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Ivahew [28]
3 years ago
11

Whats the equation for this problem?

Mathematics
2 answers:
Marina86 [1]3 years ago
7 0
I know that x is equal to 11 and i hope it helps you 
if so please mark as Brainliest
Vlad [161]3 years ago
4 0
I think they may be right
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Which ratio does not belong with the other three? Explain your reasoning. 4/10 2/5 3/5 6/15
JulsSmile [24]

Answer:

3/5

Step-by-step explanation:

7 0
3 years ago
Write an equation for the line parallel to the given line that contains C<br> C(2,6); y= -4x + 7
MrRa [10]

Answer:

y=-4x+14

Step-by-step explanation:

step 1

Find the slope of the given line

we have

y=-4x+7

This is the equation of the given line in slope intercept form

The slope is m=-4

step 2

Find the slope of the line parallel to the given line

we know that

If two lines are parallel, then their slopes are the same

therefore

The slope of the line parallel to to the given line is m=-4

step 3

Find the equation of the line in point slope form

y-y1=m(x-x1)

we have

m=-4

C\ (2,6)

substitute

y-6=-4(x-2)

step 4

Convert to slope intercept form

y-6=-4(x-2)

isolate the variable y

y-6=-4x+8

y=-4x+8+6

y=-4x+14

6 0
3 years ago
Lines p and q are crossed by transversal s. Classify 1 and 5
allsm [11]

Answer:

corresponding angles

Step-by-step explanation:

5 0
3 years ago
The center of a circle is (4, 6), and an endpoint of a diameter is (2, 5). What is the other endpoint of the diameter?
valentinak56 [21]
Consider this option:
1. if the point (4;6) is the centre of the circle and the point (2;5) is the first endpoint of its diameter, then point (4;6) is the middle point of the diameter (it means that is the middle between the 1st and the 2d endpoints of diameter).
2. using the property described above:
for x of the 2d endpoint of the diameter: x=4*2-2=6;
for y of the 2d endpoint of the diameter: y=6*2-5=7.

answer: (6;7)
3 0
2 years ago
Read 2 more answers
Simplify (−4k4+14+3k2)+(−3k4−14k2−8). Write answer in standard form.
marissa [1.9K]

Answer:

−7k^4−11k^2+6

Step-by-step explanation:

−4k^4+14+3k^2−3k^4−14k^2−8

(−4k^4+−3k^4)+(3k^2+−14k^2)+(14+−8)

−7k^4−11k^2+6

...

8 0
2 years ago
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