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3 years ago

5

You bought donuts for .60 each. In order to make a profit, you mark up the donuts 25%. What will you charge a customer for a don

ut?

1 answer:

Brut [27]3 years ago

6 0

75cents. 60cents divided by 4 is 15cents.

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Why is x = 6 a solution to the proportion StartFraction 192/36=32/x?

mars1129 [50]

Answer:

C

Step-by-step explanation:

When you cross multiply 192 times 6 it equals 1152 and if you multiply 32 times 36 you get 1152 and 1152 is equal to 1152.

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3 years ago

What is the solution to the equation 1 over 56 multiplied by x is equal to 1 over 8

Kruka [31]

Lets write the equation
x*1/56= 1/8

Lets do left side

1x/56=1/8

Now we see that in order to remove 56 and only be left with 1x on leftside we have to multipley by 56. If we do so on on side we have to do on the other side. So both sides multiplies by 56

1x/56*56= 1*56/8

From above you see that 1x= 7

x=7

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2 years ago

What do you call it when a bunch of kids throw circles at eachother

Hunter-Best [27]

It’s called a pi fight

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3 years ago

Read 2 more answers

Could somebody answer this question

Tpy6a [65]

We did this same kind of problem in school. All we did was divide the those 2 numbers to figure out the unit price. So, 20/5 = 4 $ per CD.

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3 years ago

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A thin metal plate, located in the xy-plane, has temperature T(x, y) at the point (x, y). Sketch some level curves (isothermals)

Sophie [7]

Answer:

Step-by-step explanation:

Given that:

$T(x,y) = \dfrac{100}{1+x^2+y^2}$

This implies that the level curves of a function(f) of two variables relates with the curves with equation f(x,y) = c

here c is the constant.

$c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)$

By cross multiply

$c({1+x^2+2y^2}) = 100$

$1+x^2+2y^2 = \dfrac{100}{c}$

$x^2+2y^2 = \dfrac{100}{c} - 1 \ \ -- (2)$

From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.

Now,

$\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1$

This is the equation for the family of the eclipses centred at (0,0) is :

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

$a^2 = \dfrac{100}{c} -1 \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}$

Therefore; the level of the curves are all the eclipses with the major axis:

$a = \sqrt{\dfrac{100 }{c}-1}$ and a minor axis $b = \sqrt{\dfrac{50 }{c}-\dfrac{1}{2}}$ which satisfies the values for which 0< c < 100.

The sketch of the level curves can be see in the attached image below.

7 0

3 years ago