel Template
The cash flow statement shows how a company generated and spent cash throughout a given timeframe. An important truth that is frequently neglected by inexperienced business owners is that profit does not equal cash. Every business owner and manager needs to have a clear idea of the cash flows...
Balance Sheet – Excel Template
If the P&L statement shows how profitable a company was over a given timeframe, we can say that the
Balance Sheet – Excel Template
If the P&L statement shows how profitable a company was over a given timeframe, we can say that the Balance sheet is like a momentary picture of the firm’s condition at the time of preparation. The Balance sheet shows what a company owns (assets) and what it owes (liabilities...
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# Quantum Computing and the Limits of the Efficiently Computable Scott Aaronson MIT.
## Presentation on theme: "Quantum Computing and the Limits of the Efficiently Computable Scott Aaronson MIT."— Presentation transcript:
Quantum Computing and the Limits of the Efficiently Computable Scott Aaronson MIT
Things we never see… Warp drive Perpetuum mobile GOLDBACH CONJECTURE: TRUE NEXT QUESTION Übercomputer The (seeming) impossibility of the first two machines reflects fundamental principles of physicsSpecial Relativity and the Second Law respectively So what about the third one?
Problem: Given a flight map, is every airport reachable from every other in 5 flights or less? Any specific map is an instance of the problem The size of an instance, n, is the number of bits used to specify it An algorithm is polynomial-time if it uses at most kn c steps, for some constants k,c P is the class of all problems for which theres a deterministic, polynomial-time algorithm that correctly solves every instance Complexity Theory 101
NP: Nondeterministic Polynomial Time 379765951771766953797024914793741172726275933019 504626889963674936650784536994217766359204092298 415904323398509069628960404170720961978805136508 024164948216028859271269686294643130473534263952 048819204754561291633050938469681196839122324054 336880515678623037853371491842811969677438058008 30815442679903720933 Does have a factor ending in 7?
NP-hard: If you can solve it, then you can solve every NP problem NP-complete: NP-hard and in NP Can you send your passenger on a round-trip that visits each city once?
P NP NP- complete NP-hard Graph connectivity Primality testing Matrix determinant Linear programming … Matrix permanent Halting problem … Hamilton cycle Steiner tree Graph 3-coloring Satisfiability Maximum clique … Factoring Graph isomorphism …
Does P=NP? The (literally) \$1,000,000 question If there actually were a machine with [running time] ~Kn (or even only with ~Kn 2 ), this would have consequences of the greatest magnitude. Gödel to von Neumann, 1956
Q: Hey, wasnt PNP proved this summer? A: No. Q: Why is P NP so hard to prove? A: Basically because all known approaches, if they worked, would also prove other things that are false (e.g., 2SAT is hard) Q: How sure are you that P NP? A: If we were physicists, we wouldve long ago declared it a law of nature and been done with it… Q: If almost everyone believes P NP, why bother proving it? A: Its not the destination, its what else well learn along the way Q: Could it be that P NP is Gödel-undecidable? A: Maybe, but Fermats Last Theorem took 350 years! Weve only been at this 40 years and know way more than when we started Q: What if P=NP, but the algorithm took n 10000 steps? A: Then wed change the question Q: Even if P NP, cant there be clever heuristics that often work? A: Yes, but cleverness would be provably necessary P/NP FAQ
Extended Church-Turing Thesis Any physically-realistic computing device can be simulated by a deterministic (or maybe probabilistic) Turing machine, with at most polynomial overhead in time and memory An important presupposition underlying P vs. NP is the... So how sure are we of this thesis? Have there been serious challenges to it?
Old proposal: Dip two glass plates with pegs between them into soapy water. Let the soap bubbles form a minimum Steiner tree connecting the pegsthereby solving a known NP-hard problem instantaneously
Protein folding: Can also get stuck at local optima (e.g., Mad Cow Disease) DNA computers: Just massively parallel classical computers Other Approaches
Ah, but what about quantum computing? (you knew it was coming) Quantum algorithms: The power of 2 n complex numbers working for YOU In the 1980s, Feynman, Deutsch, and others noticed that quantum systems with n particles seemed to take ~2 n time to simulateand had the amazing idea of building a quantum computer to overcome that problem Quantum mechanics: Probability theory with minus signs (Nature seems to prefer it that way) Shor 1994: A quantum computer, if built, could factor integers (and hence break RSA) in polynomial time
But remember: factoring isnt thought to be NP-complete! To this day (and contrary to popular misconception), we dont know a fast quantum algorithm to solve NP-complete problems (though not surprisingly, we also cant prove there isnt one) Bennett et al. 1997: Quantum magic wont be enough If you throw away the problem structure, and just consider an abstract landscape of 2 n possible solutions, then even a quantum computer needs ~2 n/2 steps to find the correct one (That bound is actually achievable, using Grovers algorithm!) So, is there any quantum algorithm for NP-complete problems that would exploit their structure?
Quantum Adiabatic Algorithm (Farhi et al. 2000) HiHi Hamiltonian with easily- prepared ground state HfHf Ground state encodes solution to NP-complete problem Problem: Eigenvalue gap can be exponentially small
Nonlinear variants of the Schrödinger Equation Abrams & Lloyd 1998: If quantum mechanics were nonlinear, one could exploit that to solve NP- complete problems in polynomial time No solutions 1 solution to NP-complete problem
Heres a polynomial-time algorithm to solve NP-complete problems (only drawback is that it requires time travel): Read an integer x {0,…,2 n -1} from the future If x encodes a valid solution, then output x Otherwise, output (x+1) mod 2 n Closed Timelike Curves (CTCs) If valid solutions exist, then the only fixed-points of the above program input and output them Building on work of Deutsch, [A.-Watrous 2008] defined a formal model of CTC computation, and showed that in both the classical and quantum cases, it has exactly the power of PSPACE (believed to be even larger than NP)
Anthropic Computing Foolproof way to solve NP-complete problems in polynomial time (at least in the Many-Worlds Interpretation): First guess a random solution. Then, if its wrong, kill yourself! Technicality: If there are no solutions, youre out of luck! Solution: With tiny probability dont do anything. Then, if you find yourself in a universe where you didnt do anything, there probably were no solutions, since otherwise you wouldve found one!
