The sales tax rate is 5%. if the sales tax on 10-speed car is $6 what is the purchase price?

1 answer:

5 0

Since the sales tax is 5% of the purchase price, we can just multiply it by $20 to get $120, the purchase price.
To get the total price, all we do is add $6 to $120, which is $126, the total price.

You might be interested in

Omg PLS SEND HELP i suck at math

Rina8888 [55]

Answer:

13/25

Step-by-step explanation:

13 = 8 circles + 5 hearts.

25 = all the shapes.

(It says not to simplify... so the answer is 13/25)

7 0

3 years ago

Read 2 more answers

An open rectangular box with a square bottom must have a surface area of A square inches. Write a formula for the volume of the

skelet666 [1.2K]

The surface area of the box is being referred to as A
The surface area will be the sum of the four rectangular sides' areas and the base area.
Base area = x²
Rectangular side area = xh
A = x² + 4xh
Volume is, as you said,
V = x²h
Lets substitute h from the formula of A to obtain A in our final answer. Substituting x is possible, but more difficult
(A - x²)/4x = h
V = (x²) * (A - x²)/4x
V = (Ax - x³)/4

5 0

3 years ago

Read 2 more answers

What is y equal to when x= -2 Y=-x+4

Alex17521 [72]

Y= -2+4

If X equals -2, you want to plug -2 in for X.

I hope I helped!

5 0

4 years ago

Please help me to prove this!

Sophie [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

Proof LHS → RHS:

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$