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Shkiper50 [21]
3 years ago
11

A problem states that Ursula earns $9 per hour. To write an expression that tells how much money Ursula earns h hours, Joshua wr

ote (fraction) 9/h. Sarah wrote 9h. Whose expression is correct and why?
Mathematics
2 answers:
ohaa [14]3 years ago
8 0
$9 per hour 
the answer is 9h because if you tried to find out how much she earns after 10 hours you would do 9*10=90 not the other one
wariber [46]3 years ago
3 0
Sarah, because it tells how much money she earns each hour.
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32x2 – 24x = 0<br> What’s the answer to this?
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Answer:

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Step-by-step explanation:

32x^2 – 24x = 0

Factor out 8x

8x(4x -3)

Using the zero product property

8x = 0   4x-3 =0

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<h3>The answer is 11</h3>

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C is the answer you are looking for.
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4 years ago
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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
6. Find the percent of decrease from 24 to 18.
In-s [12.5K]

Answer:

25%

Step-by-step explanation:

24-18= 6

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<u>24</u>

4 0
3 years ago
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