Relativity Computer DONE
Zenos Computer STEP 1 STEP 2 STEP 3 STEP 4 STEP 5 Time (seconds)
Includes P NP as a special case, but is stronger No longer a purely mathematical conjecture, but also a claim about the laws of physics If true, would explain why adiabatic systems have small spectral gaps, the Schrödinger equation is linear, CTCs dont exist... The No-SuperSearch Postulate There is no physical means to solve NP-complete problems in polynomial time.
Question: What exactly does it mean to solve an NP- complete problem? Example: Its been known for decades that, if you send n identical photons through a network of beamsplitters, the amplitude for the photons to reach some final state is given by the permanent of an n n matrix of complex numbers But the permanent is #P-complete (believed even harder than NP- complete)! So how can Nature do such a thing? Resolution: The amplitudes arent directly observable, and require exponentially-many probabilistic trials to estimate Lesson: If you cant observe the answer, it doesnt count! (Recently, Alex Arkhipov and I gave the first evidence that even the observed output distribution of such a linear-optical network would be hard to simulate on a classical computer but the argument was necessarily more subtle)
One could imagine worse research agendas than the following: Prove PNP (better yet, prove factoring is classically hard, implying PBQP) Prove NP BQPi.e., that not even quantum computers can solve NP-complete problems Build a scalable quantum computer (or even more interesting, show that its impossible) Determine whether all of physics can be simulated by a quantum computer Derive as much physics as one can from No- SuperSearch and other impossibility principles Conclusion
www.scottaaronson.com/talks www.scottaaronson.com/papers www.scottaaronson.com/blog
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Home / Expert Answers / Geometry / in-hyperbolic-geometry-given-two-lines-t-n-with-common-perpendicular-q-vec-ms-if-f-n-on-s-such-t-pa238
# (Solved): In hyperbolic geometry: Given two lines t,n with common perpendicular q, vec(MS). If F,N on s such t ...
In hyperbolic geometry: Given two lines
`t,n`
with common perpendicular
`q,`
`vec(MS)`
. If
`F,N`
on
`s`
such that
`M*F*N`
, then show that
`d(N,t)>d(F,t)`
.
We have an Answer from Expert
### Expert Answer
We have an Answer from Expert
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# Modules
### MAS2602 : Computing for Mathematics and Statistics (Inactive)
• Inactive for Year: 2020/21
• Module Leader(s): Dr Chris Graham
• Lecturer: Dr Lee Fawcett
• Owning School: Mathematics, Statistics and Physics
• Teaching Location: Newcastle City Campus
##### Semesters
Semester 1 Credit Value: 10 ECTS Credits: 5.0
#### Aims
To reinforce the computing in Python/R studied within MAS1801/1802, and to move towards expectations of more independent programming. To introduce a wider range of mathematical/statistical topics/techniques within Python & R, including methods that will be useful towards future project work. To introduce some basic ideas of algorithm complexity and computational complexity.
Module summary
Computing methods are of great use in a wide range of applications of pure and applied mathematics and statistics. This module builds on the methods introduced in MAS1801 and MAS1802, introducing additional techniques, some of increasing mathematical and computational sophistication. In implementing these methods, students will attain increasing competence with mathematical/statistical computing, and an increasing ability to use such methods independently, towards project-orientated goals.
#### Outline Of Syllabus
[Primarily using Python]
Plotting of vector fields (quiver plots) and trajectories (streamlines). Curve fitting (e.g. least squares fitting of known function to data). Root finding (Newton-Raphson and built-in Python solvers). Numerical derivatives through finite difference, and related techniques of numerical integration and numerical solution of ODEs.
[Primarily using R]
Simulations from univariate distributions (extending beyond those in MAS1802). Simulations from bivariate and trivariate distributions, including use of conditional distributions. Transformations of random variables. Sampling distributions. Illustrations of properties of hypothesis tests and confidence intervals.
[Using either Python or R]
Introduction to algorithm complexity and computational complexity. Illustration via sorting algorithms.
#### Teaching Methods
Please note that module leaders are reviewing the module teaching and assessment methods for Semester 2 modules, in light of the Covid-19 restrictions. There may also be a few further changes to Semester 1 modules. Final information will be available by the end of August 2020 in for Semester 1 modules and the end of October 2020 for Semester 2 modules.
##### Teaching Activities
Category Activity Number Length Student Hours Comment
Scheduled Learning And Teaching ActivitiesLecture21:002:00Revision lectures
Guided Independent StudyAssessment preparation and completion210:0020:00Revision for tests
Scheduled Learning And Teaching ActivitiesLecture111:0011:00Formal lectures
Scheduled Learning And Teaching ActivitiesPractical21:202:40Computer-based tests
Scheduled Learning And Teaching ActivitiesPractical102:0020:00Computer practicals
Scheduled Learning And Teaching ActivitiesDrop-in/surgery140:102:20Office hours
Guided Independent StudyIndependent study127:0027:00Studying, practising and gaining understanding of course material
Guided Independent StudyIndependent study115:0015:00Preparation for final project
Total100:00
##### Teaching Rationale And Relationship
Lectures are used for the delivery of theory and explanation of methods, illustrated with examples, and for giving general feedback on marked work. Practicals are used to help the students to develop their programming skills but also afford an opportunity to develop the students’ abilities at applying the theory to solving problems. Office hours (two per week) provide an opportunity for more direct contact between individual students and the lecturer: a typical student might spend a total of one or two hours over the course of the module, either individually or as part of a group.
#### Assessment Methods
Please note that module leaders are reviewing the module teaching and assessment methods for Semester 2 modules, in light of the Covid-19 restrictions. There may also be a few further changes to Semester 1 modules. Final information will be available by the end of August 2020 in for Semester 1 modules and the end of October 2020 for Semester 2 modules.
The format of resits will be determined by the Board of Examiners
##### Other Assessment
Description Semester When Set Percentage Comment
Written exercise1M20Project
Computer assessment1M40PC test 1 (80 mins, in-class, open book)
Computer assessment1M40PC test 2 (80 mins, in-class, open book)
##### Assessment Rationale And Relationship
The project allows the students to develop their problem solving techniques and to practise the methods learnt in the module. They also allows the assessment of the computational skills acquired by the students. The PC tests allow the students to assess their progress with the material. They both allow feedback to the students and so act as formative as well as summative assessment.
