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2 years ago

PLEASE ASAP I NEED HELP

Mathematics

1 answer:

Lelechka [254]2 years ago

8 0

Answer:

with wut???

Step-by-step explanation:

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Triangle DEF has coordinates D (5,2), E (5,-2), and F(-4, 0). Which of the

andreev551 [17]

Answer:

Well, I don't have further explanation to this question but I know the answer which is: The x-axis is the only line of symmetry

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3 years ago

A florist sold 14 flower bouquets yesterday, including 7 lily bouquets. Based on past data, how many of the next 20 bouquets sol

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

From past data:

Fraction of lily sold :

Total flowers / number of lily

Total flowers = 14

Number of lily = 7

Fraction of lily = 14 /7 = 1/2

Going by these ;

Expected number of lilies in the next 20 bouquets sold :

Fraction of lily * number of flower in bouquet

1/2 * 20

= 10 lilies

5 0

3 years ago

Write an equation of a line that passes through (-12, -14) with slope 6.

damaskus [11]

The slope intercept form of a line is y = mx + b

m = slope
b = y-intercept

Plug in the slope, 6, into m.

Rewrite the equation;

y = 6x + b
We need to find b, your y-intercept, to finish this equation.

Plug in your point coordinate, (x, y) ⇒ (-12, -14) into the equation.

-14 = 6(-12) + b

Solve for b to find the y-intercept.

-14 = -72 + b
58 = b

Your new equation (your answer) is<em> </em>y = 6x + 58.

6 0

3 years ago

A car dealership earns a portion of its profit on the accessories sold with a car. The dealer sold a Toyota Camry loaded with ac

WINSTONCH [101]

$24000/8 = $3000
The accessories cost $3000

4 0

3 years ago

Read 2 more answers

A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (> r) from th

jeyben [28]

Consider a circle with radius $r$ centered at some point $(R+r,0)$ on the $x$ -axis. This circle has equation

$(x-(R+r))^2+y^2=r^2$

Revolve the region bounded by this circle across the $y$ -axis to get a torus. Using the shell method, the volume of the resulting torus is

$\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx$

where $2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}$ .

So the volume is

$\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx$

Substitute

$x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt$

and the integral becomes

$\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt$

Notice that $\sin t\cos^2t$ is an odd function, so the integral over $\left[-\frac\pi2,\frac\pi2\right]$ is 0. This leaves us with

$\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt$

Write

$\cos^2t=\dfrac{1+\cos(2t)}2$

so the volume is

$\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}$

6 0

3 years ago