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Sveta_85 [38]
3 years ago
12

Prove that 3: 8 is equivalent to 12: 32

Mathematics
1 answer:
Kazeer [188]3 years ago
6 0
3:8 can also be written as 3/8

12:32 can also be written as 12/32

\frac{12}{32} = \frac{3\times 4}{8\times 4} = \frac{3}{8}

So thus 3:8 is equivalent to 12:32
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Solve each equation for m -5 =8
Diano4ka-milaya [45]

Answer:

m = 13

Step-by-step explanation:

Given

m - 5 = 8 ( isolate m by adding 5 to both sides of the equation )

m = 8 + 5 = 13

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4 years ago
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Cube root of 8 to the 2/3
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I hope this helps you

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Can some tell me the answer for 6a and b please!
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In ΔVWX, the measure of ∠X=90°, WV = 17, VX = 15, and XW = 8. What ratio represents the sine of ∠W?
Dahasolnce [82]

Answer:

Sin \ \angle W=\frac{15}{17}

Step-by-step explanation:

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Sin \angle (base)=\frac{Opposite \ Length}{Hypotenuse}

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Sin \ \angle W=\frac{Opp}{Hyp}\\\\=\frac{15}{17}

Hence, sine of  ∠W is given represented by the ratio \frac{15}{17}

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4 years ago
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Help calculus module 8 DBQ<br><br> please show work
igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of R(t) at the endpoints of each subinterval. The sum is then

\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since R is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval [a,b], it guarantees the existence of some c\in(a,b) such that

\dfrac{R(b)-R(a)}{b-a)=R'(c)

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such c for which R'(c)=0. I would chalk this up to not having enough information.

3. R(t) gives the rate of water flow, and R(t)\approx W(t), so that the average rate of water flow over [0, 8] is the average value of W(t), given by the integral

R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt

If doing this by hand, you can integrate by parts, setting

u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt

\mathrm dv=\mathrm dt\implies v=t

R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)

For the remaining integral, consider the trigonometric substitution t=\sqrt 7\tan s, so that \mathrm dt=\sqrt 7\sec^2s\,\mathrm ds. Then

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)

\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}

or approximately 3.0904, measured in gallons per hour (because this is the average value of R).

4. By the fundamental theorem of calculus,

g'(x)=f(x)

and g(x) is increasing whenever g'(x)=f(x)>0. This happens over the interval (-2, 3), since f(x)=3 on [-2, 0), and -x+3>0 on [0, 3).

5. First, by additivity of the definite integral,

\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt

Over the interval [-2, 0), we have f(x)=3, and over the interval [0, 6], f(x)=-x+3. So the integral above is

\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}

6 0
3 years ago
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