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1. ## xsin^-1 x
Hello :P
How do i simplify this?:
log4(16)*log4(4)
And this:
f(x)=xsin^(-1)x
thanks...!
2. Originally Posted by MissWonder
Hello :P
How do i simplify this?:
log4(16)*log4(4)
thanks...!
Welcome to MHF!
2 rules very handy for this type of problem are;
$log_{a}a=1$
$log_{a}(b^c)=c \cdot log_{a}b$
So we know $log_{4}4 = 1$
and we can simplify $log_{4}16=log_{4}(4^2)=2log_{4}4$.
Therefore,
$(log_{4}16)(log_{4}4)$
$=(log_{4}(4^2))(1)$
$=(2 \cdot log_{4}4)(1)$
$=(2)(1)(1)$
$=2$
Does this help?
3. Originally Posted by MissWonder
f(x)=xsin^(-1)x
you can't simplify this as written.
4. Wee yes ;D I did not know the first part with "loga(A)=1".... Thanks kasper, you're the man!...
5. $x(sin^-1)x$... better? I really don't know how to write it in a good way
6. Originally Posted by MissWonder
$x(sin^-1)x$... better? I really don't know how to write it in a good way
Yeah, if you mean $(x)sin^{-1}(x)$ as in $(x)arcsin(x)$; I'm with Krizalid there.
7. Yeah :P I really can't figure it out either! Maybe differentiate it is sufficient?..
8. Originally Posted by MissWonder
Yeah :P I really can't figure it out either! Maybe differentiate it is sufficient?..
It's not so much that it can't be figured out, there is just no way to simplify that expression further.
Keep in mind that differentiation and algebraic simplification are two very different things!
Differentiating will bring out an entirely new function that is not at all equivalent to the original.
9. I know.. But I can't simplify it.. So maybe differentiate it, and then simplify it?
10. Originally Posted by MissWonder
I know.. But I can't simplify it.. So maybe differentiate it, and then simplify it?
No, no, because simplification is essentially rewriting a function such that it is equivalent to the original function and so it looks nicer and is easier to work with.
If you differentiate, you lose that equivalency to the original function; so nothing to do with differentiation can be considered an algebraic simplification. Sure you *might* be able to simplify the derivative, but it would not be a simplification of $(x)sin^{-1}(x)$.
11. OMG! I just wrote an e-mail to my mathteacher.. She wrote that a differentiation is sufficient
12. Originally Posted by MissWonder
OMG! I just wrote an e-mail to my mathteacher.. She wrote that a differentiation is sufficient
Perhaps your math teacher knew the exact instructions that came with the expression f(x)=x sin^(-1)x. Because we certainly didn't, despite several increasingly frustrated attempts to find out what they were.
Next time post all the instructions given in the question.
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Problem of the Week Problem C and Solution Jellybean Surprise
Problem
Ru has $$100$$ jellybeans that she is placing in small boxes for a party game. She has decided that each box must contain at least one jellybean and no two boxes can contain the same number of jellybeans. As well, no box can go inside any other box.
Determine the maximum number of boxes Ru can use for her jellybeans.
Solution
In order to maximize the number of boxes, each box must contain the smallest number of jellybeans possible. However, no two boxes can contain the same number of jellybeans. The simplest way to approach this problem is to put one jellybean in the first box and then let the number of jellybeans in each box after that be one more than the number of jellybeans in the box before it, until all $$100$$ jellybeans are in boxes.
We will put $$1$$ jellybean in the first box, $$2$$ jellybeans in the second box, $$3$$ jellybeans in the third box, and so on. After filling $$12$$ boxes this way, we have used $$1+2+3+4 + 5 + 6 + 7 + 8 + 9 + 10+11+12=78$$ jellybeans. After putting $$13$$ jellybeans in the thirteenth box, we have used $$78+13=91$$ jellybeans. There are $$9$$ jellybeans left, but we already have a box containing $$9$$ jellybeans. The remaining $$9$$ jellybeans must therefore be distributed among the existing boxes while maintaining the condition that no two boxes contain the same number of jellybeans.
One way to do this is to put the $$9$$ jellybeans in the last box which already contains $$13$$ jellybeans. This would mean that the final box would contain $$13+9=22$$ jellybeans. Another solution is to increase the number of jellybeans in each of the final nine boxes by one jellybean each. This solution is summarized in the following table.
Box Number $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ $$13$$ Number of Jellybeans $$1$$ $$2$$ $$3$$ $$4$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ $$13$$ $$14$$
Either way, the maximum number of boxes required is $$13$$.
If you had $$14$$ boxes, with the first box containing $$1$$ jellybean and each box after that containing one more jellybean than the box before, you would need $$1+2+3+\cdots+13+14=105$$ jellybeans, which is more than the number of jellybeans Ru has.
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## Introduction: Cowboy Duel Game
Welcome visitor!
I made this project for school(HKU Games&Interaction, Netherlands). With the idea to mix laser cut wood with technology(Arduino Yún) and make a small game out of it.
My inspiration was a small mini game from Kirby on the NES. Which has this cowboy duel mini-game. The idea is that two players play a game where they have to push a button at the last beep sound of a countdown. The player that pushes first shoots(his/her cowboy rotates his arm) and wins. They can play another game by pressing the reset button and then the countdown start again. It's just a simple entertaining game for kids and adults.
In the next part, I'll explain how it started and how I made this project.
## Step 1: Step 1: Setting Up the Arduino Yún.
So first to begin with a confession, I'm a total noob with programming. It's like reading Japanese. So I had a big challenge in front of me. First I wrote what I needed and what the Arduino needed to do and made a list for it.
Supplies
• It needs to have 3 buttons, 1 for each cowboy(2 in total) and 1 reset button.
• It needed 2 servo's to operate the arms.
• A buzzer to make a countdown sound.
• Resistors.
• Wires.
• A printed circuit board.
