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dedylja [7]
3 years ago
8

HELP ME PLZZZ!!! I'VE BEEN TRYING TO GET THE AWNSER ON HERE FOR HOURS SO I SPENT OVER 1000 POINTS TRYING TO GET THIS AWNSER!

Mathematics
1 answer:
Annette [7]3 years ago
4 0
It would be A. 0.65
0.5x0.5+1,2(0.5+0.5+0.5+0.5)x0.4
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If the slope is 2 and the line goes through (3,-3) what is the equation?
motikmotik

Answer:

y = 2x -9

Step-by-step explanation:

To find the equation of a line using slope and a point, first use the slope to create the basic line using the slope and work from there.

For instance, the base equation here is

y = 2x

This line passes through the point (0, 0).

You can then plug in a value for x. In this case, use the value of 3, as it corresponds with your question.

y = 2 * 3

A point on the line of y = 2x would thus be (3, 6).

To make the y-value equal -3, you must then subtract from the original equation. There are 9 units between 6 and -3, so you must subtract nine units in the equation. You should get this at the end:

y = 2x -9

6 0
3 years ago
How to factor 5m2-29m+20?
ser-zykov [4K]

Question :

5m^2 - 29m + 20

Steps:

What can u factor to get 20?

you can make 5 and 4 to make 20 so,..

1. Break it into groups

(5m^2 - 4m) + (-25m + 20)

*Factor out m from (5m^2 - 4m) : m(5m - 4)

*Factor out -5 from (-25m + 20) : -5(5m - 4)

2. Then, you can factor out the common term : (5m - 4) because they both have it. And then u would move the m by the -5 so,...

= (5m - 4)(m - 5) <======= <em>Answer</em>


Hope this helped!!!

~Shane


3 0
3 years ago
What is 5/3x 1/6= -1
Ivahew [28]
0.2777777777777777777
6 0
4 years ago
The pair of square pyramids are similar. Use the given information to find the scale factor of the smaller square pyramid to the
Mrrafil [7]
\bf \qquad \qquad \textit{ratio relations}&#10;\\\\&#10;\begin{array}{ccccllll}&#10;&\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\&#10;&-----&-----&-----\\&#10;\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}&#10;\end{array}\\\\&#10;-----------------------------

\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\&#10;-------------------------------\\\\&#10;\stackrel{\stackrel{pyramids}{scale~factor}}{\cfrac{small}{large}}\qquad \qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{64}}{\sqrt[3]{343}}\implies \cfrac{s}{s}=\cfrac{4}{7}\implies 4:7
8 0
3 years ago
I Forgot All About Area and Perimeter Can Anybody Help Please?
Zinaida [17]
So you would multiply the length times width!
7 0
3 years ago
Read 2 more answers
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