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erik [133]
3 years ago
7

Find the average value of the function i = 15(1 - e1/2t) from t = 0 to t = 4. A. 7.5(1 + e-2) B. 7.5(2 - e-2) C. 7.5(2 + e-2) D.

7.5(3 - e-2)
Mathematics
1 answer:
klasskru [66]3 years ago
7 0
I think the average value of the function i = 15(1 - e1/2t) from t = 0 to t = 4 is <span>C. 7.5(2 + e-2). The correct answer is letter C.</span>
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Dmitrij [34]

Answer:

x=18

Step-by-step explanation:

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(3x+20)=(5x-16)

20+16=5x-3x

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Two identical masses are moving down a slope one masses traveling twice as fast as the other mass how does the kinetic energy of
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Answer:

One mass's kinetic energy is 4 times the other.

Step-by-step explanation:

The formula for kinetic energy is KE=\frac{1}{2}mv^2

Where

KE is the kinetic energy

m is the mass

v is the velocity

  • Now, let mass of the first be m (second is identical so mass of 2nd is m as well)
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<u>KE of 1st mass is:</u>

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Hence, the kinetic energy of the 2nd mass is 4 times the first.

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3 years ago
D/dx [tan^-1 (x)] find the derivative
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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle
Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

(x-h)^{2} +(y-k)^{2}=r^{2}

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x+3)^{2} +(y-4)^{2}=r^{2}

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

r=\sqrt{(2-4)^{2}+(-6+3)^{2}}

r=\sqrt{(-2)^{2}+(-3)^{2}}

r=\sqrt{13}\ units

The equation of the circle is equal to

(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}

(x+3)^{2} +(y-4)^{2}=13

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

(10+3)^{2} +(4-4)^{2}=13

(13)^{2} +(0)^{2}=13

169=13 -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x-1)^{2} +(y-3)^{2}=r^{2}

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

r=\sqrt{(6-3)^{2}+(2-1)^{2}}

r=\sqrt{(3)^{2}+(1)^{2}}

r=\sqrt{10}\ units

The equation of the circle is equal to

(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}

(x-1)^{2} +(y-3)^{2}=10

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

(11-1)^{2} +(5-3)^{2}=10

(10)^{2} +(2)^{2}=10

104=10 -----> is not true

therefore

The point is not on the circle

The statement is false

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