• And some LED for testing
I first needed to test if the button mechanic would work, so I tested it with some LED lights. To test if the who presses first would work. I made it so that the Arduino finds out who presses first and blocks the other button. Then the player who pressed first would see his LED would light up. After that, I began testing the servo's. They needed to turn 90 degrees so they would point at the other player when activated. I did encounter some troubles due to the right cowboy needed to be programmed in reverse. As an example; The left had to turn from 0 tot 90 so the other one had to turn from 90 to 0(both not the precise degrees though, used will be shown in code). After that I made a mechanic so the buzzer would do a countdown with 4 beeps and the players can press when the 4th beep is produced.
This is basically how it works. I'll show my code in a picture so fellow enthusiasts can use it as well.
## Step 2: Step 2: Laser Cutting
This is just a small step, but I'll talk about the laser cutting step. First of all, if you want to both laser cut and engrave the wood, you need to make each a different layer. I made the mistake and by making the file in Photoshop and import it to Illustrator. I do recommend using Illustrator right away. The line you want to laser cut just has to be a vector line and it needs to connect to each other. Everything that needs to be engraved has to be one layer of vector shapes. You can see the file I used in the pictures above(which I edited while laser cutting, so this one is correct). What I needed to be laser cut was;
• 2 cowboys, made 1 and mirrored the other.
• 2 cowboy arms, made 1 mirrored the other.
• 5 sides(the back is open), the top needed holes for the buttons and the cowboys(although something went wrong and had to glue them on).
You can see the results above in the pictures.
## Step 3: Step 3: Finnishing the Project
The last steps where to put it all together and make it work. I build in the Arduino and breadboard and installed the buttons and made sure the servo's were on the right side. Then I glued on the cowboys and attached the servo's on the right place. After that, I glued on the arms and began testing the rotation of the servo's. You can see the back of the cowboys and the inside of the box in the pictures. Unfortunately, I only got a small video due to my Arduino died on me last night(you can see it above this post).
I'm very happy with the results. It was defiantly an interesting project to work on. My uncle helped me understand the code and helped me build it. Although I liked this project I don't have any plans on working with the Arduino in the future, but laser cutting is something else. I still got some wood left and plan on doing something with it in the future. Thanks for reading my instructable and hope you liked it. Sorry for the bad English(if there is any).
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# Diagrams for categorical data
## Barplots
### Simulate data
set.seed(123)
dice <- sample(1:6, 100, replace=TRUE)
(dTab <- table(dice))
dice
1 2 3 4 5 6
19 15 18 11 16 21
### Simple barplot
barplot(dTab, ylim=c(0, 30), xlab="Result", ylab="N", col="black",
main="Absolute frequency")
barplot(prop.table(dTab), ylim=c(0, 0.3), xlab="Result",
ylab="relative frequency", col="gray50",
main="Relative frequency")
# not shown
### Barplots for contingency tables of two variables
#### Stacked barplot
roll1 <- dice[1:50]
roll2 <- dice[51:100]
rollAll <- rbind(table(roll1), table(roll2))
rownames(rollAll) <- c("first", "second"); rollAll
1 2 3 4 5 6
first 9 8 11 6 9 7
second 10 7 7 5 7 14
barplot(rollAll, beside=FALSE, legend.text=TRUE, xlab="Result", ylab="N",
main="Absolute frequency in two samples")
#### Grouped barplot
barplot(rollAll, beside=TRUE, ylim=c(0, 15), col=c("red", "green"),
legend.text=TRUE, xlab="Result", ylab="N",
main="Absolute frequency in two samples")
## Spineplot
N <- 100
age <- sample(18:45, N, replace=TRUE)
drinks <- c("beer", "red wine", "white wine")
pref <- factor(sample(drinks, N, replace=TRUE))
xRange <- round(range(age), -1) + c(-10, 10)
lims <- c(18, 25, 35, 45)
spineplot(x=age, y=pref, xlab="Age class", ylab="drink", breaks=lims,
main="Preferred drink by age class")
## Mosaic-plot
ageCls <- cut(age, breaks=lims, labels=LETTERS[1:(length(lims)-1)])
group <- factor(sample(letters[1:2], N, replace=TRUE))
cTab <- table(ageCls, pref, group)
mosaicplot(cTab, cex.axis=1)
## Pie-chart
Think hard about better alternatives, maybe a bar chart.
dice <- sample(1:6, 100, replace=TRUE)
dTab <- table(dice)
pie(dTab, col=c("blue", "red", "yellow", "pink", "green", "orange"),
main="Relative frequencies from rolling dice")
dTabFreq <- prop.table(dTab)
angles <- dTabFreq * 2 * pi
csAngles <- cumsum(angles)
csAngles <- csAngles - angles/2
text(x=textX, y=textY, labels=dTabFreq)
## Conditional density plot
N <- 100
X <- rnorm(N, 175, 7)
Y <- 0.5*X + rnorm(N, 0, 6)
Yfac <- cut(Y, breaks=c(-Inf, median(Y), Inf), labels=c("lo", "hi"))
myDf <- data.frame(X, Yfac)
cdplot(Yfac ~ X, data=myDf)
## Useful packages
More plot types for categorical data are available in packages vcd and vcdExtra.
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View the step-by-step solution to:
# '3 2 Surface Area and Volume 7 Directions: Type the correct answer in each box. Spell all words correctly, and use numerals instead of words for...
Can somebody please help me with this problem for my daughter and thank you? We have made several attempts at it and need help with locating the answer.
’3 2 Surface Area and Volume 7 Directions: Type the correct answer in each box. Spell all words correctly, and use numerals instead of words for
numbers. If necessary, use 1 for the fraction barls). 1
Sandra starts to pack cubes with side length i inch in o Rectangular Prism A. Then, she measures the dimensions of Rectangular Prism B. ___________ T 1% in ____________________
1 in
T_ 4 In
5-in
Rectangular Prism A Rectangular Prism B
Use the dragram to answer the duestlens,
1
Which rectangular prism wrll hold more cubes with side length i Inch?| Enter A or B
Which rectangular prism has a greater volume? Enter A or B,
1
How many more cubes can be lit into the larger prism than the smaller prism? cubes with side length i inch. @- 6 of 10 Answered Session Timer: 9:11 Session Score: 67% (416)
1) A will hold more 1/2 inch... View the full answer
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### Home > CC1MN > Chapter 5 Unit 5B > Lesson CC1: 5.3.3 > Problem5-92
5-92.
