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3 years ago

Use the points (1,-3) and (5,9) to determine the slope between the two points using the slope formula

1 answer:

5 0

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{9}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{9-(-3)}{5-1}\implies \cfrac{9+3}{4}\implies \cfrac{12}{4}\implies 3$

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Answer:

A) 12 miles per inch

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3 years ago

What is the y-intercept and x-intercept of 5x-y=35

Veronika [31]

First lets put this into a nicer format
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4 years ago

15a+65b<150 Please help

son4ous [18]

Answer:a<−13/3

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Step-by-step explanation:

Step 1: Add -65b to both sides.

15a+65b+−65b<150+−65b

15a<−65b+150

Step 2: Divide both sides by 15.

15a

−65b+150

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b+10

Answer:

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3 years ago

Which expression is equivalent to -2 1/4÷ (-2/3)

sergejj [24]

They asked which was equivalent not whats the value of the expression -_-

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3 years ago

Read 2 more answers

Elena bikes 20 minutes each day for exercise. Write an equation to describe the relationship between her distance in miles, D, a

kompoz [17]

Answer:

The equation to describe the relationship between Elena distance:

a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

Solution:

Given, Elena bikes 20 minutes each day for exercise.

We have to write an equation to describe the relationship between her distance in miles, D, and her biking speed, in miles per hour,

We know that, distance travelled = speed $\times$ time

a. At a constant speed of 13 miles per hour for the entire 20 minutes

Her speed is 13 miles per hour.

Then, distance D miles = 13 miles per hour $\times$ 20 minutes

$\mathrm{D}=13 \text { miles per hour } \times \frac{20}{60} \text { hours } \rightarrow \mathrm{d}=13 \times \frac{1}{3} \rightarrow \mathrm{d}=4.34 \text { miles approximately. }$

b. At a constant speed of 15 minutes per hour for the first 5 minutes, then at 12 miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with 15 mph + distance travelled with 12 mph

$\begin{array}{l}{\mathrm{D}=15 \mathrm{mph} \times 5 \text { minutes }+12 \mathrm{mph} \times 15 \text { minutes }} \\\\ {\mathrm{D}=15 \mathrm{mph} \times \frac{5}{60} \text { hours }+12 \mathrm{mph} \times \frac{15}{60} \text { minutes }} \\\\ {\mathrm{D}=15 \times \frac{1}{12}+12 \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{5}{4}+3 \rightarrow \mathrm{D}=3+1.25 \rightarrow \mathrm{D}=4.25 \mathrm{miles}}\end{array}$

c. At a constant speed of M miles per hour for the first 5 minutes, then at N miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with M mph + distance travelled with N mph

$D = M mph \times 5 minutes + N mph \times 15 minutes$

$\mathrm{D}=\mathrm{M} \mathrm{mph} \times \frac{5}{60} \text { hours }+\mathrm{N} \mathrm{mph} \times \frac{15}{60} \text { minutes }$

$\mathrm{D}=\mathrm{M} \times \frac{1}{12}+\mathrm{N} \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{N}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{3 N}{12} \rightarrow 12 \mathrm{D}=\mathrm{M}+3 \mathrm{N} \text { miles }$

Hence, a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

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4 years ago