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3 years ago

You purchase 5 pounds of cheese for $28. How much would 1 quart of cheese cost?

Mathematics

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

You would get 1.4

Step-by-step explanation:

First you divide 5 pounds by $28 your answer will be :<em> </em><em>5</em><em>.</em><em>6</em>

Than you divide 5.6 by 4 because 4 quarters make 1 whole

So, you divide 5.6 by 4

Than you get the answer 1.4.

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Kai is making a chessboard using pieces of wood that measure 1sq in when she finishes the chessboard well have an area of 64 sq

Kay [80]

Answer:

8 pieces of wood should be put along each side of the chessboard

Step-by-step explanation:

Here in this question, we are interested in knowing the the number of pieces of wood that should be put along each side of the cheese board.

Firstly, to know the number of pieces of wood to use, we need to divide the area of the board by the area of each piece of wood

Mathematically that would be 64 square inches/ 1 square inch = 64 pieces of wood

64 pieces of wood would be needed to complete the cheeseboard.

The number of pieces to put along each side of the cheese board will be √64.

This is because the shape of a chessboard is square.

Thus the number of pieces to put on each side will be √64 = 8

3 0

3 years ago

a coyote can run up to 43 mph while a rabbit can run up to 35 mph. Write two equivalent expressions and then find how many more

Dafna11 [192]

The coyote can run 48 more miles than the rabbit

8 0

3 years ago

Whats the answer please help

Mice21 [21]

Answer:

On what?

Step-by-step explanation:

6 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Which graph best represents the relationship between time and the number of teams registered?

alexdok [17]

<h2>Explanation:</h2><h2></h2>

The complete question is in the attached file. So we have to choose between two graphs. On of them is a linear model while the other is an exponential model. From the statements, we have a relationship between time and the number of teams registered. So we can establishes variables in the following form:

$x:\text{Time} \\ \\ y:\text{Number of teams registered}$

We also know that each week 6 teams register to participate, so:

$\bullet \ \text{For week 0:} \rightarrow \text{0 teams registered} \\ \\ \bullet \ \text{For week 1:} \rightarrow \text{6 teams registered} \\ \\ \bullet \ \text{For week 2:} \rightarrow \text{12 teams registered (Because 6+12)} \\ \\ \bullet \ \text{For week 3:} \rightarrow \text{18 teams registered (Because 12+6)}$

As you can see, as x increases one week, y increases at a constant ratio of 6. Therefore, this can be modeled by a linear function given by the form:

$y=6x$

In conclusion, <em>the linear model (first graph below) is the one that bests represents the relationship between time and the number of teams registered.</em>

7 0

3 years ago