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3 years ago

Use the distributive property to remove the parentheses.

Mathematics

1 answer:

nlexa [21]3 years ago

8 0

Answer:

Step-by-step $12c^{3} -40c^{8}$ explanation:

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Please help!

Ganezh [65]

Answer:

The measures of ∠BOC is 25.4° and that of ∠AOB is 15.8°.

Step-by-step explanation:

Given that,

∠AOC = 31°

∠BOC = 2x+10

∠AOB = 4x-15

As we can seen from the figure that,

∠AOC = ∠AOB + ∠BOC

Put all the values,

31 = 4x-15+2x+10

31 = 6x-15

31+15 = 6x

46 = 6x

x = 7.7

∠BOC = 2(7.7)+10 = 25.4°

∠AOB = 4(7.7)-15 = 15.8°

So, the measures of ∠BOC is 25.4° and that of ∠AOB is 15.8°.

4 0

3 years ago

Ama deposited her babysitting money into her savings account, which already had a

Natali5045456 [20]

Answer:

295= 210+X

Step-by-step explanation:

If she has a total of 295 dollars deposited and she had 210 before then you could just subtract 295-210 to find how much she deposited. The answer would be that she deposited 85 dollars in her savings account and the equation is listed above.

4 0

3 years ago

Plss helo with this question!!! i will mark brainliest!

pychu [463]

Answer:

the answer for 2 is 10, 4 is 20, 15 is 3

Explanation:

1:5=1x5

2:10=2x5 and so on

7 0

3 years ago

Read 2 more answers

Jivesh also has a more powerful Model B rocket. For this rocket, he uses the equation h=-490t^2+1260t. When is the height of the

kkurt [141]

Answer:

1.29 s

Step-by-step explanation:

h = -490t² + 1260t

810 = -490t² + 1260t

490t² - 1260t + 810 = 0

49t² - 126t + 81 = 0

(7t - 9)² = 0

7t - 9 = 0

t = 9/7

t ≈ 1.29

4 0

3 years ago

Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B

telo118 [61]

In order to prove

$\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)$

Let's write both sides in terms of $\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)$ only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$

and

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

So, their multiplication is

$\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)$

So, the left hand side simplifies to

$\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now, on with the right hand side. We have

$1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now simply make this expression one fraction:

$1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

And as you can see, the two sides are equal.

6 0

4 years ago