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2 years ago

I dont really understand this

Mathematics

1 answer:

galina1969 [7]2 years ago

4 0

Answer:

1. a^-b

2.a^b/a^c

3.a^c/b^c

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katrin [286]

Answer:

0.75 + 0.25 = t

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3 years ago

Determine the percentage increase from 10 yards to 13 yards

julia-pushkina [17]

Answer:

30%

Step-by-step explanation:

In this case, we have a positive change (increase) of 30% because the new value is greater than the old value. Using this tool you can find the percent increase for any value.

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Brain-list?

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3 years ago

Two vertical poles of length 10 and 12 feet, respectively, stand 17 feet apart. A cable reaches from the top of one pole to the

Marat540 [252]

Answer:

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433} ft$

Step-by-step explanation:

let length of calble = L

distance from 10-foot pole = x ft

Moving from the top of the 10-foot pole to point x, we have

From the diagram that

$x^{2} +10^{2}=L_{1} ^{2} \\x^{2} + 100 = L_{1} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}+100}=\sqrt{L_{1} ^{2}}\\ \sqrt{x^{2}+100}= L_{1}$

Moving from point x to the top of the 12-foot pole, we have

From the diagram that

$(17-x)^{2}+12^{2}=L_{2} ^{2} \\289-34x+x^{2} +144=L_{2} ^{2}\\x^{2} -34x+433=L_{2} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}-34x+433}=\sqrt{L_{2} ^{2}} \\\sqrt{x^{2}-34x+433}=L_{2}$

Amount of cable used, L = L1 + L2

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433}$

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3 years ago

Please graph y=−1/2. thank you!

Dominik [7]

Answer:

There would be a straight line at -0.5 on the graph.

Step-by-step explanation:

Hope this helps

4 0

3 years ago

Help Me Please I will give brainliest if correct!

il63 [147K]

Answer:

9 $c^{7}$

Step-by-step explanation:

$c^{-4}$ is equal to $\frac{1}{c^{4} }$ . $\frac{9c^{3} }{\frac{1}{c^{4} } }$ is equal to $9c^{7}$ .

3 0

2 years ago

Read 2 more answers