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Alja [10]
3 years ago
5

A cashier has only one-dollar bills, quarters, and dimes. List 9 ways you could receive $2.50 in change.

Mathematics
1 answer:
musickatia [10]3 years ago
7 0
1. 2 one dollar bills and 2 quarters
2. 2 one dollar bills and 5 dimes
3. 8 quarters and 5 dimes
4. 10 quarters
5.25 dimes
6.20 dimes and 2 quarters
7. 4 quarters and 15 dimes
8. 6 quarters and 10 dimes
9. 1 one dollar bill and 6 quarters

Hope this helped
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Simplify the expression where possible.<br><br> (-5x2)^3
Paul [167]

Answer:

-1000

Step-by-step explanation:

1. Do the Parenthesis problem first. (PEDMAS/PEMDAS)

You should then have -10^3

2. Solve with the exponent. (-10 x 10 x 10)

You should then get your final answer, which is -1000.

Any Questions/Concerns?

6 0
3 years ago
Can some one please tell me the answer<br>to this 2(x+4)-1=2x+7<br>​
dangina [55]

Answer:

2(x+4)-1=2x+7 is true for everything

Step-by-step explanation:

2(x+4)-1

2(x+4)

=2x+2*4

=2x+8

=2x+8-1

=2x+7

2x+7=2x+7

2x=2x

2x-2x=2x-2x

0=0

true for all x

7 0
3 years ago
Hey guys please help me
Rina8888 [55]
Answer is 20. -4 x 1 is -4 so it doesn’t change. -4 -8 makes it -4 + 8 as there’s two negatives. 9 divided by 3 is 3. So it’s -4 +8 x 3. 8x3 is 24. So -4 + 24 = 20. Therefore final answer is 20
8 0
2 years ago
If a = 20 mm, b = 12 mm, and m∠C = 80°, what is the approximate area of ABC?
Lana71 [14]

Answer:

118.18 mm^2

Step-by-step explanation:

Sides <em>a</em> and <em>b</em> form angle C, so the information given is a side-angle-side (SAS) problem.

Use the formula A=\frac{1}{2}ab\sin{C}.

A=\frac{1}{2}(12)(20)\sin{80^\circ} \approx 118.18 \rm \,mm^2

8 0
3 years ago
First-order linear differential equations
kkurt [141]

Answer:

(1)\ logy\ =\ -sint\ +\ c

(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

Step-by-step explanation:

1. Given differential equation is

  \dfrac{dy}{dt}+ycost = 0

=>\ \dfrac{dy}{dt}\ =\ -ycost

=>\ \dfrac{dy}{y}\ =\ -cost dt

On integrating both sides, we will have

  \int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}

=>\ logy\ =\ -sint\ +\ c

Hence, the solution of given differential equation can be given by

logy\ =\ -sint\ +\ c.

2. Given differential equation,

    \dfrac{dy}{dt}\ -\ 2ty\ =\ t

=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty

=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})

=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt

On integrating both sides, we will have

   \int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}

=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c

=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

Hence, the solution of given differential equation is

log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

8 0
4 years ago
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