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2 years ago

12

The expression below represents the cost, c, of an item that is on sale for 25% off. c-0.25c what equation is equivalent ?

2 answers:

alexira [117]2 years ago

5 0

Answer:

The answer is 0.75c.

Step-by-step explanation:

i just had the question and got it correct!

hope it helps :)

Tcecarenko [31]2 years ago

3 0

Another way of solving for the price when the item is on sale for 25% off is to subtract the rate from 100% and multiply it by C.

So for this problem, 100% - 25% = 75%.

The equivalent equation will be 0.75c.

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Given: EL- tangent, EK- secant <br> Prove: EJ·LK = EL·LJ

drek231 [11]

Answer:

This is possible.

Step-by-step explanation:

We can say that m<E=m<E, because of the Reflexive Property

Then, we have angles JKL and ELJ, which are equal through the peripheral angle theorem.

With these two angles, we can say that triangles ELK and EJL are similar, by the Angle-Angle Postulate (AA).

Then we can create this ratio through the Corresponding Parts of Similar Triangles Theorem, (CPST), $\frac{LK}{LJ} =\frac{EL}{EJ}$ .

With this ratio, we can cross multiply to get the desired result

$EJ$ · $LK=EL$ · $LJ$

Hope this helps with your RSM problem

Yup, i caught ya.

3 0

2 years ago

5. Anthony has 80 pine tree seedlings to plant in his meadow. He first plants one row of 12

inna [77]

D. 16 clusters
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7 0

1 year ago

Show and explain how replacing one equation by the sum of that equation and a multiple of the other produces a system with the s

telo118 [61]

Answer:

(1/2, 5)

Step-by-step explanation:

8x + 7y = 39

-2(4x - 14y = -68)

-8x + 28y = 136

8x + 7y = 39

add the two equations together

35y = 175

y = 5

8x + 7(5) = 39

8x = 4

x = 1/2

3 0

1 year ago

What logarithmic function represents the data in the table?

alina1380 [7]

216 =6^3
1,296 = 6^4
7,776 = 6^5
x = 6^f(x)
log x = log 6^f(x) = f(x)log 6
f(x) = log x / log 6 = $log_6x$

8 0

2 years ago

Read 2 more answers

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

2 years ago