1....1/4 y + 1 1/2 + 2-1 3/4 y -12 2....5=1/5x-3 3....1/2 h + 1 from 3/4 h +4

At 10 A.M the outside temperature was -5 F. By noon, the temperature was -17 F. Ricardo said that it was getting warmer outside.

Natasha_Volkova [10]

Ricardo is not correct because the temperature is actually getting colder (idk if I need to explain it more). The difference is -12 F

8 0

3 years ago

Read 2 more answers

A zoo train ride costs $3 per adult and $1 per child. On a certain day, the total number of adults (a) and children (c) who took

bearhunter [10]

$a+c=30$
$3a+c=50$
$c=30-a$
$3a+30-a=50$
$2a=20$
$a=10$
$a+c=30$
$10+c=30$
$c=20$

10 Adults and 20 Children

8 0

2 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

You have seventy-nine $10 chips. The chips have a value of $ ??

rewona [7]

1 chip is $10
So, for 79 chips,
= 79 x $10
= $790

5 0

2 years ago

A rectangle has a width of 12.5m and a length of 105m. What is the effect on the perimeter when the dimensions are multiplied by

vladimir1956 [14]

It is two times bigger

7 0

2 years ago