PLEASE HELP WITH QUESTION C

Chris and his brother leave the same point and run in opposite directions on a straight road. After running for 3 Hours they are

ikadub [295]

Chris' average speed is 3.5mph: Lets take Chris' brother speed = B then Chris speed = (B+ 1) mph-------given in the question and: The average speed for covering 18 miles= chris' average speed + his brother's average speed = (B+ 1) + B since Speed = Distance covered / time taken then average speed for 18 miles can also be found by: 18 miles/3 hours = 6 miles per hour then we have; (B+ 1) + B = 6mph (B +1) = (6-B) B+B = 6-1=5 2B=5 B=5/2=2.5 mph (this is Chris' brother average speed) Hence Chris' average speed is = B+1= 2.5+1= 3.5mph

8 0

3 years ago

what is the standard deviation of the data set given below 2 3 6 9 10 A: 10 B: square root of 10 C: square root of 12.5 D: squar

AleksAgata [21]

The standard deviation of the set of values (2, 3, 6, 9, 10) is square root of 12.5 ( $\sqrt{12.5}$ ) therefore OPTION C

6 0

3 years ago

Read 2 more answers

Which of the following is in standard form of quadratic?

shutvik [7]

Answer:

I think C and D options can be correct

Step-by-step explanation:

PLEASE MARK ME BRAINLIEST IF MY ANSWER IS CORRECT PLEASE

7 0

2 years ago

Which shows equivalent expressions a^2+3b-2a^2=3^a+3b-4a^2 a^2+3b-2a^2=3^a+4b-4a^2 -a^2+3b-2a^2=3a^2+3b -a^2+3b+2a^2=3a^+4b-2a^2

scoray [572]

I don’t think this is formatted correctly, but I think it may be the 2nd one.

6 0

3 years ago

You are designing a miniature golf course and need to calculate the surface area and volume of many of the objects that will be

laiz [17]

a. To solve the first part, we are going to use the formula for the surface area of a sphere: $A=4 \pi r^2$
where
$A$ is the surface area of the sphere
$r$ is the radius of the sphere
We know from our problem that $r=5ft$ ; so lets replace that value in our formula:
$A=4 \pi (5ft)^2$
$A=314.16ft^2$

To solve the second part, we are going to use the formula for the volume of a sphere: $V= \frac{4}{3} \pi r^3$
Where
$V$ is the volume of the sphere
$r$ is the radius
We know form our problem that $r=5ft$ , so lets replace that in our formula:
$V= \frac{4}{3} \pi (5ft)^3$
$V=523.6ft^3$

We can conclude that the surface area of the sphere is 314.16 square feet and its volume is 523.6 cubic feet.

b. To solve the first part, we are going to use the formula for the surface area of a square pyramid: $A=a^2+2a \sqrt{ \frac{a^2}{4} +h^2}$
where
$A$ is the surface area
$a$ is the measure of the base
$h$ is the height of the pyramid
We know form our problem that $a=8ft$ and $h=12ft$ , so lets replace those value sin our formula:
$A=(8ft)^2+2(8ft) \sqrt{ \frac{(8ft)^2}{4} +(12ft)^2}$
$A=266.39ft^2$

To solve the second part, we are going to use the formula for the volume of a square pyramid: $V=a^2 \frac{h}{3}$
where
$V$ is the volume
$a$ is the measure of the base
$h$ is the height of the pyramid
We know form our problem that $a=8ft$ and $h=12ft$ , so lets replace those value sin our formula:
$V=(8ft)^2 \frac{(12ft)}{3}$
$V=256ft^3$

We can conclude that the surface area of our pyramid is 266.39 square feet and its volume is 256 cubic feet.

c. To solve the first part, we are going to use the formula for the surface area of a circular cone: $A= \pi r(r+ \sqrt{h^2+r^2}$
where
$A$ is the surface area
$r$ is the radius of the circular base
$h$ is the height of the cone
We know form our problem that $r=5ft$ and $h=8ft$ , so lets replace those values in our formula:
$A= \pi (5ft)[(5ft)+ \sqrt{(8ft)^2+(5ft)^2}]$
$A=226.73ft^2$

To solve the second part, we are going to use the formula for the volume os a circular cone: $V= \pi r^2 \frac{h}{3}$
where
$V$ is the volume
$r$ is the radius of the circular base
$h$ is the height of the cone
We know form our problem that $r=5ft$ and $h=8ft$ , so lets replace those values in our formula:
$V= \pi (5ft)^2 \frac{(8ft)}{3}$
$V=209.44ft^3$

We can conclude that the surface area of our cone is 226.73 square feet and its surface area is 209.44 cubic feet.

d. To solve the first part, we are going to use the formula for the surface area of a rectangular prism: $A=2(wl+hl+hw)$
where
$A$ is the surface area
$w$ is the width
$l$ is the length
$h$ is the height
We know from our problem that $w=6ft$ , $l=10ft$ , and $h=16ft$ , so lets replace those values in our formula:
$A=2[(6ft)(10ft)+(16ft)(10ft)+(16ft)(6ft)]$
$A=632ft^2$

To solve the second part, we are going to use the formula for the volume of a rectangular prism: $V=whl$
where
$V$ is the volume
$w$ is the width
$l$ is the length
$h$ is the height
We know from our problem that $w=6ft$ , $l=10ft$ , and $h=16ft$ , so lets replace those values in our formula:
$V=(6ft)(16ft)(10ft)$
$V=960ft^3$

We can conclude that the surface area of our solid is 632 square feet and its volume is 960 cubic feet.

e. Remember that a face of a polygon is a side of polygon.
- A sphere has no faces.
- A square pyramid has 5 faces.
- A cone has 1 face.
- A rectangular prism has 6 faces.
Total faces: 5 + 1 + 6 = 12 faces

We can conclude that there are 12 faces in on the four geometric shapes on the holes.

f. Remember that an edge is a line segment on the boundary of the polygon.
- A sphere has no edges.
- A cone has no edges.
- A rectangular pyramid has 8 edges.
- A rectangular prism has 12 edges.
Total edges: 8 + 20 = 20 edges

Since we have 20 edges in total, we can conclude that your boss will need 20 brackets on the four shapes.

g. Remember that the vertices are the corner points of a polygon.
- A sphere has no vertices.
- A cone has no vertices.
- A rectangular pyramid has 5 vertices.
- A rectangular prism has 8 vertices.
Total vertices: 5 + 8 = 13 vertices

We can conclude that there are 0 vertices for the sphere and the cone; there are 5 vertices for the pyramid, and there are are 8 vertices for the solid (rectangular prism). We can also conclude that your boss will need 13 brackets for the vertices of the four figures.

7 0

3 years ago