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Elza [17]
2 years ago
15

PLZ HELP FAST (multiple choice)​

Mathematics
2 answers:
Nataly_w [17]2 years ago
8 0
The answer is 0 (my sentence needs to have 20 characters so I’m typing extra)
never [62]2 years ago
6 0

Answer:

C. 0

Step-by-step explanation: I think if i am wrong im sorry

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In a jar of coins, 18 out of the 40 coins are dimes. Express the fraction of the coins
Step2247 [10]

Answer:

Percent: 20%

Fraction: 1/5

Decimal: 0.20

Step-by-step explanation:

8:40*100 =

( 8*100):40 =

800:40 = 20%

Percent to fraction:

20%=20/100

= 0.2

=0.2×10/10

=2/10

=1/5

Percent to decimal:

20/100 = 0.20

8 0
3 years ago
Which system of inequalities is shown? A. Y>x y<2 b. Y 2 D. Y>x y>2
AURORKA [14]

Answer: y > x, y > 2

The horizontal line goes through 2 on the y axis. This boundary line is represented by the equation y = 2, since every point on this line has a y coord of 2. The shading above it means that the inequality is y > 2. Every point in the shaded region of y > 2 has a y coord that is larger than 2.

The other inequality is y > x because we shade above the dashed boundary line y = x, which is that slanted dashed line.

Combining the two regions of y > 2 and y > x leads to what is shown.

6 0
1 year ago
A tractor can plant a field at a rate of 2.5 acres per 5 minutes. If a mammoth farm measuring 4 square miles needs planting, how
REY [17]

Step-by-step explanation:

hey I hope to the mall is to be a part the highest in a few days and reaction is the area of parallel with you and I hope you feel better soon I have

7 0
2 years ago
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
PLEASE HELP ME ASAP
pickupchik [31]

lol we have the same question on our test besides i’m in middle school Answer: idfk

Step-by-step explanation: don’t ask me i’m failing math

7 0
3 years ago
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