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pishuonlain [190]
3 years ago
13

Find the distance between (-3,-1) and (4, 2). (Round up to the nearest hundredths.)

Mathematics
1 answer:
yanalaym [24]3 years ago
4 0

Answer:

The answer is

<h2>7.61 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

<h3>d =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  }  \\</h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-3,-1) and (4, 2).

The distance is

d =  \sqrt{ ({ -  3 - 4})^{2}  +  ({ - 1 - 2})^{2} }  \\  =  \sqrt{ ({  - 7})^{2}  +  ({ -   3 })^{2} }  \\  =  \sqrt{49 + 9}  \\  =  \sqrt{58}  \\  = 7.615773

We have the final answer as

<h3>7.61 units to the nearest hundredth</h3>

Hope this helps you

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Luke bought 256256256 ounces of strawberries. He divided them evenly between 888 different pies. How many pounds of strawberries
SIZIF [17.4K]

Answer:

2 pounds.

Step-by-step explanation:

We have been given that Luke bought 256 ounces of strawberries. He divided them evenly between 8 different pies. We are asked to find the pounds of strawberries that Luke put on each pie.

First of all, we will convert 256 ounces into pounds.

We know that 1 pound equals 16 ounces. To convert 256 ounces into pounds, we need to divide 256 by 16.

\text{256 ounces}=\frac{256}{16}\text{ pounds}=16\text{ pounds}

Now we will divide 16 pounds by 8 to find number of strawberries put on each pie.

\text{Pounds of strawberries on each pie}=\frac{16}{8}

\text{Pounds of strawberries on each pie}=2

Therefore, Luke put 2 pounds strawberries on each pie.

4 0
3 years ago
To make 5 Apple pies you need about 2 pounds of apples. How many pounds of apples do you need to make 20 Apple pies
ollegr [7]
You will need 40 pounds of apples
5 0
3 years ago
Read 2 more answers
2. You are given the hyperbolic relation modeled by xy = -2. Do the following: a) Rewrite the relation such that the dependent v
N76 [4]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The given equation is :

  • xy =  - 2

1. The relationship such that dependent variable (y) is isolated is :

  • y =   \dfrac{ - 2}{x}

2. The table accompanying this equation :

  • x = -4 and y = 0.5

  • x = -1 and y = 2

  • x = - 0.5 and y = 4

  • x = 0 and y = undefined

  • x = 0.5 and y = -4

  • x = 1 and y = -2

  • x = 4 and y = -0.5

3. graph of the given equation is in attachment ~

4 0
3 years ago
3-18 even problems algebra 1 math
Tresset [83]

The answers for problems 13-18 are -9, 8, -2, 5, -2, and 15 after equating the same functions.

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

It is given that:

From problems 13 - 18, some functions are shown,

13) h(x) = -7x, h(x) = 63

Equate both the function:

-7x = 63

x = -9

14) t(x) = 3x, t(x) = 24

Equate both the function:

3x = 24

x = 8

15) m(x) = 4x + 15, m(x) = 7

4x + 15 = 7

4x = -8

x = -2

16) 6x - 12 = 18

x = 5

17) 1/2 x - 3 = -4

1/2 x = -1

x = -2

18) j(x) = -4/5 x + 7, j(x) = -5

-4/5 x + 7 = -5

-4/5 x = -12

x = 15

Thus, the answers for problems 13-18 are -9, 8, -2, 5, -2, and 15 after equating the same functions.

Learn more about the function here:

brainly.com/question/5245372

#SPJ1

7 0
1 year ago
Denise and Jacqueline are training for a swimming and running race.
bezimeni [28]

Answer:

Jacqueline runs at 6.25 minutes per mile

Step-by-step explanation:

Here we have that the total training time for Jacqueline = 54 minutes

The time for which Jacqueline swims = 30 minutes

Therefore, the time in which Jacqueline runs = 54 min. - 30 min. = 20 minutes

Therefore the equation for Jacqueline's speed = Distance/Time we have

Jacqueline's pace = 3.2 miles/20 minutes = 0.16 miles/minute

Therefore Jacqueline's pace = 0.16 miles/minute

The equation for Jacqueline's pace in miles/minute is presented as follows

Jacqueline's pace in miles/minute = Time of running/Distance

Jacqueline's pace in miles/minute  = 1/0.16 miles/minute = 6.25 minutes/mile.

5 0
3 years ago
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