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3 years ago

Given n//p m<2=50 degrees what is m<7?

Mathematics

2 answers:

erica [24]3 years ago

4 0

ANSWER

$m \angle7 = 130 \degree$

EXPLANATION

$m \angle2 + m \angle3 = 180 \degree$

Sum of angles on a straight line.

$50 \degree+ m \angle3 = 180 \degree$

$m \angle3 = 180 \degree - 50 \degree$

Simplify:

$m \angle3 = 130 \degree$

$m \angle3 = 130 \degree = m \angle7$

Corresponding angles are equal.

Bogdan [553]3 years ago

3 0

Answer: m∠7=130°

Step-by-step explanation:

If two parallel lines are cut by another line (transversal), each one will have four angles surrounding the intersection.

By definition, the adjacent angles of each parallel line, are supplementary, this means that they add up 180°.

Then, if m∠2=50°, then m∠3 is:

m∠3 $=180-50=130$ °

By definition, the angles that are in the same relative position are know as corresponding angles and they are congruent.

Therefore:

m∠3=m∠7

m∠7=130°

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3 years ago

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Let $x$ be the width of the rectangle, so the length will be $12+x$ .
Now, to find the perimeter of our rectangle, we are going to use the formula for the perimeter of a rectangle formula: $P=2(w+l)$
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We want <span>values of the width that will make the perimeter less than 96 feet, so lets set up our inequality:
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3 years ago

A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is

Brums [2.3K]

Answer:

$\frac{259}{54}\text{ or }4.8\text{feet}$

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one $\frac{1}{6}$ as high as the preceding one.

TO FIND: If this ball is dropped from a height of $12$ feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces $\frac{1}{6}$ of preceding one and comes down the same distance.

When the ball is dropped from $12$ feet height

after first hit $=12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4$

new height $=12\times\frac{1}{6}=2\text{feet}$

after second hit $=2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}$

new height $=2\times\frac{1}{6}=\frac{1}{3}\text{ feet}$

after third hit $=\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}$

new height $=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}$

after fourth hit $=\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}$

adding all distance $=4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}$

$=\frac{259}{54}$ feet

Hence the ball will travel $\frac{259}{54}$ feet before it hits the surface fifth time.

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3 years ago