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Eva8 [605]
3 years ago
10

Find the slope of the line that passes through the points shown in the table.

Mathematics
2 answers:
Semenov [28]3 years ago
8 0

Answer:

it would be -2/7

Step-by-step explanation:

Fed [463]3 years ago
4 0
Answer: -2/7
This value is a fraction.

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

Work Shown:

The first two rows in the table produce these two points (x1,y1) = (-14,8) and (x2,y2) = (-7,6)

So, x1=-14, y1=8, x2=-7, and y2=6

Plug these four values into the slope formula and simplify
m = (y2 - y1)/(x2 - x1)
m = (6 - 8)/(-7 - (-14))
m = (6 - 8)/(-7 + 14)
m = -2/7
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3 years ago
Please help me with this please answer this correctly
sergij07 [2.7K]

30 and 13


x-y=17 and x+y=43

x=17+y and x=43-y, so 17+y=43-y

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8 0
3 years ago
Find BD.<br> help me pls
Anna [14]

Answer:

16

Step-by-step explanation:

We can see that Ar is the perpendicular bisector of chord BD. Since A is the center of the circle, AR is the radius of the circle, which is 10 (6+4)

Next, we can see that when we connect point A to point D, it is also a radius. Thus, AD is also equal to 10 as the radius of the circle remains the same.

Using Pythagoras theorem, a^2 + b^2 = c^2, we can make a right angled triangle of ACD.

AC = 6 = a

CD = ? = b

AD = 10 = c

10^2 = 6^2 + b^2

b^2 = 10^2 - 6^2 = 64

b = CD = 8

Now, since Ar is the perpendicular bisector of chord BD, BD = CD x 2

BD = 8 x 2 = <u>16</u>

5 0
2 years ago
A\b=c for bsolve for b and show your steps.show all steps.
Rufina [12.5K]
A/b=c
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6 0
3 years ago
Suppose lines EF and GH are reflected over the line Y equals X to form the lines JK and LM which statement would be true about t
katrin [286]

Answer:

A) JK and LM will be parallel to each other.

Step-by-step explanation:

On reflection on y=x line the x co-ordinate changes with y co-ordinate and y co-ordinate changes with x co-ordinate

(x,y)\rightarrow (y,x)

Points on line EF

(0,6) , (-5,-2)

On reflection of this line on y=x the new points we get for line JK are

(6,0),(-2,-5)

Points on line GH

(-4,9),(-9,1)

On reflection on y=x line the new points we get for line LM are

(9,-4),(1,-9)

Slope of line JK

m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-5)-0}{(-2)-6} \\m=\frac{-5}{-8}=\frac{5}{8}

Slope of line LM

m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-9)-(-4)}{1-9} \\m=\frac{-9+4}{-8}=\frac{-5}{-8}\\m=\frac{5}{8}

For two line to be parallel, their slopes will be same.

m_{JK} =\frac{5}{8} , m_{LM}=\frac{5}{8}

Since slopes of lines JK and LM are same therefore we can say that these are parallel to each other.

5 0
3 years ago
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