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xz_007 [3.2K]
3 years ago
9

What is the equation of the line?

Mathematics
2 answers:
iris [78.8K]3 years ago
6 0
1352282847738199$:)3782737
Sholpan [36]3 years ago
6 0
Y=1/2x+2
hence y=mx+b
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Pls Help! Which equation matches the given points?
ira [324]

Answer:

  D) y = −2/5x

Step-by-step explanation:

In the equation for a line ...

  y = mx +b

the constant 'b' is the y-intercept of the line. The y-intercept is the value of y when x = 0.

The point (0, 0) tells you the y-intercept is 0, or b=0. That reduces the equation to the form ...

  y = mx

The value of m is most easily found if you have a point where x=1. In that case, ...

  y = 1m

  y = m

In the given list of points, there is one that is helpful here: (1, -0.4). That is, the equation for the given points is ...

  y = -0.4x

The answer choices are given as fractions, so you must convert -0.4 to the fraction -2/5.

  y = -2/5x . . . . describes the given points

5 0
3 years ago
-6(5x - 3) < -36x + 6
Alona [7]
-6(5x-3) -36x+6
-30x+18<-36x+6
-30x+36x<6-18
6x<6-18
6x<-12
x<-2
Make at brainliest
5 0
3 years ago
Please help I’m terrible at math
chubhunter [2.5K]

Answer:

25 1/2 feet

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the radius and diameter of these circles?
Step2247 [10]

Answer:

a radius 5mmm diameter 10mm

b radius 6cm diameter 12cm

c radius 9m diameter 18m

d radius 8km diameter 16km

e radius 11m diameter 22

f radius 15mm diameter 30mm

g radius 13km diameter 26km

h radius 7 diameter 14

Step-by-step explanation:

When looking for radius you divide your diameter by two

When looking for diameter you multiply your radius by 2

3 0
3 years ago
Someone help me please!!!!!
goldfiish [28.3K]
First you are using the pothagorean theorem which is a^2+b^2=c^2. You have a and b, so you should plug them in to get 4^2+4^2=c^2. Then simplify it some more to 16+16=c^2. Now finish simplifying to c^2=32. Now take the square root of each side and you have your answer. So your answer will be c=√32 or c is about 5.66.
6 0
3 years ago
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