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3 years ago

7a+(8b) +(-15a)+9b + (-4c)+8=

2 answers:

8 0

-8a+17b-4c+8

steps:
when there is a "+" in front of the parenthesis the expression remains the same.

7a+8b-15a+9b-4c+8

Collect the like terms

7a+8b-15a+9b-4c+8

7a and 15a
8b and 9b

15a-7a= -8a
9b+8b= 17b

Final answer:

-8a+17b-4c+8

Arte-miy333 [17]3 years ago

6 0

7a+(8b) +(-15a)+9b + (-4c)+8

7a+8b-15a+9b-4c+8 ( changing some of the symbols)

7a-15a+8b+9b-4c+8

=-8a+17b-4c+8

Answer: -8a+17b-4c+8

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In the coordinate plane three vertices of rectangle ABCD are A(0,0), B(0,a) and D(b,0). What are the coordinates of point C?

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2 years ago

A quantity P is an exponential function of time I, such that P = 160 when t = 6 and P = 150 when I = 4. Use the given informatio

Klio2033 [76]

Answer:

k = 0.032
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Step-by-step explanation:

Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...

$P(t)=P_0\cdot e^{kt}$

We know from the given points that we can write the equation as ...

$P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t$

Comparing this to the desired form, we see that ...

$P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693$

So, the approximate equation for P is ...

$P(t)=131.836\cdote^{0.032t}$

And the parameters of interest are ...

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3 years ago

Read 2 more answers

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3 years ago