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2 years ago

Find the answers to these two problems. PLEASE HELP!!

Mathematics

1 answer:

AleksAgata [21]2 years ago

6 0

Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.

1st question

angles 6 and 7
or
angles 2 and 3

2nd question

Angles 3 and 2
or
Angles 5 and 7

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Solve by elimination

inna [77]

2x - 2y = -4
-3x -3y = 0
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-1x - 5y = -4

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if the diameter of a circle is 15, its radius is half that or 7.5.

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lakkis [162]

Circumference of a circle = 2πr

So,

$circumference = 2\pi \: r \\ = > circumference = 2 \times \frac{22}{7} \times \frac{63 \: cm}{2} \\ = > circumference = 2 \times \frac{11}{7} \times 63 \: cm \\ = > circumference = 2 \times 11 \times 9 \: cm \\ = > circumference = 198 \: cm$

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If the point (a, b) is equidistant from the points (-a, 2) and (2, -b), then prove that 3(a+b)+4=0

erastova [34]

Answer: see proof below

Step-by-step explanation:

Since (a, b) is equidistant from (-a, 2) and (2, -b), then it is the midpoint of the those two points. Use Midpoint formula to find (a, b).

$M_x=\dfrac{x_1+x_2}{2}\qquad \qquad \qquad M_y=\dfrac{y_1+y_2}{2}\\\\\\a=\dfrac{-a+2}{2}\qquad \qquad \qquad \quad b=\dfrac{2-b}{2}\\\\\\2a=-a+2\qquad \qquad \qquad \quad 2b=2-b\\\\\\3a=2\qquad \qquad \qquad \qquad \qquad 3b=2\\\\\\a=\dfrac{2}{3}\qquad \qquad \qquad \qquad \qquad b=\dfrac{2}{3}$

3(a + b) - 4 = 0

$3\bigg(\dfrac{2}{3}+\dfrac{2}{3}\bigg)-4=0\\\\\\3\bigg(\dfrac{4}{3}\bigg)-4=0\\\\\\4-4=0\\\\0=0\qquad \text{TRUE!}$

Notice that I changed the equation to "negative 4" because the equation you provided did not make a true statement.

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