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3 years ago

6

If 8.45% life of each mobile phone is used in a day by a typical user for which mobile phone is 2.1 hours of battery life used i

n a day?
A)20

B)25

C)10

D) 18

1 answer:

mamaluj [8]3 years ago

7 0

8.45% = 0.0845

0.0845(25) = 2.1...ur answer is 25

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Is it possible to factor x cubed?

liberstina [14]

I dont think so.
factoring means make simpler
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3 years ago

Y= -0.25x + 4.7<br> Y=4.9x - 1.64

Andreas93 [3]

Answer:

The answer would be (1.23, 4.39)

Step-by-step explanation:

Because they are both equal to y, we can set the equations equal to each other and then solve.

4.9x - 1.64 = -0.25x + 4.7

5.15x - 1.64 = 4.7

5.15x = 6.34

x = 1.23

Now that we have the value for x, we can plug into either equation and find y.

y = -0.25x + 4.7

y = -0.25(1.23) + 4.7

y = -.31 + 4.7

y = 4.39

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3 years ago

(w - 3)2 = -5w - 4 + 9w<br><br>

alukav5142 [94]

Answer:

-1

Step-by-step explanation:

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3 years ago

Which inverse operation should be performed first?: m = 7 = 32.

Mumz [18]

The answer is multiplication

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3 years ago

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