2 years ago

Write AB−⇀− with A(−7,5) and B(−3,−7) in component form.

Mathematics

1 answer:

Sladkaya [172]2 years ago

8 0

Answer:

ANSWER: (4, -12)

Step-by-step explanation:

Lol

You might be interested in

What is the product of written in simplest form?

storchak [24]

Sorry but we need the numbers

6 0

3 years ago

Five cats ate 1/4 cup of food how much food did five cats eat

Maslowich

If it was EACH cat they had 1.25 cups otherwise they all shared the 1/4 cup offered.

7 0

3 years ago

Write twenty-four minus the product of three and two as a numerical expression

JulijaS [17]

Step-by-step explanation:

kajakpplallalallalallala

6 0

3 years ago

Write a ratio for the following description: Kaleel made three times as many baskets as John during a basketball practice.

Hoochie [10]

Answer:

3:1

Step-by-step explanation:

If John makes 3, Then Kaleel makes 9, If John makes 1, Then Kaleel makes 3.

8 0

3 years ago

If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$

4 0

3 years ago