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Oksi-84 [34.3K]
3 years ago
11

28 ft2 at $3.89 per square foot $71.98 $111.44 $108.92 $106.40

Mathematics
1 answer:
Alinara [238K]3 years ago
4 0
The answer would be $108.92. You can solve this by multiplying @* by 3.89, getting 108.92.
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A news site offers a subscription that costs $28.50 for 6 months. What is the unit price per month.
crimeas [40]
Divide the total cost by the amount of months
28.5 divided by 6=4.75
7 0
3 years ago
Axel follows these steps to divide 40 by 9: Start with 40. 1. Divide by 9 and write down the remainder. 2. Write a zero after th
yarga [219]
4 r4
9⬜️40
-36
——
4
“Divide 40 by 9” means 40÷9
Axel did 9÷40


9 goes outside the dog house (⬜️)
9 can’t go into 4 so move to the next number
9 goes into 40 4 times.
9 times 4 equals 36
40 minus 36 equals 4
There are no other numbers under the house can’t go into four and 9 can’t go into 4. That means there is a remained of four.
7 0
3 years ago
The points (2, 4) and (6, 7) lie on a line. What is the slope of the line?
fgiga [73]

Answer: 1.5

Step-by-step explanation:

=7-4/6-4

=3/2

=1.5

5 0
2 years ago
!!!!!!!!!!!! Help please
SOVA2 [1]

Answer:

Yes it is a solution

Step-by-step explanation:

2*0=0

3*2=6

4 0
3 years ago
Read 2 more answers
If sinx = p and cosx = 4, work out the following forms :<br><br><br>​
Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16} $

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

\boxed{\sin(-x)=-\sin(x)}

\boxed{\cos(-x)=\cos(x)}

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

\sin(x)=p

\cos(x)=4

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]

Using the sum and difference identities:

\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}

\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}

\sin(\pi -x) = \sin(x)

\sin(2\pi +x)=\sin(x)

\cos(2\pi-x)=\cos(x)

Therefore,

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]

\implies [p+4] \cdot [p \cdot 4]=4p^2+16p

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16} $

8 0
3 years ago
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