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Liono4ka [1.6K]

3 years ago

9

4. Consider the mailbox shown below. It consists of a square prism for the main pole, a rectangular prism for the main box and h

alf a cylinder for top portion. You are going to provide the dimensions for each of these parts. When you provide the dimensions, they must meet the following criteria:
Use these dimensions to determine the overall volume of the mailbox. This should include both the pole, main box and cylindrical top portion. If necessary, use 3.14 for π and round your answer to the nearest hundredths.

1 answer:

IrinaK [193]3 years ago

4 0

Answer:

Too many answers could be correct

Step-by-step explanation:

This one is hard because it will depend on what numbers you pick for each measurement.

-----------------------------------------------------------------------------

if a = 1, b = 2, c = 3, d = 4, and e = 5

find the volume for all the shapes

The post

V = lwh = (1)(1)(2) = 2

The bottom of the mailbox

V = lwh = (3)(4)(5) = 60

The top of the mailbox

V = πr²h = (3)(5)π = 15π = 48.2

Then add them all together:

2 + 60 + 48.2 = 110.2 is the total Volume

You might be interested in

A circular clock face has a radius of 3.5 inches. What is the area of the clock face? Round to the nearest tenth. Use 3.14 for p

gtnhenbr [62]

The area is about 316 in2
Step-by-step Divide 63 by π and by 2 to determine the radius of the clock. 63÷2÷π≈10. To find the area of the face of the clock multiply π by 102

4 0

3 years ago

13 points if you can answer the problem AND teach me how to answer the problem

Harman [31]

<h3>Answer:</h3>

Y = 3X + 12

<h3>Explanation:</h3>

You are given several points on the line, so you can use a couple of them to find the slope, then write the equation in point-slope form and rearrange it to the desired form.

... slope = (change in Y)/(change in X)

Using the first two points, this is ...

... slope = (69 -51)/(19 -13) = 18/6 = 3

<em>Point-Slope Form</em>

The point-slope form of the equation of a line is usually written ...

... y -y1 = m(x -x1) . . . . . . where m = slope, (x1, y1) = point

This can be put into a y= form by adding y1:

... y = m(x -x1) +y1

<em>Your Equation</em>

Filling in m=3, (x1, y1) = (13, 51), the equation becomes

... y = 3(x -13) +51

... y = 3x -39 +51 . . . . . eliminate parentheses

... y = 3x +12 . . . . . . . . . simplify to slope-intercept form

_____

<em>Alternate Solution</em>

Having found the slope to be 3, you can write the slope-intercept form and fill in what you know.

... y = mx + b . . . . . slope-intercept form

... 51 = 3·13 +b . . . . slope-intercept form with slope and first point filled in

... 51 -39 = b = 12 . . . . equation solved for <em>b</em> by subtracting 39

Now you know that the equation is

... y = 3x +12

3 0

3 years ago

Can yall help me please?

eduard

Answer:

5/6 - 2/5 = 13/30

Step-by-step explanation:

6 0

3 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

3 years ago

Jenny multiplies the square root of her favorite positive integer by $\sqrt{2}$. Her product is an integer. a) Name three number

Dmitry [639]

Answer:

Part a) $2,50,18$

Part b) When Jenny divides the square root of her favorite positive integer by $\sqrt{2}$ , she gets an integer

Step-by-step explanation:

Let

x-------> the favorite positive integer

Part a)

1) For $x=2$

$\sqrt{2}*\sqrt{2}=\sqrt{4}=2$ -----> the product is an integer

so

The number $x=2$ could be Jenny favorite positive integer

2) For $x=50$

$\sqrt{50}*\sqrt{2}=\sqrt{100}=10$ -----> the product is an integer

so

The number $x=50$ could be Jenny favorite positive integer

3) For $x=18$

$\sqrt{18}*\sqrt{2}=\sqrt{36}=6$ -----> the product is an integer

so

The number $x=18$ could be Jenny favorite positive integer

Part B)

1) For $x=2$

$\sqrt{2}/\sqrt{2}=\sqrt{1}=1$ -----> the result is an integer

2) For $x=50$

$\sqrt{50}/\sqrt{2}=\sqrt{25}=5$ -----> the result is an integer

3) For $x=18$

$\sqrt{18}/\sqrt{2}=\sqrt{9}=3$ -----> the result is an integer

Therefore

When Jenny divides the square root of her favorite positive integer by $\sqrt{2}$ , she gets an integer

4 0

3 years ago