All categories

3 years ago

What equation would have the points containing (8,3) and (13,5)

Mathematics

2 answers:

vichka [17]3 years ago

8 0

Answer:

y = (2/5)x -1/5

Step-by-step explanation:

Find the slope between the two points using the slope formula:

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

$m = \frac{5 - 3}{13 - 8}$

$m = \frac{2}{5}$

Insert the slope and a random point, (8,3) into the equation for a line:

y = mx + b

3 = (2/5)8 + b

3 = 16/5 + b

b = -1/5

Insert m=2/5 and b = -1/5 into the equation of a line:

y = (2/5)x -1/5

Novay_Z [31]3 years ago

7 0

The answer would be y= 2/5x because 5-3 =2 so you put that on top and 13-8 =5 and you put that on bottom

You might be interested in

. 18 convert to a fully reduced fraction

Taya2010 [7]

Answer:

9/50 im pretty sure

Step-by-step explanation:

you should vote me brainliest pleaseeeeeeeeeeeee

4 0

3 years ago

Read 2 more answers

The width of a container is 5 feet less than its height. Its length is 1 foot longer than its height. The volume of the containe

juin [17]

The height of container is 8 feet

Solution:

Let "w" be the width of container

Let "l" be the length of container

Let "h" be the height of container

The width of a container is 5 feet less than its height

Therefore,

width = height - 5

w = h - 5 ------ eqn 1

Its length is 1 foot longer than its height

length = 1 + height

l = 1 + h ---------- eqn 2

The volume of container is given as:

$v = length \times width \times height$

Given that volume of the container is 216 cubic feet

$216 = l \times w \times h$

Substitute eqn 1 and eqn 2 in above formula

$216 = (1 + h) \times (h-5) \times h\\\\216 = (h+h^2)(h-5)\\\\216 = h^2-5h+h^3-5h^2\\\\216 = h^3-4h^2-5h\\\\h^3-4h^2-5h-216 = 0$

Solve by factoring

$(h-8)(h^2+4h+27) = 0$

Use the zero factor principle

If ab = 0 then a = 0 or b = 0 ( or both a = 0 and b = 0)

Therefore,

$h - 8 = 0\\\\h = 8$

Also,

$h^2+4h+27 = 0$

Solve by quadratic equation formula

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\mathrm{For\:} a=1,\:b=4,\:c=27:\quad h=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \:27}}{2\cdot \:1}$

$h = \frac{-4+\sqrt{4^2-4\cdot \:1\cdot \:27}}{2}=\frac{-4+\sqrt{92}i}{2}$

Therefore, on solving we get,

$h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i$

Thus solutions of "h" are:

h = 8

$h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i$

"h" cannot be a imaginary value

Thus the solution is h = 8

Thus the height of container is 8 feet

8 0

3 years ago

an athlete completes a race in 55.72 seconds how many times greater is the digit in the tens place than the digit in the ones pl

Inessa05 [86]

7 is 5 greater than the 2

7 0

3 years ago

Read 2 more answers

Hello fellow Brainliers. I was wondering if anyone out there would like to go to school for me during the next school year and a

deff fn [24]

Answer:

I would not like to take your place but I can help you find someone. I don't have a laughing emoji. (On my computer)

6 0

2 years ago

3. Two quantities, x and y, are in a

frez [133]

Answer:

y = x + 1.8

Step-by-step explanation:

6 0

2 years ago