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Levart [38]
3 years ago
8

2x – yl. when x = 3 and y = 10​

Mathematics
1 answer:
iragen [17]3 years ago
5 0

Answer:

Step-by-step explanation:

Here you go mate

Step 1

2x-y  Equation/Question

Step 2

2x-y  Evaluate for x and y

2(3)-10

Step 3

2(3)-10  Simplify

6-10

answer

-4

Hope this helps

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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
  • P0 = 131.836

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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4 0
3 years ago
What is the geometric mean of 5 and 8?
Lisa [10]

Answer:

About 6.325.

Step-by-step explanation:

The geometric mean between two numbers is the smaller number multiplied by the square root of the quotient of the two numbers. In this case, the square root of 8/5 multiplied by 5 is about 6.325. Hope this helps!

7 0
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Answer: Roughly 111.97

Step-by-step explanation:

1. 39 divided by 3 is 13.  

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3. 1455.66 divided by 13 is roughly 111.97 (Again, this does not distribute evenly)

4. To check your work, multiply 111.97 times 13 and you should get somewhere around 1455.66 which is (4367 divided by 3).

Hopefully this helps! Feel free to mark brainliest! :)

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Find the length of the radius. Type a numerical value in the space provided. Do not type spaces in your answers,
IgorC [24]
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