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3 years ago

I need help what’s the answer?

Mathematics

1 answer:

KonstantinChe [14]3 years ago

3 0

Answer:

Step-by-step explanation:

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A line goes through the points (4, 5) and (2, -6). Write the equation of the line in point-slope

Lerok [7]

Answer:

m (slope) = $\frac{11}{2}$

Step-by-step explanation:

point-slope form: m= $\frac{y2-y1}{x2-x1}$

(4, 5) (2, -6)

x1 y1 x2 y2

-------------------

$\frac{-6 - 5\\}{2 - 4}$ = m

-6 - 5 = -11

2 - 4 = -2

m = $\frac{11}{2}$

5 0

3 years ago

Sandra bought 120 square feet of carpet for a room in her new house. If the room is square, what is the length of each side of t

BARSIC [14]

Answer:
length of each side of the room is 2√30 which is approximately 10.9545 ft

Explanation:
The area of the square can be computed as follows:
Area = l²
where l is the length of the side

We are given that:
area = 120 ft²

Substitute with the area in the above equation and solve for l as follows:
area = l²
120 = l²
l = √120
l = 2√30 ft
This is approximately equal to 10.9545 ft

Hope this helps :)

3 0

3 years ago

Read 2 more answers

No Spam. I will mark Brainliest, if you get it right. A tank's length is 4 feet longer than its width. The height is 2 feet more

dmitriy555 [2]

Answer: 3

Step-by-step explanation:

Let say x the width of the tank,

x+4 its length,

x+2 its height

$Volume=105=x(x+4)(x+2)\\\\P(x)=x^3+6x^2+8x-105=0\\\\105=3*5*7\\\\P(1)\neq 0\\P(-1)\neq 0\\P(3)=0\\\\\begin{array}{|c|cccc|}&x^3&x^2&x&1\\&1&6&8&-105\\x=3&&3&27&105\\----&--&--&--&--\\&1&9&35&0\\\end{array}\\\\\\P(x)=x^3+6x^2+8x-105=(x-3)(x^2+9x+35)\\And \ x^2+9x+35\ is\ not \ factorizable\ (\Delat=9^2-4*35 < 0)\\$

width=3

length=3+4=7

height=3+2=5

4 0

3 years ago

How many solutions does the system of equations have? 0 1 3 4

myrzilka [38]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Evaluate the difference quotient for the given function. Simplify your answer.

nasty-shy [4]

I suppose you mean

$f(x) = \dfrac{x+5}{x+1}$

Then

$f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2$

and the difference quotient is

$\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}$

If it's the case that x ≠ 3, then (x - 3)/(x - 3) reduces to 1, and you would be left with

$\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}$

4 0

3 years ago