Ask question

Ask question

All categories

2 years ago

8

I need help plssssssss

1 answer:

vagabundo [1.1K]2 years ago

7 0

Answer: The answer is C

Step-by-step explanation:

It shows every possible outcome

You might be interested in

A training field is formed by joining a rectangle and two semicircles. The semicircles are attached on opposite sides of the rec

Anarel [89]

Step-by-step explanation:

There is figures in you question. Please check your question.

5 0

2 years ago

The product of two exponentail expressionss is x5 y2 z. One of the expressions is x2 y. What is the other expression?

Yakvenalex [24]

I hope this helps you

x^5.y^2.z/x^2.y

x^5-2.y^2-1.z

x^3.y.z

8 0

2 years ago

Read 2 more answers

Help is appreciated. Easy I just am always confused

saveliy_v [14]

Answer:

BA=BC

Step-by-step explanation:

8 0

3 years ago

Find an equation of the circle that has center (-1,2) and passes through (5,-4).

kherson [118]

General equation for a circle
(x - a)² + (y - b)² = r²
with (a,b) represents the center, (x,y) represents one of the points lie on the circle, and r represents the radius

Determine r² by substituting the points into the general equation
(x - a)² + (y - b)² = r²
(5 - (-1))² + (-4 - 2)² = r²
(5 + 1)² + (-6)² = r²
6² + 36 = r²
36 + 36 = r²
72 = r²

Determine the equation of the circle
(x - a)² + (y - b)² = r²
(x - (-1))² + (y - 2)² = 72
(x + 1)² + (y - 2)² = 72 (This is the equation of the circle)

8 0

2 years ago

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago