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3 years ago

How do I get y equals by itself in the equation 2x+5y=20

Mathematics

2 answers:

Likurg_2 [28]3 years ago

6 0

Answer:

Subtract 2x to bring it to the other side

Step-by-step explanation:

2x+5y=20

-2x. -2x

5y=-2x+20

ruslelena [56]3 years ago

3 0

Answer: subtract 2x from that side to get 5y= 20-2x

Step-by-step explanation:

You subtract to get 5y = 20-2x. If you want y completely by itself divide it all by 5. 5y divides by 5, 20 divided by 5 and -2x by 5.

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Simplify -5 to the 2nd power, + 8 |-1| + (-3).

madreJ [45]

Answer:

-19

Step-by-step explanation:

7 0

3 years ago

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A box contains plain pencils and pens. A second box contains color pencils and crayons. One item from each box is chosen at rand

pochemuha

Answer:

1/2

Step-by-step explanation:

its 1 out of 2 chances

6 0

3 years ago

What is the constant of proportionality in the equation y=2xy

dybincka [34]

Answer:

2 is the constant of proportionality in the equation y = 2x .

Step-by-step explanation:

Definition of constant of proportionality

When two variables are directly proportional to each others .

Let us assume that u and v .

u \propto vu∝v

Than the equation becomes u= kv

Where k is called the constant of proportionality .

Thus in the question x and y are proportional variables .

i.e

y \propto xy∝x

y = kx

Where k is called the constant of proportionality .

Compare the equation y = kx with y=2x .

Thus

k = 2

Therefore 2 is the constant of proportionality in the equation y = 2x

the explanation is not mine only the answer

6 0

2 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

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

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

$\tan(x+y)$ goes with $\sqrt{3}-2$

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:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ 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.

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

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

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

Divide top and bottom by 4 to reduce fraction:

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

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

Distribute:

$\sqrt{3}-2$

6 0

3 years ago

15 POINTS<br> 1.) name two planes that intersect in WX in the figure to the right

SOVA2 [1]

The correct answer is C

6 0

3 years ago