Janelle earned $253.75 for 5 days of work. how do I find her unit rate of pay?

Is this a function please help I’m failing

igomit [66]

Answer:

Yes

Step-by-step explanation:

The relation is a function. For a relation to be a function there must be a unique x value for each y value. So this means x's can not repeat, and in this relation, the x-values never repeat. Therefore this is a function.

7 0

2 years ago

State the open intervals over which the function is (a) increasing, (b) decreasing or (c) constant

ipn [44]

Answer:

a. [-3, 4]

b. (-inf, -3]

c. [4, inf)

Step-by-step explanation:

Our intervals will represent the x-values

We know that since there's an arrow pointing to the left of the line that it goes on infinitely

Same thing when the arrow is going to the right

Then we can just looking at the x-values on the graph for the intervals where it starts and stops

Hope this helps

Best of luck

3 0

1 year ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

What is the product of the sum of eight and two and the difference between eight and two? Show all work.

meriva

<h2>Answer:</h2><h3>=60</h3><h2>Step-by-step explanation:</h2><h3>=Solution:</h3><h3> (8+2) × (8-2)</h3><h3> =10×6</h3><h3> =60#</h3>

3 0

2 years ago

Two numbers have a sum of 124

andrew-mc [135]

Answer:

78 and 46

Step-by-step explanation:

7 0

3 years ago

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Janelle earned $253.75 for 5 days of work. how do I find her unit rate of pay?​

Janelle earned $253.75 for 5 days of work. how do I find her unit rate of pay?