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FromTheMoon [43]
3 years ago
10

Can some one please help me i need it or i will fail the class please...

Mathematics
2 answers:
Oliga [24]3 years ago
7 0
1)
x^-3y^-12 / x^15y^-15
answer is D. last choice

2)
answer is <B'C'D' (third choice)

4)
equivalent 3/9 = 0.33333(3 repeat)
answer is B. second choice

5)
1.7 x 3.5 = 5.95
10^-13 x 10^25 = 10^12
so 
= 5.95 x10^12

answer is D. last choice


frozen [14]3 years ago
5 0
 the frist page D next page is C the thrid page is B and fouth page is D
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The sum of two consecutive odd numbers is 88 find the numbers​
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Answer:

43,45

Step-by-step explanation:

43+45=88

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What is the difference between a theorem and an axion ?​
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Find two consecutive odd integers whose product is 143
Dennis_Churaev [7]
Two consecutive odd integers whose product is 143 are 11 and 13.
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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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