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3 years ago

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A set of problems on Algebra, Geometry, and Combinatorics consists of 52, 70, and 88 problems. If 2 more problems from each cate

gory were added to the set, what is the largest number of tests that can be made containing an equal amount of problems from each category?

2 answers:

Talja [164]3 years ago

6 0

Anwser: 36
explanation: i take geometry so ik

MatroZZZ [7]3 years ago

4 0

Answer:

The largest number of tests that can be made is 36

Step-by-step explanation:

Let us list the information

Algebra consists of 52 problems
Geometry consists of 70 problems
Combinatorics consists of 88 problems
2 more problems from each category were added to the set

That means each category will increase by 2

∴ Algebra = 52 + 2 = 54

∴ Geometry = 70 + 2 = 72

∴ Combinatorics = 88 + 2 = 90

We need to find the largest number of test that can be made containing an equal amount of problem

∵ 54 = 6 × 9

∵ 72 = 6 × 12

∵ 90 = 6 × 15

6 is the number of problem in each test because we need an equal amount of problems in each category and the other numbers are the number of tests in each category

∴ The largest number of Algebra tests is 9

∴ The largest number of Geometry tests is 12

∴ The largest number of Combinatorics test is 15

Add them to find the largest number of tests that can be made

∵ 9 + 12 + 15 = 36

∴ The largest number of tests that can be made is 36

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Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

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Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

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$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

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$-(2-\sqrt{3})$

Distribute:

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