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3 years ago

IS THIS GROWTH OR DECAY

Mathematics

1 answer:

Maksim231197 [3]3 years ago

4 0

Answer:

It is a growth.

Step-by-step explanation:

If the number in parenthesis (exponential base) is greater than 1, then it is a growth. If it is between 1 and 0, then it is a decay.

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-3m - 4 = -28 please help it’s missing work and it was due a week ago

Sladkaya [172]

Answer:

m = 8

Step-by-step explanation:

-3m - 4 = -28

+4 +4

-3m = -24

÷-3 ÷-3

m = 8

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2 years ago

Factor the GCF: 12a3b + 8a2b2 − 20ab3

Harrizon [31]

<span>12a^3b + 8a^2b^2 − 20ab^3

</span>12a^3b = 4ab(3a^2)
8a^2b^2 = 4ab(2ab)
20ab^3 = 4ab(5b^2)

GCF = 4ab

12a3b + 8a2b2 − 20ab3 = 4ab(3a^2 + 2ab - 5b^2)

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3 years ago

Read 2 more answers

Identify the lengths of sides m and n.

juin [17]

Answer:

m = 26

n = 13

Step-by-step explanation:

Use the sine rule

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3 years ago

What is the slope of the line shown below?

liraira [26]

Answer:

C) 3

Step-by-step explanation:

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2 years ago

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A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago