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3 years ago

In this figure prove that cd//ef

Mathematics

1 answer:

MaRussiya [10]3 years ago

8 0

Answer:

1) 36 + 50 = 86 : Addition property of equality

2) m<D = m<ADB : If two angles have the same measure, then they are equal.

3) <D congruent to <ADB : if two angle are equal, then they are congruent.

4) <D, <ADB are alternate interior angles: assumed from diagram

5) <u>CD // AB</u> : if alternate interior angles are congruent, then the 2 lines cut by a transversal are //.

6) 130 + 50 = 180 : addition property of equality

7) m<E, m<BAE are supplements : if the sum of two angles equal to 180, then they are supplements.

8) <E, <BAE are same-side interior angles : assumed from diagram

9) <u>EF // AB</u> : if same-side interior angles are supplements, then the 2 lines cut by a transversal are //.

10) <u>CD // EF </u>: if two lines are // to the same line, then they are //.

im pretty sure this is the proof

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$HEY THEIR!!$

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3 years ago

Solve equation using square roots (x-1)^2=4

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3 years ago

the sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger

Reil [10]

<h2>Answer:</h2>

$x\geq 3 \ and \ x+1 \geq 4$

<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable $x$ which is the smaller number. Accordingly, the lager number will be $x+1$ given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

$\leq \\ which \ means \ less \ than$

So:

1. The sum of 2 consecutive integers can be written as:

$v+(v+1)$

2. Nine times the smaller and 5 times the larger can be written as:

$9v-5(v+1)$

Finally, the whole statement:

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

$x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\$

$x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ \frac{6}{2} \leq \frac{2x}{2} \\ \\ 3 \leq x \\ \\ x\geq 3 \\ \\ and \\ \\ x+1 \geq 4$

The two numbers are:

$x\geq 3 \ and \ x+1 \geq 4$

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3 years ago

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3 years ago

In this figure prove that cd//ef​

In this figure prove that cd//ef