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3 years ago

Dave rented a limousine for his wife's birthday. The hourly rate is $80. They used the limousine for 4 hours, plus Dave gave the

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

3 0

He would spend $, he spent $320 for just renting the limousine then spent an extra $32 for the 10% tip

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Rectangle pqrs is shown with its diagonals, pr and qs. what type of triangle is △str?right triangleequilateral triangleisosceles

Rainbow [258]

Since you did not provide a diagram, I made one for easier reference. You are asking what kind of triangle is the one shaded in red. Since a rectangle has two sets of congruent sides, their diagonals are also equal. When two diagonals intersect each other, they make 4 congruent sides. Thus, the triangle is an isosceles triangle, since it has two equal sides ST and TR.

8 0

3 years ago

Read 2 more answers

Iliana is painting a picture. She has green, red, yellow, purple, orange, and blue paint. She wants her painting to have four di

ivanzaharov [21]

The number of ways for which she could pick four colours if green must be one of them is; 10 ways.

<h3>How many ways can she picks four colours if green must be there?</h3>

It follows from the task that there are 6 colours in total that she could pick from.

Hence, since she needs four colours with green being one of them, it follows that she only has 3 colours to pick from 5.

Hence, the numbers of possible combinations is; 5C3 = 10 ways.

Read more on combinations;

brainly.com/question/2280043

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5 0

1 year ago

If G = {(-1, 7),(-8, 2),(0, 0),(6, 6)), then the range of G is

lana66690 [7]

Answer:

$\{0, 2, 6, 7\}$

Step-by-step explanation:

You are given the relation $G = \{(-1, 7),(-8, 2),(0, 0),(6, 6)\}$

This means that

$f(-1)=7\\ f(-8)=0\\f(0)=0\\f(6)=6$

The input values of this relation are $-1,\ -8,\ 0,\ 6$ , so the domain of this relation is $\{-8,-1,0,6\}$

The output values of this relation are $7,\ 2,\ 0,\ 6$ , so the range of this relation is $\{0, 2, 6, 7\}$

8 0

3 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago

HELP FOR BRAINLEST!!

Tju [1.3M]

Answer:

I believe D is the correct answer.

Step-by-step explanation:

8 0

2 years ago