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2 years ago

It’s algebra 2! help asap plzz

Mathematics

1 answer:

san4es73 [151]2 years ago

3 0

This is the difference of two perfect cubes.

(6y-4)(36y^2+24y+16)

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What's the expanded form to write 16,107,320

Romashka [77]

(10,000,000) + (6,000,000) + (100,000) + (7,000) + (300) + (20)

6 0

2 years ago

What is the equation of the line that passes through the points (-3,4) and (6,7)

inn [45]

Y=mx+b, m=(y2-y1)/(x2-x1)

m=(7-4)/(6--3)=3/9=1/3

y=x/3 +b, now use either point to solve for b, I'll use (6,7)

7=6/3 +b

7=2+b

b=5 so our line is:

y=x/3 +5 or more neatly...

y=(x+15)/3

6 0

3 years ago

Read 2 more answers

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

Can someone answer my question please

Charra [1.4K]

Answer:

Step-by-step explanation:

Arc length=∅/360×2πr (can't find the symbol of theta, use ∅ instead LOL)

$\frac{\theta }{360}\cdot 2\pi \left(11\right)=16$

$\frac{\theta }{360}=\frac{16}{22\pi }$

$\theta \:=\frac{16}{22\pi \:}\cdot 360$

${\theta}=83.34$

8 0

2 years ago

Read 2 more answers

Pls I need help geomtry

Basile [38]

Answer:

1st option

i know for sure and....

Last one totally wrong

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers