What’s the length of EF?

Mathematics

1 answer:

blsea [12.9K]3 years ago

5 0

Answer:

Step-by-step explanation:

If the whole length of DEF is given as 7, and segment DE is 2, then segment EF is 7 - 2 = 5

You might be interested in

Why is the product of two rational numbers always rational?

Mice21 [21]

Let a/b and c/d represent two rational numbers. This means a, b, c,and d are INTEGERS , and b and d are not 0. The product of the numbers is ac/bd , where bd is not 0. Both ac and bd are INTEGERS , and bd is not 0. Because ac/bd is the ratio of two INTEGERS, the product is a rational number.

$\frac{a}{b}\cdot \frac{c}{d}= \frac{a\cdot c}{b\cdot d}=\frac{e}{f},\,\,\,\,e,f\in\cal{Z}$

7 0

4 years ago

Read 2 more answers

What is the undefined variable in this equation (6x2 +3x) ÷ (3x) simplified 3x + 1

ozzi

X is the undefined variable hope it helps

4 0

3 years ago

I BET NO ONE KNOWS THIS QUESTION. Lower income tends to predict less parental involvement. IN this example, lower income is:

jenyasd209 [6]

I think it is D too.

7 0

4 years ago

Read 2 more answers

Please help me to prove this!

Sophie [7]

Answer: see proof below

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

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$