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quester [9]
3 years ago
12

Find the expression below that is equivalent to 5^11/5^13

Mathematics
2 answers:
Mekhanik [1.2K]3 years ago
4 0
\dfrac{5^{11}}{5^{13}}

= 5^{11-13}

= 5^{-2}

=  \dfrac{1}{5^2}

=  \dfrac{1}{25}
zalisa [80]3 years ago
3 0
Hi there!

Answer:
\frac{ 5^{11} }{5 ^{13} } = 5 ^{11-13} =  5^{-2} = \frac{1}{ 5^{2} } = \frac{1}{25} = 0.04

Additional notes:
In the first step, I used the algebraic rules for the exponents in a fraction, which says the following:
\frac{ x^{a} }{ x^{b} } =  x^{a-b}

In the third step, I used the algebraic rules for negative exponents, which says the following:
x^{-a} =  \frac{1}{ x^{a} }
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Anna works as a painter. She charges S130 for paint supplies and $25 for each hour, h, she
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$25(h) + $130

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Nicoli jokic plays for the Denver nuggets in one game, his field-goal total was 26 points made up of 2 point and 3 point baskets
mel-nik [20]

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Answer:

  • 4 3-point baskets
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Step-by-step explanation:

Let x represent the number of 2-point baskets Nicoli made. Then (x -3) is the number of 3-point baskets. The total number of points is ...

  2x +3(x -3) = 5x -9 = 26

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Jokic made 7 2-point baskets and 4 3-point baskets.

7 0
2 years ago
I need help with Q9 c, I’ve done a and b if it helps, b=325 but I am just stuck on question 9C, I know the answer is 500N (as it
777dan777 [17]

Step-by-step explanation:

I know you've already done parts a and b, but I'll show the work for that before I do c.

Draw two free body diagrams, one for the car and one for the trailer.  The car pulls the trailer forward with a tension force T, so the trailer pulls backward on the car with an equal and opposite force T.

The car also has a 1200 N forward force from the engine, and a 200 N backwards force from resistance.

The trailer has a backwards resistance force of 100 N.

Sum of forces on the car:

∑F = ma

1200 − 200 − T = 900a

1000 − T = 900a

Sum of forces on the trailer:

∑F = ma

T − 100 = 300a

To solve the system of equations, first add the equations together.

1000 − 100 = 1200a

900 = 1200a

a = 0.75 m/s²

Plug back into either equation to find the tension force:

T = 325 N

Now for part c, draw new free body diagrams for the car and trailer.  This time, the car is pushing back on the trailer to slow it down.  So the trailer is pushing forward on the car with an equal and opposite force.  The magnitude of that tension force is given to be 100 N.

The car also has a backwards 200 N force from resistance, and a backwards brake force F.

The trailer has a backwards 100 N force from resistance.

Sum of forces on the car:

∑F = ma

100 − 200 − F = 900a

-100 − F = 900a

Sum of forces on the trailer:

∑F = ma

-100 − 100 = 300a

-200 = 300a

a = -⅔

Plugging into the first equation:

-100 − F = 900 (-⅔)

-100 − F = -600

F = 500 N

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3 years ago
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Answer:

AA only

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