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Viktor [21]
4 years ago
8

What is the solution to the system of equations?

Mathematics
2 answers:
gizmo_the_mogwai [7]4 years ago
8 0
The solution to the system of equation is
(- 3,5,6)
vladimir1956 [14]4 years ago
6 0

Answer: A) -3,5,6

Step-by-step explanation: Took the test

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∆ABC is plotted on a coordinate plane. If ∆ABC rotates 90° clockwise around point C, what are the coordinates of A'?
Volgvan
4,6 is what I think, see if you have a plain that has a90 deg around c.

3 0
3 years ago
Please help! Due tonight!
BigorU [14]
A^2 + b^2 = c^2 (pythagoras theorem)
A = 5, b = 8
(5)^2 + (8)^2 = c^2
25 + 64 = 89
c^2 = 89
The answer would be D.
4 0
3 years ago
Read 2 more answers
Can someone help me please
Vera_Pavlovna [14]

Answer:

B) Y = 5x - 10

Step-by-step explanation:

m = y2-y1 / x2-x1

Hope this helps. Pls give brainliest.

6 0
3 years ago
Find the slope of the line going through the point (0,3) and (0,4)
telo118 [61]

Answer:

undefined

Step-by-step explanation:

To find the slope of the line between two points, use the slope formula \frac{y_2-y_1}{x_2-x_1}. x_1 and y_1 represent the x and y values of one point the line passes through, and x_2 and y_2 represent the x and y values of another point the line also passes through. Therefore, use the x and y values of the points (0,3) and (0,4) to find the slope. Substitute them into the formula in the right order:

\frac{(4)-(3)}{(0)-(0)} \\= \frac{4-3}{0-0}\\=\frac{1}{0}

However, we can't divide 1 by 0. Therefore, the slope is undefined. (You could also graph the points to see that they form a vertical line, and all vertical lines have an undefined slope.)

5 0
3 years ago
Read 2 more answers
A student solved the following problem and made an error: Triangles ABC and DEF. Angles A and F are congruent and measure 135 de
nignag [31]

The first false statement in the proof as it stands is in Line 5, where it is claimed that a line of length 2.83 is congruent to a line of length 4.47. This mistake cannot be corrected by adding lines to the proof.

_____

The first erroneous tactical move is in Line 4, where the length of DE is computed, rather than the length of FD. This mistake can be corrected by adding lines to the proof.

A correct SAS proof would use segment FD in Line 4, so it could be argued that the first mistake is there.

7 0
3 years ago
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