Find the GCF of each pair of monominals 54gh,72g

Mathematics

1 answer:

wel3 years ago

7 0

18 is the answer I believe hopefully this is correct

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Sinx=cos19∘ What is the value of x? Enter your answer in the box.

rusak2 [61]

Remember that cos is just a translation of sin and vice versa.
So:
Sin(x) = Cos(90 - x)
Cos(x) = Sin(90 - x)

Therefore, to answer your question:
Cos(19) = Sin(90 - 19)
= Sin(71)

6 0

3 years ago

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kati45 [8]

Answer:

I really do not know.

Step-by-step explanation:

5 0

2 years ago

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The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet

vladimir1956 [14]

$\bf s(t)=\sqrt{t^3+1}
\\\\\\
\cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t)
\\\\\\
\cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies 
\cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}$

$\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies 
\boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\
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3 years ago

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Suppose the lengths of two sides of a right triangle are represented by 2x and 3 (x + 1), and the longest side is 17 units. Find

densk [106]

Answer:

x=4

Step-by-step explanation:

Step 1:-

given the lengths of two sides of a right angle are represented by 2x and 3(x+1) and longest side is 17 units.

AB = 2x and BC = 3(x+1) and longest side AC= 17

by using Pythagoras theorem