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3 years ago

A store sells 6 T-shirts for $58.08. What is the unit cost per T-shirt?

Mathematics

2 answers:

Semmy [17]3 years ago

5 0

Answer:

$9.68

Step-by-step explanation:

Divide 58.08 / 6 and you get 9.68, this is 1st grade math (:

Veronika [31]3 years ago

5 0

Answer:

$9.68

Step-by-step explanation:

To find the unit cost per t-shirt (how much each t-shirt costs), we need to divide 58.08 by 6. This tells us how much each shirt costs. From this, we get 9.68, this means that each shirt is $9.68. I hope this helps!

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bixtya [17]

Answer:

28.4 divided by 6.5

Step-by-step explanation:

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3 years ago

1) Graph by making a<br> table.<br> y=1/2x+2

nadya68 [22]

Answer:

Below.

Step-by-step explanation:

You can use any 4 x values, so we could have:

- 2 0 2 4

Now find the corresponding y values by substitution in the equation:

For x = -2, y = 1/2 *-2 + 2 = 1.

For x = 0, y = 1/2 * 0 + 2 = 2.

For x = 2 , y = 1/2* 2 + 2 = 3.

For x = 4 , y = 1/2 * 4 + 2 = 4.

So the table is:

-2 0 2 4

1 2 3 4

You then plot the points (-2, 1), (0, 2), (2, 3) AND (4, 4) on the grid and draw a line through them The graph will be a straight line.

3 0

3 years ago

Write the fact family. for 6, 4, 24

GarryVolchara [31]

$4 \times 6 = 24 \\ 6 \times 4 = 24 \\ 24 \div 6 = 4 \\ 24 \div 4 = 6$
That's it :3

8 0

3 years ago

Read 2 more answers

Solve the equation for x. show each step of the solution. name the justifications for each step of the solutions.

UkoKoshka [18]

Answer:

What grade is this?

Step-by-step explanation:

7 0

3 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

3 0

3 years ago