Given the function, f(x)= 5(x+4)^3/2, which of the following represents its y-intercept

Please help with all 3 questions.

mixas84 [53]

Answer:

1. *2/multiplied by 2, since each number gets higher by *2.

2. *3/multiplied by 3, since each number gets higher by *3

3. ÷2/divided by 2, since each number gets lower by ÷2.

5 0

1 year ago

Quadrilateral ACEG ≅ Quadrilateral NMRP

3241004551 [841]

You line the segments up in order from the first quad to the second.
The only one that lines up is AG and NP the first option.

4 0

2 years ago

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B) The number of volume of a cylinder is half of its number of the total surface area.

faust18 [17]

Answer:

$V=179.6cm^3$

Step-by-step explanation:

the volume of a cylinder is given by:

$V=\pi r^2h$

where r is the radius and h is the height.

and the surface area:

$SA=2\pi r^2+2\pi rh$

the first term is the area of the circles and the second term is the area of the body.

since "The number of volume of a cylinder is half of its number of the total surface area." we will have that:

$V=\frac{1}{2}SA$

substitutig the equivalent expressions on each side:

$\pi r^2 h = \frac{1}{2} (2\pi r^2+2\pi r h)$

and we simplify and solve for the height (since is the value we don't know of the cylinder):

$\pi r^2 h = \pi r^2+\pi r h\\\pi r^2h-\pi rh=\pi r^2\\h(\pi r^2-\pi r)=\pi r^2\\h=\pi r^2/(\pi r^2-\pi r)$

we substitute the value of the radius $r=7cm$ , and we get:

$h=\pi (7cm)^2/(\pi (7cm)^2-\pi (7cm))\\h=153.938/(153.938-21.991)\\h=153.938/131.947\\h=1.1666cm$

thus the volume is:

$V=\pi r^2h$

$V=\pi (7cm)^2(1.1666cm)\\V=179.6cm^3$

3 0

2 years ago

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100 points please help

Eva8 [605]

V= b x h = 5 x 18 = 90ft

6 0

2 years ago

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a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$

$\sigma = 3.5795$

<em>See attachment for result of 1-VarStats</em>

8 0

2 years ago