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kotykmax [81]
3 years ago
8

Help me solve this..! Please!

Mathematics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

x=2

Solve for x by simplifying both sides of the equation, then isolating the variable.


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62.5is what I got but i think you round it

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A flying squirrel jumped from a tree 11 feet in the air at an initial velocity of 9 feet per sencond. The equation h=-2t^2+9t-11
vovikov84 [41]
This must be on the moon as the acceleration due to gravity in this equation must be around 1/8 that on earth. :)  Anyway...

h=-2t^2+9t+11

A)

dh/dt=-4t+9, when velocity, dh/dt=0, it is the maximum height reached

dh/dt=0 only when 4t=9, t=2.25 seconds

h(2.25)=21.125 ft  (21 1/8 ft)

B)

As seen in A), the time of maximum height was 2.25 seconds after the squirrel jumped.

C)

The squirrel reaches the ground when h=0...

0=-2t^2+9t+11

-2t^2-2t+11t+11=0

-2t(t+1)+11(t+1)=0

(-2t+11)(t+1)=0, since t>0 for this problem...

-2t+11=0

-2t=-11

t=5.5 seconds.

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HELP!!! What is the MAD for these numbers? 5 22 14 13 18 7 9 12
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Find the length of the curve y = 3/5x^5/3 - 3/4x^1/3 + 6 for 1 < = x < = 8. The length of the curve is . (Type an exact an
Mashutka [201]

Answer:

\sqrt\frac{387}{20}

Step-by-step explanation:

Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx

y=\dfrac{3}{5}x^{\frac{5}{3}}-  \dfrac{3}{4}x^{\frac{1}{3}}+6

\frac{dy}{dx} =x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}}

1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})

=\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}}

For the Interval 1\leq x\leq 8

Length of the Curve =\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx\\

Using T1-Calculator

=\sqrt\frac{387}{20}

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4 years ago
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