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katen-ka-za [31]

2 years ago

9

What is the length of any side of equilateral triangle TUV? a. 7 b. 15 c. 20 d. 24

1 answer:

NikAS [45]2 years ago

6 0

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Which two quadrants have negative x-values?

mart [117]

Answer:

II and III ( 2 and 3)

Step-by-step explanation:

The quadrants are numbered counter clockwise, with the top right being number I

II I

III IV

The ones having negative x values are II and III ( 2 and 3)

3 0

2 years ago

Verify that the roots of 5x²- 6x -2 = 0 are <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%20%2B%20%5Csqrt%7B19%7D%20%7D%7B5%7D%2

Mice21 [21]

Answer:

Proof below.

Step-by-step explanation:

<u>Quadratic Formula</u>

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

<u>Given quadratic equation</u>:

$5x^2-6x-2=0$

<u>Define the variables</u>:

a = 5
b = -6
c = -2

<u>Substitute</u> the defined variables into the quadratic formula and <u>solve for x</u>:

$\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}$

$\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{76}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}$

$\implies x=\dfrac{6 \pm2\sqrt{19}}{10}$

$\implies x=\dfrac{3 \pm \sqrt{19}}{5}$

Therefore, the exact solutions to the given <u>quadratic equation</u> are:

$x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}$

Learn more about the quadratic formula here:

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3 0

1 year ago

Read 2 more answers

-30-0.6k <br>what does k equal?

Fed [463]

Answer:

K would equal to:

K= 50

Hope that helps! :)

Step-by-step explanation:

3 0

2 years ago

HELP ME FASTTTTTTT !!!!!!!!!!!!!!!!!!!!!!!!!

DedPeter [7]

Answer:

50 degrees ( Vertically opposite)

Hope it helps

Step-by-step explanation:

5 0

2 years ago

What is the slope of the line whose equation is y−4= StartFraction 5 Over 2 EndFraction.(x−2)?

Yuki888 [10]

For this case we must find the slope of the following equation:

$y-4 = \frac {5} {2} (x-2)$

By definition, the point-slope equation of a line is given by:

$y-y_ {0} = m (x-x_ {0})$

Where:

$(x_ {0}, y_ {0})$ : It is a point through which the line passes

m: It's the slope

So, the slope of the given equation is:

$m = \frac {5} {2}$

Answer:

Option D

8 0

3 years ago

Read 2 more answers