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dsp73
3 years ago
6

Someone please answer this and try hard. I selected the highest amount of points to give. It was 90, but it doesn't always come

out as 90. Thanks.

Mathematics
2 answers:
azamat3 years ago
5 0
Volume of a cylinder = PI x r^2 x H

r = 4 so r^2 = 4^2 = 16

h = 3

Volume = 3.14 x 16 x 3 = 150.72 ft^3

 it doesn't say how to round the answer so you may need to round it

Naily [24]3 years ago
5 0
Yea, thats the right answer 
VV
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Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th
zalisa [80]
We have the following equations:

(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

y \leq -0.5x+5 \\ y \leq -1.25x+8

Region B.

Let's take a point that is in this region, that is:

P(0,6)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5  \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\  \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8

So the inequalities are:

(1) \ y  \geq  -0.5x+5 \\  (2) \ y  \leq  -1.25x+8

Region C.

A point in this region is:

P(0,10)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \  \ 8

So the inequalities are:

(1) \ y  \geq  -0.5x+5 \\ (2) \ y  \geq  -1.25x+8

Region D.

A point in this region is:

P(8,0)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \  \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2

So the inequalities are:

(1) \ y  \leq  -0.5x+5 \\ (2) \ y  \geq  -1.25x+8

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\  \\ Solving \ for \ y: \\ y=-0.5(4)+5=3

Accordingly, the point is:

\boxed{p(4,3)}

Point q:

This point is the x-intercept of the line:

y=-0.5x+5

So let:

y=0

Then

x=\frac{5}{0.5}=10

Therefore, the point is:

\boxed{q(10,0)}

Part b) 

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

O(0,0)

Substituting in the inequalities:

y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}

It is true.

Region B.

Let's take a point that is in this region, that is:

P(0,7)

Substituting in the inequalities:

y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y  \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}

It is true

Region C.

Let's take a point that is in this region, that is:

P(0,11)

Substituting in the inequalities:

y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}

It is true

Region D.

Let's take a point that is in this region, that is:

P(9,0)

Substituting in the inequalities:

y  \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}

It is true

7 0
3 years ago
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