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riadik2000 [5.3K]
3 years ago
13

Round 38,237 to nearest ten thousand

Mathematics
2 answers:
dimaraw [331]3 years ago
4 0
The answer is 40,000 
hope it helps!
Alik [6]3 years ago
3 0
I think its something like 20,000 Im sorry i might be wrong
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I strongly disagree bc it was an corporation for the public in 1995
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Jina is saving money to buy a bike. She has $63 and is going to save an additional $9 each week. The bike costs $198. In how man
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Answer:

Jina will have enough money to buy the bike in 15 weeks

Step-by-step explanation:

y = mx + b

198 = 9x + 63

Subtract 63 from both sides;

135 = 9x

Divide both sides by 9;

x = 15

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What decimals are equivalent to (5×10)+(2×1/100)?
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The decimals are equivalent to given expression are 50.020 and 50.02.
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2 years ago
1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.
xxTIMURxx [149]

First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}

(If you were to plot the actual curve, you would have both y=x^{2/3} and y=-x^{2/3}, but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx

We have

y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}

Then in the integral,

\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx

Substitute

u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx

This transforms the integral to

\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du

and computing it is trivial:

\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)

We can simplify this further to

\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}

7 0
3 years ago
The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of
Lilit [14]

Check the picture below.

\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill

\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}

\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1

7 0
3 years ago
Read 2 more answers
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