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3 years ago

Suppose c varies jointly with a and b. When c = 25, a = 2, and b = 2.5. What is the value of c when b = 4/3 and a = 6?

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

8 0

You have to find the value of c when b=4/3and a=6

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A store prices tapes by raising the wholesale price 50% and adding 25 cents .What must a tape's wholesale price be if the tape s

Andre45 [30]

Let's call the store value as s and the wholesale price as w. A store prices tapes by raising the wholesale price 50%(0.5 in decimals) and adding 25 cents, writing this as an equation, we have

$s(w)=w+0.5w+0.25$

If we invert the equation we're going to find the the wholesale price as a function of the store price.

$\begin{gathered} s=w+0.5w+0.25 \\ s=1.5w+0.25 \\ s-0.25=1.5w \\ w(s)=\frac{s-0.25}{1.5} \end{gathered}$

Now, to find the wholesale price if the sales price is $1.99, we just need to evaluate s = 1.99 on the function we created.

$w(1.99)=\frac{1.99-0.25}{1.5}=1.16$

The wholesale price is $1.16.

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1 year ago

This is for math and I need help plz

ollegr [7]

Answer:

A i think

Step-by-step explanation:

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3 years ago

Read 2 more answers

A box contains plain pencils and pens. A second box contains color pencils and crayons. One item from each box is chosen at rand

pochemuha

Answer:

1/2

Step-by-step explanation:

its 1 out of 2 chances

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3 years ago

What is x for the the problem y=6x + 8

Misha Larkins [42]

If Y is 8, we solve by subtracting 8 on both sides,

0 = 6x

Divide by 6,

0 = x

x is equal to 0

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3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .

The area (in square inches) of the circle that the hexagon did not cover would be:

$\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}$ .

3 0

3 years ago