Find the area of the following triangles. Show all your work.
Two of these triangles make a parallelogram, what is the area of the parallelogram?
To find the area of a parallelogram, multiply the base by the height. What does the area of the triangle have to be?
What is the area of the rectangle at right? Use it to find the area of the original triangle.
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# The Unapologetic Mathematician
## Modules
The analogue in ring theory for the idea of a group action is that of a module. Again we want every element of the ring to behave like a function on a set and for multiplication to correspond to composition of functions, but now we have the addition floating around and we’d like to include it in the structure as well. We’ll handle this by letting the ring act on an abelian group.
So let’s take a ring $R$ and an abelian group $M$. We say that $M$ has the structure of a left $R$-module if each element $r\in R$ acts as a linear function from $M$ to itself. We write $r\cdot m$ for the effect of this function on an element $m\in M$. Linearity here means that $r\cdot(m_1+m_2)=r\cdot m_1+r\cdot m_2$. We also require that the ring structure play its part
• $(r_1+r_2)\cdot m=r_1\cdot m+r_2\cdot m$
• $(r_1r_2)\cdot m=r_1\cdot(r_2\cdot m)$
• $1\cdot m=m$ if the ring has an identity element $1$
We can say all this another way. Since this action of $R$ on $M$ is linear in both the ring and the abelian group we get a linear function from the tensor product $\alpha:R\otimes M\rightarrow M$ just like we had one for the multiplication of the ring: $\mu:R\otimes R\rightarrow R$. Not every such function will work, though. We also need that the following diagram commutes:
Around the top of the diagram we use the module action twice, while around the bottom we first multiply in the ring and then use the action once. If the ring has an identity we also require that the following diagram commute
where on the top we use the unique linea function sending the integer $1$ to the identity in $R$, and the diagonal is the canonical isomorphism from $\mathbb{Z}\otimes M$ to $M$.
And there’s another way to say it. Remember that every abelian group $M$ comes equipped with the endomorphism ring ${\rm End}(M)$. An $R$-module structure on $M$ is a ring homomorphism $R\rightarrow{\rm End}(M)$.
All three ways of defining a module — from the raw axioms, as a linear function $R\otimes M\rightarrow M$, or as a homomorphism $R\rightarrow{\rm End}(M)$ — are useful in various situations, and it’s important to be able to slide back and forth between the different pictures.
Of course, the fact that I said left $R$-module above should immediately lead you to think about right $R$-modules. These are the same, except that the module action is written $m\cdot r$ and satisfies $(m\cdot r_1)\cdot r_2=m\cdot(r_1r_2)$ — the order in which the factors are applied is reversed. Over a commutative ring we often ignore this distinction, since we can always switch the order of any product.
Every ring $R$ immediately has the structure of a left and a right $R$-module, using multiplication as the action. This is most clearly seen from the tensor-product definition, where the commuting square just expresses the associativity of ring multiplication. On the other hand, every abelian group $M$ immediately has the structure of an ${\rm End}(M)$-module. This is trivial in the homomorphism picture: just pick the identity homomorphism.
Somewhat more importantly, every abelian group is naturally a $\mathbb{Z}$-module. We define $n\cdot m=m+...+m$ — adding up $n$ copies of the group element $m$. Everything we say for modules in general will apply to abelian groups, and in fact much of what we’ve already said about abelian groups will extend to other modules.
At this point I want to make a point about language. The terms “abelian group” and “$\mathbb{Z}$-module” are interchangeable, but there are cases in which I feel the context clearly indicates using one or the other. Later we’ll come to (and many reader have already seen) “homology groups”, which are more naturally modules than groups. Similarly, the one-dimensional circle, torus, and real projective space are all semantically very different, even though they happen to be equivalent in many situations. I think that this imprecision can lead to confusion on the part of a student, so I’ll try my best to use the most apropriate of equivalent terms.
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April 21, 2007 - Posted by | Ring theory
## 10 Comments »
1. […] important to understand what sort of functions connect different instancees of the structure. For modules we have module […]
Pingback by Homomorphisms of modules « The Unapologetic Mathematician | April 23, 2007 | Reply
2. […] algebras as well, substituting “-module” for “abelian group”. An -module is a left -module if there is an -linear function , and a similar definition works for right […]
Pingback by Algebras « The Unapologetic Mathematician | May 8, 2007 | Reply
3. […] example: let be a right module over a ring , and be a left -module. Now for any abelian group we can consider the set of all […]
Pingback by Universal arrows and universal elements « The Unapologetic Mathematician | June 25, 2007 | Reply
4. […] of linear maps — homomorphisms — between abelian groups, and particularly between modules or vector spaces, which are just modules over a field. In particular we’ll focus on vector […]
Pingback by Linear Algebra « The Unapologetic Mathematician | May 19, 2008 | Reply
5. […] Well, it turns an algebra element into an endomorphism. And the most important thing about an endomorphism is that it does something to vectors. So given an algebra element , and a vector , we get a new vector . And this operation is -linear in both of its variables. So we have a linear map , built from the representation and the evaluation map . But this is just a left -module! […]
Pingback by Algebra Representations « The Unapologetic Mathematician | October 24, 2008 | Reply
6. […] check any algebra text or here for more information on […]
Pingback by Generalized HT90 « A Mind for Madness | December 12, 2008 | Reply
7. […] the context of graded algebras we also have graded modules. A -graded module over the -graded algebra can also be written down as a direct […]
Pingback by Graded Objects « The Unapologetic Mathematician | October 23, 2009 | Reply
8. […] A (left) module is like a group action for rings. Actually, UM mentions three equivalent […]
Pingback by Relearning algebra « Information Flow | August 15, 2010 | Reply
9. […] geometry: Lie algebras. These are like “regular” associative algebras in that we take a module (often a vector space) and define a bilinear operation on it. This much is covered at the top of […]
Pingback by Lie Algebras « The Unapologetic Mathematician | May 17, 2011 | Reply
10. […] already defined them over a year ago. So let’s pick up with a recap: a Lie algebra is a module — usually a vector space over a field — called and give it a bilinear operation which […]
Pingback by Lie Algebras Revisited « The Unapologetic Mathematician | August 6, 2012 | Reply
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# questions
plz finish the last four questions (10-13)
write the report for those 4 questions. Estimated length: 1.5 pages (excluding tables/figures).Make sure to explain intermediate steps of each calculation in your report and clearly indicate what parts refer to each of the individual questions of the case study.
