3 years ago

Fifth root of thirteen

Mathematics

1 answer:

oee [108]3 years ago

3 0

Answer:

1.67027765233

this is the answer mate

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When a 40-foot pole casts a shadow of

Semenov [28]

Set up a ratio with height over shadow height:

40/30 = x/15

Cross multiply:

40 x 15 = 30x

600 = 30x

Divide both sides by 30:

X = 20

The tree is 20 feet tall.

5 0

3 years ago

Read 2 more answers

√47 rounded to the hundredth place

ankoles [38]

Answer:

6.855

Step-by-step explanation:

I don't know how to explain that, you can use a calculator.

6 0

3 years ago

I need help I can’t figure it out

dusya [7]

Answer:

$\large\boxed{\dfrac{41p}{70q^4(r-9)^2}}$

Step-by-step explanation:

$\dfrac{2p^4}{5q^5(r-9)^3}\cdot\dfrac{41q(r-9)}{28p^3}\\\\=\dfrac{2\!\!\!\!\diagup^1\cdot41}{5\cdot28\!\!\!\!\!\diagup_{14}}\cdot\dfrac{q\!\!\!\!\diagup}{q^{5\!\!\!\!\diagup^4}}\cdot\dfrac{(r\!\!\!\!\!\!{--}-9)\!\!\!\!\!\!{--}}{(r-9)^{3\!\!\!\!\diagup^2}}\cdot\dfrac{p^{4\!\!\!\!\diagup}}{p^3\!\!\!\!\!\!\diagup}\\\\=\dfrac{41}{70}\cdot\dfrac{1}{q^4}\cdot\dfrac{1}{(r-9)^2}\cdot\dfrac{p}{1}\\\\=\dfrac{41p}{70q^4(r-9)^2}$