All categories

3 years ago

A line has the points (3,7) and (5,19) on it. What is the slope of that line?

Mathematics

1 answer:

Volgvan3 years ago

6 0

Answer:

C) 6

Step-by-step explanation:

19 - 7 = 12

5 - 3 = 2

12/2 = 6

hope this helps :) please mark brainliest if you see fit :)

You might be interested in

I'll give brainliest

Igoryamba

Answer:

7 oof

Step-by-step explanation:

7 0

3 years ago

An algebra test contains 38 problems. Some of the problems are worth 2 points each. The rest of the questions are worth 3 points

Naily [24]

Let 'x' be the problems worth 2 points.

Let 'y' be the problems worth 3 points.

Since, there are 38 total problems.

So, $x+y =38$ (equation 1)

x = 38-y

Since, a perfect score is 100 points.

So, $2x+3y = 100$ (equation 2)

Substituting the value of 'x', we get

$2(38-y)+3y=100$

$76-2y+3y=100$

$76+y = 100$

y = 24

x+y = 38

x = 38-24 = 14

So, 14 problems are worth 2 points and 24 problems are worth 3 points.

5 0

3 years ago

Read 2 more answers

Marvin can do one load of laundry with 1/16 box of laundry detergent. How many loads of laundry can he wash with 4 boxes of dete

Tcecarenko [31]

Answer:

64 loads of laundry, your welcome

3 0

3 years ago

Read 2 more answers

Consider the linear expression 3x -15

Jobisdone [24]

Answer:

In linear expresiion 3 is the coefficient

X will be the variable and -15 will be the constant term.

5 0

3 years ago

If one root is the square of the other in the equation 8+mx+27x²=0,find the value of m

aleksklad [387]

Answer:

The value of $m$ is -30.

Step-by-step explanation:

Let $27\cdot x^{2}+m\cdot x +8 = 0$ , we notice that expression is a second order polynomials. We can rearrange the polynomial and use factorization to calculate all roots:

$x^{2}+\frac{m}{27}\cdot x + \frac{8}{27} = 0$ (1)

$-x^{2}-x = \frac{m}{27}$ (2)

$x^{3} = \frac{8}{27}$ (3)

From (3) we find the least root:

$x = \sqrt [3]{\frac{8}{27}}$

$x = \frac{2}{3}$

By (2) we have the value of $m$ :

$m = -27\cdot (x^{2}+x)$

$m = -30$

3 0

3 years ago