Don't use plagiarized sources. Get Your Custom Essay on
questions
Just from \$13/Page
1.
1.Aftaer you merged both datasets as described in the Instructions: Compute the market capitalization as [abs(PRC) * SHROUT] for each stock in each month. Why are some prices negative? (You might need to lookup the definition of PRC in WRDS, for example with an internet search, to answer this question.)
Why can closing prices at time “t” be negative? [select every answer that you think is correct]:
10 points
Question 2
1. For each stock, compute the average excess return and the standard deviation over the period from January 2000 to December 2010 (to calculate such summary statistics the PivotTable is very useful). Which stock has the highest average excess return? (Reminder: excess return is the difference between the return and the risk-free rate)
10 points
Question 3
1. Now compute the Sharpe ratio for each stock over this period. Which stock looks most attractive to you in terms of the tradeoff between return and risk over this period?
10 points
Question 4
1. Do you observe any pattern regarding the Sharpe ratio of a stock and its average market cap (computed per PERMNO over the whole sample period)? Run a simple cross-sectional regression to check.
10 points
Question 5
1. Now run a formal CAPM (“market model”) time-series regression for each of the 100 firms (you can use the “LINEST” function in EXCEL to do so). Estimate the market model over the entire sample period. Which stock has the largest beta estimate? Insert the PERMNO of the stock with the largest Beta coefficient below.
10 points
Question 6
1. Make a scatter plot with on the vertical axis the historical average excess return of all stocks and on the horizontal axis the stock’s beta estimates over the full sample period. Also add a linear trend line, which is the security market line (SML). Does this plot support the CAPM predictions?
Indicate every correct statement from the list below.
10 points
Question 7
1. Please indicate the Coefficient of determination (R2) from the regression in Question 6 (i.e. the estimated SML) that is closest to what you find:
10 points
Question 8
1. Now run a cross-sectional regression. Explain the wholesample average excess returns of each stock by the wholesample average MARKETCAP of each stock and the estimated beta (from Question 5). What do you find? Assume statistical significance is indicated by a t-statistic below -2 or above 2. Select every answer that corresponds with your findings.
10 points
Question 9
1. Now run a so called Fama-MacBeth regression (2nd step), that is: each month regress excess returns on MarketCap only, then compute the average coefficient on Market Cap as well as the Fama-MacBeth t-statistic, which is avg(X)/[stddev(X)/sqrt(T)], where: X is the monthly estimated slope coefficient when explaining Returns by MarketCap and T is the number of observations (the number of months in the sample).
10 points
Question 10
1. As before, run a so called Fama-MacBeth regression and compute the Fama-MacBeth t-statistic. But this time explain Returns by MarketCap as of January for each year, i.e. in February to December you use the market cap estimated in January of each year for each stock.
10 points
Question 11
1. Of course, smaller stocks are also associated with higher risk. Hence, redo the Fama-MacBeth regressions, use the MarketCap as of January for each year, and the CAPM-beta (estimated over the whole sample) as a control variable, to explain monthly returns (as before).
10 points
Question 12
1. Are your results so far consistent with the data you received from Ken French’s website (using data from Jan-2000 to Dec-2010)?
10 points
Question 13
1.
Report the Fama-MacBeth test statistic, i.e. sqrt(N)*avg(X)/stddev(X), where N is the number of observations (the number of months), and X is the monthly estimated slope coefficient on MarketCap when explaining Returns by MarketCap and CAPM-Beta (i.e. the slope coefficients from the previous regression).
Round the value to two decimal digits, and use the dot to separate decimal from non-decimal digits, i.e. enter like:
12.23
Use all slope coefficients from 2005 (i.e. N=12).
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# A license plate has 3 ltrs & 3 digit in tht order. A witness
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A license plate has 3 ltrs & 3 digit in tht order. A witness to a hit & run accident saw the first 2 ltrs. & the last digit. If the letter & the digit can be repeated, how many number license plates must be checked by the police to find the culprit?
Also, pls. let me know how many problems involving statistics, probabilities, permutation & combination are in GMAT? If you have any question bank for this or others, pls. do forward.
thanks,
sheetal
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thanks,
SS
Last edited by sheetalsanjana on 30 Apr 2010, 14:10, edited 1 time in total.
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Kudos [?]: 70816 [1] , given: 9857
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30 Apr 2010, 14:04
1
KUDOS
Expert's post
sheetalsanjana wrote:
Pls help me with this problem:-
A license plate has 3 ltrs & 3 digit in tht order. A witness to a hit & run accident saw the first 2 ltrs. & the last digit. If the letter & the digit can be repeated, how many number license plates must be checked by the police to find the culprit?
Also, pls. let me know how many problems involving statistics, probabilities, permutation & combination are involved in GMAT? If you have any question bank for this, pls. do forward.
thanks,
sheetal
26 choices for the third letter;
10 choices for the first digit;
10 choices for the second digit.
26*10*10=2600.
P.S. I think at max 1-2 questions from each area (statistics, probabilities, permutation & combination).
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Posts: 14
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30 Apr 2010, 14:07
but why 10 choice for 1st & 10 choice for 2nd?
_________________
thanks,
SS
Math Expert
Joined: 02 Sep 2009
Posts: 33061
Followers: 5773
Kudos [?]: 70816 [0], given: 9857
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30 Apr 2010, 14:26
Expert's post
sheetalsanjana wrote:
but why 10 choice for 1st & 10 choice for 2nd?
Witness saw first two letters and last digit: LLX-YZD, LL and D is known, X, Y and Z are not.
X can be ANY letter from 26;
Y can be any digit from 10 (0,1,2,..9);
As digits on license can be repeated Z can also be any digit from 10.
Hence 26*10*10=2600.
Hope it's clear.
