Toni bought 5 t-shirts for 35 if each T-shirt cost the same amount .how much tony would pay for20 shirts

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

8 0

Answer: The answer is Toni can buy 20 tee shirts for 140 dollars

Step-by-step explanation:

First you find out how many times 5 can go into 20

Next you see that 5 can go into 20, 4 times so you take the 35 and multiply 4 and you get 140 so you can buy 20 tee shirts for 140 dollars.

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You are paid $8.50/hr at your part-time job. Your deductions are FICA (7.65%), federal tax withholding (9.15%), and state tax wi

dexar [7]

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3 years ago

Can somebody help me with this ?? :)

Alex777 [14]

Answer:

Your answer would be D, the equation is a function because it is a polynomial and all polynomials are functions.

Step-by-step explanation:

I like to remember that polynomials have no negative , fraction exponents and no division. In this case everything is positive so I knew it was a polynomial , and like the answer says all polynomials are functions !

Hope this helped

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3 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$