Someone help me?Please

How to find average value of a function over a given interval?

Strike441 [17]

f(x)=8x−6f(x)=8x-6 , [0,3][0,3]

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.(−∞,∞)(-∞,∞){x|x∈R}{x|x∈ℝ}f(x)f(x) is continuous on [0,3][0,3].f(x)f(x) is continuousThe average value of function ff over the interval [a,b][a,b] is defined as A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dx.A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dxSubstitute the actual values into the formula for the average value of a function.A(x)=13−0(∫308x−6dx)A(x)=13-0(∫038x-6dx)Since integration is linear, the integral of 8x−68x-6 with respect to xx is ∫308xdx+∫30−6dx∫038xdx+∫03-6dx.A(x)=13−0(∫308xdx+∫30−6dx)A(x)=13-0(∫038xdx+∫03-6dx)Since 88 is constant with respect to xx, the integral of 8x8x with respect to xx is 8∫30xdx8∫03xdx.A(x)=13−0(8∫30xdx+∫30−6dx)A(x)=13-0(8∫03xdx+∫03-6dx)By the Power Rule, the integral of xx with respect to xx is 12x212x2.A(x)=13−0(8(12x2]30)+∫30−6dx)

3 0

3 years ago

Find the perimeter of the triangle with vertices A(negative 2,2,3), B(4,negative 4,3), and C(7,6,4).

gayaneshka [121]

Answer:

$P\approx 28.872\,\text{units}$

Step-by-step explanation:

We are given coordinates of each of the vertices of the triangle ABC, hence we can use the distance formula to find the lengths of each of the side.

the distance formula is generally written as:

$r = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

for the side AB: A(-2,2,3) and B(4,-4,3)

$AB = \sqrt{(-2-4)^2+(2-(-4))^2+(3-3)^2}$

$AB = \sqrt{(-6)^2+(6)^2+(3-3)^2}$

$AB = \sqrt{72}$

for the side BC: B(4,-4,3) and C(7,6,4)

$BC = \sqrt{(4-7)^2+((-4)-6)^2+(3-4)^2}$

$BC = \sqrt{110}$

for the side CA: C(7,6,4) and A(-2,2,3)

$CA = \sqrt{(7-(-2))^2+(6-2)^2+(4-3)^2}$

$CA = \sqrt{98}$

Now we all the sidelengths:

to find the perimeter we need to just sum the three side lengths:

$P = AB + BC + CA$

$P = \sqrt{72} + \sqrt{110} + \sqrt{98}$

$P\approx 28.872\,\text{units}$

this is the perimeter of triangle ABC

6 0

3 years ago

Yusra's dog is overweight. The vet says the dog needs to lose 7 pounds to get down to the desirable weight of 45 pounds. How muc

irga5000 [103]

Answer:

the dog weighs 52lb

7 0

2 years ago

one rectangle measures 2 units by 7 units. a second rectangle measures 11 units by 37 units. are these two figures scaled versio

Amanda [17]

Answer: The both triangle are not scale version to each other.

Explanation:

It is given that the one rectangle measures 2 units by 7 units. a second rectangle measures 11 units by 37 units.

The two shapes are scaled version of each other if their corresponding sides are in same proportion. In other words the if two shapes scaled version of each then the shapes are similar.

The sides or first rectangle are 2 and 7. The measures of second rectangle are 11 and 37.

The proportion of small side is,

$\frac{2}{11}$

The proportion of large side is,

$\frac{7}{37}$

So,

$\frac{2}{11}\neq \frac{7}{37}$

Since the proportion are not equal, therefore both triangle are not scale version to each other.

6 0

3 years ago

Use the Pythagorean Theorem to find the distance to the nearest tenth, between F(9, 5) and G(-2, 2). (Hint: Place F and G on the

artcher [175]

Put a point H on (9, 2). Sketch a triangle out of the three points. Distance between (-2, 2) and (9, 2) is going to be 11. Distance between (9, 2) and (9, 5) is going to be 3. These correspond to the a and b of the Pythagorean Theorem
c^2=a^2+b^2
c=√11^2+3^2=√130
Square root of 130 is 11.4

6 0

3 years ago