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allochka39001 [22]
4 years ago
14

Can someone help me with these two questions

Mathematics
1 answer:
GenaCL600 [577]4 years ago
5 0
7. 16: 3/8*x+1=5, 3/8*x=6, x=16
8.  -32: 7-1/4*x=15, -1/4*x=8, x=-32
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I have been stuck on the problem for so long. please help me:
Xelga [282]

Answer:

D-2

Step-by-step explanation:

You can use elimination to get the value of y. Move numbers to one side and x and ys to the other side. To use elimination, multiply the first equation with 2. You will get 2x+2y=24. add it with the equation below. y will be eliminated and you will get the value of x. X=10. Plug in x and get the y value. Y=2

6 0
3 years ago
9) Simplify 5^12 ÷ 5^3 A. 5^4 B. 5^9 C. 1^4 D. 1^9
Mashcka [7]
When you divide powers, you subtract them.

12-3=9

B. 5^9 is the answer.

I hope this helps!
~kaikers
5 0
3 years ago
I need help plz. Its DeMoivres Theorem
vagabundo [1.1K]

\bf \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ z=\stackrel{a}{-1}\stackrel{b}{-\sqrt{3}~i}~~ \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right)\\[-0.5em] \hrulefill\\ r=\sqrt{(-1)^2+(\sqrt{3})^2}\\ \qquad \sqrt{4}\\ \qquad 2\\ \theta =tan^{-1}\left( \frac{-\sqrt{3}}{-1} \right)\\\\ \qquad \frac{4\pi }{3} \end{cases}

\bf z=2\left[ cos\left( \frac{4\pi }{3} \right) -i~sin\left( \frac{4\pi }{3} \right)\right]\implies z^6=\left[ 2\left[ cos\left( \frac{4\pi }{3} \right) -i~sin\left( \frac{4\pi }{3} \right)\right] \right]^6 \\\\\\ z^6=2^6\left[ cos\left( 6\cdot \frac{4\pi }{3} \right) -i~sin\left( 6\cdot \frac{4\pi }{3} \right)\right]\implies z^6=64[cos(8\pi )-i~sin(8\pi )] \\\\\\ z^6=64[(-1)-i~(0)]\implies z^6=\stackrel{\stackrel{a}{\downarrow }}{-64}~~\stackrel{\stackrel{b}{\downarrow }}{-0i}

4 0
4 years ago
When a certain prescription drug is taken orally by an adult, the amount of the drug (in mg/L) in the bloodstream at t hours is
castortr0y [4]

Answer:

1. y = f(8)

2. So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.

Step-by-step explanation:

The concentration of the drug in the bloodstream at t hours is:

y = f(t)

1. What is the concentration of the drug in the bloodstream at t= 8 hours?

At t hours, y = f(t)

So at 8 hours, y = f(8)

2. During what time interval is the drug level in the bloodstream increasing? Decreasing?

A function f(t) is increasing when

f'(t) > 0

And is decreasing when

f'(t) < 0

So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.

5 0
3 years ago
List all the pairs of adjacent sides in the quadrilateral
34kurt
I just answered this question for someone else. You should really look before you ask. 

x w z y
6 0
3 years ago
Read 2 more answers
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