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VashaNatasha [74]
3 years ago
5

Lynne is driving at a constant rate of 48 miles per hour.how far will she drive in 4 hours

Mathematics
1 answer:
Kryger [21]3 years ago
7 0

48*4=192mph

Answer is 192mph

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I need help please it would really help
Softa [21]

Answer:

B

Step-by-step explanation:

V = 4/3 * radius^3 * pi = 4/3 * 4^3 * pi = 256 /3 * pi

7 0
3 years ago
A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a
kkurt [141]

Answer:

b=h=\sqrt{6} m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box=2b^2+4bh

2b^2+4bh=36

b^2+2bh=18

2bh=18-b^2

h=\frac{18-b^2}{2b}

Volume of box, V=b^2h

Substitute the values

V=b^2\times \frac{18-b^2}{2b}

V=\frac{1}{2}(18b-b^3)

Differentiate w. r.t b

\frac{dV}{db}=\frac{1}{2}(18-3b^2)

\frac{dV}{db}=0

\frac{1}{2}(18-3b^2)=0

\implies 18-3b^2=0

\implies 3b^2=18

b^2=6

b=\pm \sqrt{6}

b=\sqrt{6}

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

\frac{d^2V}{db^2}=-3b

At  b=\sqrt{6}

\frac{d^2V}{db^2}=-3\sqrt{6}

Hence, the volume of box is maximum at b=\sqrt{6}.

h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}

h=\frac{18-6}{2\sqrt{6}}

h=\frac{12}{2\sqrt{6}}

h=\sqrt{6}

b=h=\sqrt{6} m

7 0
3 years ago
Find atleast four equivalent fractions for each fraction 8/18
sergiy2304 [10]

16/36, 24/54, 32/72 and 40/90.

these are the equivalent fractions of 8/18

PLEASE MARK MY ANSWER AS BRAINLIEST

5 0
3 years ago
Read 2 more answers
A band is receiving $650.00 for playing at a festival. Their manager takes 15% of the money received for each show, and the band
Dafna1 [17]
650-15%= 552.50 (they pay the manager 97.50) which leaves 552.50 profit 10% goes to ST(55.25 to the ST) which is 497.25 left for the band
5 0
3 years ago
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Find the volume of the solid obtained by rotating the region bounded by y=4x and y=2sqrt(x) about the line x=6.
Pie

Check the picture below.

so by graphing those two, we get that little section in gray as you see there, now, x = 6  is a vertical line, so we'll have to put the equations in y-terms and this is a washer, so we'll use the washer method.

y=4x\implies \cfrac{y}{4}=x\qquad \qquad y=2\sqrt{x}\implies \cfrac{y^2}{4}=x~\hfill \begin{cases} \cfrac{y}{4}=x\\\\ \cfrac{y^2}{4}=x \end{cases}

the way I get the radii is by using the "area under the curve" way, namely, I use it to get R² once and again to get r² and using each time the axis of rotation as one of my functions, in this case the axis of rotation will be f(x), and to get R² will use the "farthest from the axis of rotation" radius, and for r² the "closest to the axis of rotation".

\stackrel{R}{\stackrel{f(x)}{6}-\stackrel{g(x)}{\cfrac{y^2}{4}}}\qquad \qquad \stackrel{r}{\stackrel{f(x)}{6}-\stackrel{g(x)}{\cfrac{y}{4}}}~\hfill \stackrel{R^2}{\left( 6-\cfrac{y^2}{4} \right)^2}-\stackrel{r^2}{\left( 6-\cfrac{y}{4} \right)^2} \\\\\\ \stackrel{\textit{doing a binomial expansion and simplification}}{3y-3y^2-\cfrac{y^2}{16}+\cfrac{y^4}{16}}

now, both lines if do an equation on where they meet or where one equals the other, we'd get the values for y = 0 and y = 1, not surprisingly in the picture.

\displaystyle\pi \int_0^1\left( 3y-3y^2-\cfrac{y^2}{16}+\cfrac{y^4}{16} \right)dy\implies \pi \left( \left. \cfrac{3y^2}{2} \right]_0^1-\left. y^3\cfrac{}{} \right]_0^1-\left. \cfrac{y^3}{48}\right]_0^1+\left. \cfrac{y^5}{80} \right]_0^1 \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{59\pi }{120}~\hfill

7 0
2 years ago
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