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30 Apr 2010, 23:02
Another way to think about it, but it gets the same answer:
Think of the 2 digits as any single two digit number. This would be 0-99, or 100 different possibilities.
You know for the missing letter there are 26 possibilities. Each one can be match with any of the 100 two digit numbers, so the answer is 26*100 or 2600.
I realize this may take a bit more time for some people, but I am still mixing up my combs/perms/probs formulas, so sometimes I benefit from working it out this way. Still studying!
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01 May 2010, 09:37
thanks you guys...got it
_________________
thanks,
SS
Re: combination problem [#permalink] 01 May 2010, 09:37
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Friday
February 12, 2016
Homework Help: calculus
Posted by Nolan on Monday, April 30, 2012 at 7:35pm.
3. Your neighbor puts tiger on his flat roof shed. You and another neighbor argue over the ht of the tiger. At a distance of 80 ft from the shed your angle of elevation from the ground level to the top of the shed is 5.74 degree. The angle of elevation to the top of the tiger from the same point is 13.37 degrees. To the nearest 10th foot how tall is the tiger
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Energy in Thermal System
ENERGY IN THERMAL SYSTEM OBJECTIVE 1. To gain thermal equilibrium. 2. To determine final temperature. INFERENCE Thermal
Citation preview
ENERGY IN THERMAL SYSTEM OBJECTIVE 1. To gain thermal equilibrium. 2. To determine final temperature. INFERENCE Thermal equilibrium is achieved when two substance reach the same temperature and exchange to heat energy. HYPOTHESIS The final temperature obtained from experiment will be the state in which the temperature of two substance reached the equal temperature at a certain time. APPARATUS 1. Calorimeter 2. Mercury thermometer 3. Beaker 4. Bunsen burner 5. Balance 6. Hot hand protector 7. Tripod stand and wire gauze 8. Lighter gun
1 1 2 1 1 1 1 1
PROCEDURE 1. Weight the mass of an empty beaker and record as . 2. Fill 200ml tap water into the beaker. Measure mass of the beaker with tap water and record as . 3. Read the initial temperature of tap water and record as . 4. Fill another empty beaker with 250ml water. Weight the mass of the beaker and record as . 5. Heat the beaker for 10 minutes. 6. Read the temperature of the hot water and record as as Figure 1. 7. Pour slowly cold water into the beaker containing hot water. 8. Using thermometer, stir slowly the mixture about 10 seconds. Record the final temperature as .
Figure 1
RESULT Table 1 Mass Empty beaker, Beaker with 200ml tap water, Beaker with 250ml hot water, Mass of tap water = = Mass of hot water = = Mass of water mixture =
Table 2 m (gram) 202.3
Temperature
388.7 424
Tap water,
28
Hot water,
60
The mixture,
45 186.4 g 221.7 g 408.1 g
+
CALCULATION Calculate
by using following equation:(
) (
(from experiment) (from calculation) % Error
(
) ) 45 44.8 0.4%
QUESTION 1. Does the final temperature in this experiment is equal with the final temperature in calculation? - No.
2. If both of the final temperatures are not same, explain why and relate to thermal equilibrium principle? -
CONCLUSION
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Skill 2D
Picture Problems: Understanding "How Many in All?"
The picture on the screen shows 3 big hamburgers and 2 little hamburgers. The question I want you to answer is - How Many in All?
But before you tell me the answer I want you to look at the number sentence below the hamburgers and tell me what operation sign (plus or minus) to put into the box to answer this problem correctly. So, what operation symbol goes in the box?
Here, we want to see if your student immediately knows to add when hearing the phrase
"how many in all" as well as understanding of operation symbols in a number sentence.
If there is hesitation regarding whether to add or subtract you should print and practice this skill.
Remember, along with understanding the words "how many in all," we want to determine if your student also understands the operation symbol for addition (as being a plus sign) and then being able to come to an answer from looking at the number sentence. If any one of these are wrong,
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# Association Analysis
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### Association Analysis
1. 1. Association Analysis<br />
2. 2. Association Analysis-Definition<br />Association Analysis is the task of uncovering relationships among data.<br />Association rules:<br />It is a model that identifies how the data items are associated with each other.<br />Ex:<br /> It is used in retail sales to identify that are frequently purchased together.<br />
3. 3. What is a rule? <br /><ul><li>Structure of rule:</li></ul>If (condition) then (result) <br />Example: IF a customer purchases coke, then the customer also purchases orange juice <br />The first part is the rule body and the second part is the rule head <br />
4. 4. Strength of a rule <br />How certain is the rule? <br />Confidence measures the certainty of a rule <br />It is the percentage of transactions containing all items stated in the condition that also contain the items in result <br />Confidence (A ,B) = P(B | A) <br />Example: The rule "If Coke then Oranje Juice" has a confidence of 100% <br />
5. 5. Strength of a rule <br />How often is the rule occurred? <br />Support measures the usefulness of a rule <br />It is the percentage of transactions that contains all items in the rule <br />Support (A , B) = P(A ,B) <br />Example: For the rule If Coke then Oranj juice <br />In all 5 transactions, 2 contains both coke and OJ <br />The support of the rule is 40% <br />
6. 6. Association Rule Mining<br />Two-step process <br />Find all frequent k-item sets, k=1, 2, 3, … <br />All items in a rule is referred as an itemset<br />Rules that contains k item forms a k-itemset<br />The occurrence frequency of an k-itemset is the number of transactions that contain all k items in the itemset<br />An itemset satisfies a minimum support (or minimum occurrence frequency) is called a frequent itemset<br />
7. 7. Association Rule Mining<br />2.Generate strong association rules from the frequent k-itemsets<br />Rules satisfy both a minimum support threshold and a minimum confidence threshold are called strong rules<br />
8. 8. Apriori Algorithm: Find all frequent k-item sets<br />Apriori principle:<br />If an itemset is frequent, then all of its subsets must also be frequent<br />
9. 9. Illustrating Apriori Principle<br />
