Ask question

Ask question

All categories

1 year ago

15

A proportional relationship is graphed on the coordinate plane. The line that represents the relationship passes through points

(0, 0) and (5, 10).
What is the unit rate of the proportional relationship?

1 answer:

Marrrta [24]1 year ago

4 0

Answer: I think its D. 1:5

Step-by-step explanation:

but im not surr

You might be interested in

Hannah spends $4.50 each day to buy lunch at her school. After school each day, she buys the same snack. Hannah spends $28.75 fo

Masja [62]

Answer:

The solution x=1.25 represents the cost of snacks per day.

Step-by-step explanation:

7 0

1 year ago

A plumber charges a $45 fee to make a house call and then $25 for each hour of labor. The plumber uses an equation in the form e

sasho [114]

M=slope
In this case the slope would be $25
y=mx+b
y=25x+45

Answer: m=25

5 0

2 years ago

Read 2 more answers

Tom supports North Storm Football Club. At the match between North Storm and West Stanford there were 71,167 spectators. If 41,1

dangina [55]

Answer:
30,058 spectator

Explanation:
The total number of spectators is equal to the sum of West Stanford's and North Storm's supporters.
We are given that:
Total number of spectators = <span>71,167 spectator
North Storm spectators = </span><span>41,109 spectator
So, to get the number of West Stanford spectators, all we have to do is subtract North Storm spectators from the total spectators as follows:
West Stanford spectators = </span>71,167 - 41,109 = 30,058 spectator

Hope this helps :)

8 0

1 year ago

Consider a single spin of the spinner.

alex41 [277]

Answer:

"landing on a shaded portion and landing on a 3"

"landing on an unshaded portion and landing on a number less than 2 "

Step-by-step explanation:

Mutually exclusive means the events will have no intersection.

Let's look at your first choice:

"landing on a shaded portion and landing on an even number"

Landing on a shaded portion would be 1 or 4.

Landing on an even number would be 2 or 4.

There is an intersection (they contain a common element), the 4.

These events are not mutually exclusive.

Let's look at your second choice:

"landing on a shaded portion and landing on a number greater than 3"

Landing on a shaded portion would be 1 or 4.

Landing on a number greater than 3 would be just 4.

There is an intersection; they both contain 4.

These events are not mutually exclusive.

Let's look at your third choice:

"landing on a shaded portion and landing on a 3"

Landing on a shaded portion would be 1 or 4.

Landing on 3 would just be 3.

There is no common elements in the lists listed. These events have no intersection.

These events are mutually exclusive.

Let's look at your fourth choice:

"landing on an unshaded portion and landing on an odd number"

Landing on a unshaded portion would be 2 or 3.

Landing on an odd number would be 1 or 3.

There is an intersection; they both have 3 in common.

These events are not mutually exclusive.

Let's look at your fifth choice:

"landing on an unshaded portion and landing on a number less than 2 "

Landing on an unshaded portion would 2 or 3.

Landing on a number less than 2 would be 1.

There is no intersection.

These events are mutually exclusive.

6 0

2 years ago

Read 2 more answers

A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) <u>Find the volume before the storm</u>

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

<u>Part b)</u> Estimate the number of lane miles that were covered with salt

5) <u>Find the amount of salt that was used during the storm</u>

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) <u>Find the pounds of road salt used</u>

$104,358.59*80=8,348,687.2\ pounds$

7) <u>Find the number of lane miles that were covered with salt</u>

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) <u>Find the pounds of road salt </u>

$4,013.79*80=321,103.20\ pounds$

9) <u>Find the number of lane miles </u>

$321,103.20/350=917.44 \ lane\ miles$

therefore

<u>the answer part c) is</u>

the number of lane miles is $917 \ lane\ miles$

7 0

2 years ago