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2 years ago

If a car was sold at an original price of $18,950, and the sales tax was 6%, what was the total price of the car?

Mathematics

2 answers:

Nadya [2.5K]2 years ago

5 0

Answer: $20,087

Step-by-step explanation: $18,950 times 0.06 is sales tax equals $1137 add that to 18,950 and you get 20,087.

Mashutka [201]2 years ago

4 0

Answer:

$20,087.

Step-by-step explanation:

Find out what 6% of $18,950 is. You will end up with $1,137. Next, add $1,137 to $18,950.

<em>Hope this helps!</em>

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Step-by-step explanation:

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Name the remainder for the quotient and describe or show how you found the remainder.

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Step-by-step explanation:

The expression x^3 - x^2 - 17x -15) ÷ (x-5) shows that x+5 is one of the factor if the polynomial function. According to factor theorem, we will equate the factor to zero and find x;

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Next is to generate the remainder.

Substitute x = -5 into the polynomial function

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3 years ago

Write the slope-intercept form of the equation that passes through the point (-3, 5) and is perpendicular to the line y = 1/5x +

aleksley [76]

Answer:

Step-by-step explanation:

$\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\=========================$

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2 years ago

What is the Distributive property of 18times 4 =72

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2 years ago

Find the area of the region which is inside the polar curve r=5sin(θ) but outside r=4. Round your answer to four decimal places

natka813 [3]

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

$\rm 5 \sin \theta = 4\\\\\theta = 0.927 , 2.214$

Then by the integration, we have

$\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\$

$\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\$

On solving, we have

$\rightarrow \dfrac{1}{2} \times 7.499\\\\\rightarrow 3.75$

Thus, the area of the region is 3.75 square units.

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

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