10. 10. Apriori Algorithm<br />Method: <br />Let k=1<br />Generate frequent itemsets of length 1<br />Repeat until no new frequent itemsets are identified<br />Generate length (k+1) candidate itemsets from length k frequent itemsets<br />
11. 11. Contd…<br />Prune candidate itemsets containing subsets of length k that are infrequent <br />Count the support of each candidate by scanning the DB<br />Eliminate candidates that are infrequent, leaving only those that are frequent<br />
12. 12. Generate strong association rules from the frequent k-itemsets<br />For each frequent k-itemset, generate all non-empty subsets <br />Fore every nonempty subset, generate the rule and the associated confidence <br />Output the rule if the minimum confidence threshold is satisfied <br />
13. 13. Multilevel association rules<br />Difficult to find strong associations at very low or primitive levels of data <br /> <br />Few people may buy "IBM desktop computer" and "Sony b/w printer" together <br />Many people may purchase "computer" and "printer" together <br />
14. 14. Concept hierarchy<br />defines a sequence of mappings from a set of low level concepts to higher level<br />EX: IBM <br /> Microsoft Hp<br /> ………<br /> computer software printer accessory <br />
15. 15. Steps to be followed<br />Top-down, progressive deepening approach <br />First mine high-level frequent items <br />Then mine their lower level frequent items and so on <br />At each level, Apriori algorithm is used <br />Use uniform minimum support for all levels, or <br />Use reduced minimum support at lower levels <br />
16. 16. Sequential Association Rule <br />Concerns sequences of events <br />New homeowners purchase shower curtains before purchasing furniture <br />When a customer goes into a bank branch and ask for an account reconciliation, there is a good chance that he or she will close all his or her accounts <br />
17. 17. Sequential Association Rule <br />Transaction must have two additional features: <br />a time stamp or sequencing information to determine when transactions occurred relative to each other <br />identifying information, such as account number or id number <br />
18. 18. Some important parameters <br />Duration <br />duration may be the entire available sequence in the database, or a user selected subsequence, such as year 1999 <br />Event folding window <br />a set of events occurring within a specified period of time, such as within the same day, can be viewed as occurring together.<br />
19. 19. Some important parameters <br />Interval <br />between events in the discovered pattern <br />0 interval means to find strictly consecutive sequences <br />min_int <= interval <= max_int means to find patterns that are separated by at least min_int at most max_int<br />interval = c, to find patterns carrying an exact interval <br />
20. 20. Some Practical Issues <br />Time window of transactions <br />Level of aggregation <br />Level of support and confidence <br />
21. 21. Time window of transactions <br />Select a time window for the transaction covers at least 2 product cycles <br />e.g. customer purchases a product with a frequency of six month or less, select a 12-month window of customer transaction data <br />For frequently purchased products, a short time window is sufficient <br />For low frequency items, a longer time window is necessary.<br />
22. 22. Level of aggregation <br />If product codes in the data are too specific (such as based on product details such as size and flavour), few associations will be discovered <br />Group products into categories according to the product hierarchy or create new level manually <br />
23. 23. Level of support and confidence <br />Start with a high support and gradually reduce it <br />Set confidence to around 50% to reduce the number of permutation <br />
24. 24. Conclusion<br />Association analysis rules such as multidimensional and sequential association rules are studied.<br />Apriori algorithm is described in detail<br />Various practical issues in association rules are analyzed.<br />
25. 25. Visit more self help tutorials<br />Pick a tutorial of your choice and browse through it at your own pace.<br />The tutorials section is free, self-guiding and will not involve any additional support.<br />Visit us at www.dataminingtools.net<br />
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## B.B.A. (Sem.-1) BUSINESS MATHEMATICS Subject Code : BB-102 Paper ID : [C0202]
• Friday, March 01, 2013
punjabtechnicaluniversity.blogspot.in
Roll No........................ Total No. of Pages : 03
Total No. of Questions : 07
B.B.A. (Sem.-1) BUSINESS MATHEMATICS Subject Code : BB-102 Paper ID : [C0202]
Time : 3 Hrs. Max. Marks : 60
INSTRUCTION TO CANDIDATES :
SECTION-A (10
x 2 = 20 Marks) l. (a) If five times the 5th term of an A.P. is equal to six times the 6th term, show that 11th term is zero.
1. SECTION-A is COMPULSORY.
2. Attempt any FOUR questions from SECTION-B
show that 11th term is zero.
(b) Which term of the series 18 - 12 + 8 ....... is 512/729?
(c) A sum amounts to Rs. 8820 in two years and Rs. 9261 in three years. Find the rate of compound interest.
75 5 32
(d) Prove that log ---- - 2 • log -- + log --- = log 2.
16 9 243
(e)Find the value of ' r' if the coefficients of (2r + 4)th and (r - 2)th terms in the expansion of (1 + x)18 are equal.
(f) Find the domain of the function ________ \______
(* -1)(2 " x)
(g) Evaluate the limit, Lt (1 - 4x)x .
x— 0
(h) Prove that (52logx5) = 2x, x > 0. Dx
(i) Find the values of x for which the function f (x) = |7x-3| is maximum/ minimum.
(j) Find the values of'm' for which the equation
x2 - 2x (1 + 3m) + 7 (3 + 2m) = 0 has equal roots.
SECTION-B (4 x 10 = 40 Marks)
2. (a) Using Cramar's Rule/^vethe following system of equations for x, y
1 -1+1=10
x y z ^^^^
xy
(b) Solve the given equations for x, y and z by Gauss Elimination method.
x + 2y - z = 6
3x - y - 2z = 3
4x + 3y + z = 9 (5,5)
3. (a) Discuss the continuity of the function
x > 3
(b) Sum the series 7 + 77 + 777 + ... to n terms. (5,5)
4. (a) There are 15 points in a plane, no three of which are in a straight line excepting four, which are collinear. Find the number of (/) straight lines (//) triangles formed by joining them.
(b) Find the term independent of 'x' in the expansion of
r 3 x2 - — '9
2x 3x
5. (a) Prove the logical expression p
(b) Find — where y " ax + xa + aa + xx - log x. (5,5)
6. (a) y = x x + (1 + x)x , find .
(b) Define a set.
State and prove De Morgan's Laws. *V^^ (5,5)
2x + 3 V<V
7. (a) If y " f(x) " 7—-, prove that f(y) " x.
(b) If xjl + y + y .ijl + x = 0,
Show that d- " - (i + x)-2. (5,5)
[A-12]
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