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Lelu [443]
3 years ago
10

What is the slope of the line?

Mathematics
2 answers:
valina [46]3 years ago
7 0

Answer:

1/3

Step-by-step explanation:

PilotLPTM [1.2K]3 years ago
5 0
The slope of the line is 1/3
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Which expression is equivalent to 2 (×+7)-18×+4/5
Ksivusya [100]

Answer:

14.8-16x

Step-by-step explanation:

2x+14-18x+0.8

14.8-16x

4 0
3 years ago
Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer​
Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
3 years ago
What is the distance between point A(-3,2) and point B(5,-1)
Gnoma [55]

Remember the distance formula

S=\sqrt{(5+3)^{2}+(-1-2)^{2} }

S=\sqrt{8^{2}+(-3)^{2} }

S=\sqrt{64+9}

S=\sqrt{73}


8 0
3 years ago
Mrs. Reed bought 48 items for her classroom. She bought some glue bottles. She bought 5 times as many pencils as she did bottles
uysha [10]
She bought 40 pencils
6 0
3 years ago
Read 2 more answers
If <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdt%7D%20%3Dky" id="TexFormula1" title=" \frac{dy}{dt} =ky" alt=" \frac
sladkih [1.3K]

This ODE is separable; we have

\dfrac{\mathrm dy}{\mathrm dt}=ky\implies\dfrac{\mathrm dy}y=k\,\mathrm dt

Integrating both sides gives a general solution of

\ln|y|=kt+C\implies y=e^{kt+C}=Ce^{kt}

B is the only choice that is applicable.

4 0
3 years ago
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