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3 years ago

Finding Slope On a coordinate plane, a line goes through points (0, 1) and (4, 2). What is the slope of the line? m =

Mathematics

1 answer:

ivolga24 [154]3 years ago

7 0

Answer:

slope = $\frac{1}{4}$

Step-by-step explanation:

Calculate the slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (0, 1) and (x₂, y₂ ) = (4, 2)

m = $\frac{2-1}{4-0}$ = $\frac{1}{4}$

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Determine whether each of the following products is rational or irrational.

ankoles [38]

Answer:

-irrational

-rational

-irrational

__________________________________________________________quick and easy way is to see if the squares are perfect squares, like square root seven is irrational, so the answer is irrational. square root 9 is 3, which is rational, so its rational.

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3 years ago

1. Find 1/8 x 2/3 *

mash [69]

1. is 1/12. 2. is 1 3/7

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3 years ago

7 + 54 / 3 (2) Solve the problem

vampirchik [111]

Answer:

Step-by-step explanation:

7 + 54/3*2

simplify the fraction

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7 0

4 years ago

Read 2 more answers

Solve AABC. Round your answers to the nearest hundredth, if necessary

Aloiza [94]

Answer:

$C=25^{\circ},\\a\approx 10.72,\\b\approx 11.83$

Step-by-step explanation:

The sum of the interior angles of a triangle is 180 degrees. Thus, angle C must be $180-90-65=25^{\circ}$ .

In any triangle, the Law of Sines is given by $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Therefore, we have:

$\frac{\sin 90^{\circ}}{b}=\frac{\sin 25^{\circ}}{5},\\\\b=\frac{5\sin90^{\circ}}{\sin 25^{\circ}}=11.8310079158\approx \boxed{11.83}$

$\frac{a}{\sin 65^{\circ}}=\frac{5}{\sin25^{\circ}},\\a=\frac{5\sin 65^{\circ}}{\sin25^{\circ}}=10.7225346025\approx \boxed{10.72}$

8 0

3 years ago

Below is the five-number summary for 144 hikers who recently completed the John Muir Trail (JMT). The variable is the amount of

bearhunter [10]

Answer:

The IQR is given by:

$IQR = Q_3 -Q_1 = 28-18= 10$

If we want to find any possible outliers we can use the following formulas for the limits:

$Lower= Q_1 - 1.5 IQR$

$Upper= Q_3 + 1.5 IQR$

And if we find the lower limt we got:

$Lower= Q_1 - 1.5 IQR= 18-1.5*10 = 3$

$Upper= Q_3 + 1.5 IQR= 28+1.5*10= 43$

So then the left boundary for this case would be 3 days

Step-by-step explanation:

For this case we have the following 5 number summary from the data of 144 values:

Minimum: 9 days

Q1: 18 days

Median: 21 days

Q3: 28 days

Maximum: 56 days

The IQR is given by:

$IQR = Q_3 -Q_1 = 28-18= 10$

If we want to find any possible outliers we can use the following formulas for the limits:

$Lower= Q_1 - 1.5 IQR$

$Upper= Q_3 + 1.5 IQR$

And if we find the lower limt we got:

$Lower= Q_1 - 1.5 IQR= 18-1.5*10 = 3$

$Upper= Q_3 + 1.5 IQR= 28+1.5*10= 43$

So then the left boundary for this case would be 3 days

6 0

3 years ago