The health food store that you manage employs 7 clerks. Individual hourly wages are $8.10, $8.25, $8.45, $8.75, $9.00, $9.25, and $10.50. If each employee has 2 weeks unpaid vacation and works 8 hours a day , 5 days per week and is paid for all holidays , what the total annual payroll for the clerks ?

Answer:

x = 16

Step-by-step explanation:

3x - 10 = 2(x+3)

First step is solve this:

2(x+3) = 2*x + 2*3 = 2x + 6

then:

3x - 10 = 2x + 6

3x - 2x = 6 + 10

x = 16

Check:

3*16 - 10 = 2(16+3)

48 - 10 = 2*19 = 38

Answer:

x = 16

Step-by-step explanation:

you start off by isolating the variable

Answer:

Tablets.

Step-by-step explanation:

You didn't want an explanation so....here you go....no explanation.

Hope that helps and maybe earns a brainliest.

Have a great day!

Answer:

3(4)to the 3

Step-by-step explanation:

Answer:

[tex]\boxed{y = -2}[/tex]

Step-by-step explanation:

Hey there!

To find y when x is -5 we go to -5 on the x-axis.

When at -5 find where the blue line is vertical to -5,

which is -2.

Hope this helps :)

This is assuming that p is not in the denominator. To make it more clear that p is not in the denominator, I recommend you write (1/3)p.

Taking 1/3 of a number is the same as dividing by 3.

Example:

27/3 = 9

27*(1/3) = 9

Answer:

[tex]\mathbf{V = [\dfrac{ 8 \pi }{3}] }[/tex]

Step-by-step explanation:

Given that:

y = 6x - x² , y = 8  about  x = 2

To find the volume of the region bounded by the curves about x = 2; we have the radius of the cylindrical shell to be x - 1, the circumference to be 2 π (x -2 ) and the height to be 6x - x² - 8

6x - x² - 8

6x - x² - 8 = 0

-x² + 6x - 8 = 0

x² - 6x + 8 = 0

(x -4) (x - 2  ) = 0

So;

x = 2 , x = 4

Thus, the region bound of the integral  are from a = 2 and b = 4

Therefore , the volume of the solid can be computed as :

[tex]V = \int \limits ^b _a \ 2x \times f(x) \ dx[/tex]

[tex]V = \int \limits ^4_2 2 \pi (x -2) (6x -x^2 -8) \ dx[/tex]

[tex]V = 2 \pi \int \limits ^4_2 (6x^2 - x^3 -8x -12 x - 2x^2 +16) \ dx[/tex]

[tex]V = 2 \pi \int \limits ^4_2 (8x^2 -x^3-20x +16) \ dx[/tex]

[tex]V = 2 \pi \int \limits ^4_2 ( -x^3+8x^2-20x +16) \ dx[/tex]

[tex]V = 2 \pi [\dfrac{ -x^7}{4}+\dfrac{8x^3}{3} -\dfrac{20x^2}{2} +16x]^4_2[/tex]

[tex]V = 2 \pi [\dfrac{ -(4^4-2^4)}{4}+\dfrac{8(4^3-2^3)}{3} -\dfrac{20(4^2-2^2)}{2} +16(4-2) ]^4_2[/tex]

[tex]V = 2 \pi [\dfrac{ -(256-16)}{4}+\dfrac{8(64-8)}{3} -10(16-4)} +16(2) ][/tex]

[tex]V = 2 \pi [\dfrac{ 4}{3}][/tex]

[tex]\mathbf{V = [\dfrac{ 8 \pi }{3}] }[/tex]