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djverab [1.8K]
3 years ago
15

PLS HELP!! :// i rlly need it!

Mathematics
1 answer:
Svetlanka [38]3 years ago
8 0

Answer:

the value of b is 74 it won't let me help

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Find the slope of the line between(5,-2)(-8,-5)
Hoochie [10]

Answer:

Step-by-step explanation:

Use the difference of coordinates to find the slope.

Slope of line passing through (x₁, y₁) and (x₂, y₂) = (y₂-y₁)/(x₂-x₁)

Slope of line passing through (5, -2) and (-8, -5) = (-5-(-2))/(-8-5) = -3/(-13) = 3/13

7 0
3 years ago
Please answer quick please
GaryK [48]
Lin is older by 3 years. 5-2=3
3 0
3 years ago
HELP ME LOOK AT THE PHOTO
Assoli18 [71]

Answer:

The answer to the question provided is 7.

Step-by-step explanation:

7x + 9 = 8x + 2 \\  \frac{ - 8x \:  \:  \:  \:  =  - 8x}{ - 1x + 9 = 2 }  \\  \frac{ \:  \:  \:  \:  \:  \:  \:  \:  - 9 =  - 9}{ \frac{ - 1x}{ - 1} =  \frac{ - 7}{ - 1}  }  \\ x = 7

[I apologize if this isn't the answer]

6 0
3 years ago
Gregory knows that Triangle A B C is reflected onto Triangle A prime B prime C prime. Which statement about the figures is true?
ziro4ka [17]
Answer: if Gregory draws the segment with endpoints A and A’, then the midpoint will lie on the line of reflection.

Explanation:

Given that a triangle ABC is reflected in triangle A'B'C'

Here reflection is done on a line

If you imagine the line as a mirror then ABC will have image on the mirror line as A'B'C'

Recall that in a mirror the object and image would be equidistant from the mirror and also the line joining the image and object would be perpendicular to the mirror

But note that corresponding images will only be perpendicular bisector to the line

So A and A' only will be corresponding so AA' will have mid point on line

Option 1 is right
5 0
3 years ago
The circumference of a circle is 15 m. Find the area. Use 3.14 for pie
zloy xaker [14]

Answer: Its 94.2

Step-by-step explanation:all you do to get the area is double 15 meaning 15 times 2 equal 30 times 3.14 equal 94.2 *that's crazy your in high school*

7 0
3 years ago
Read 2 more answers
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