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3 years ago

Can someone help me again pleaseeee :')

Mathematics

1 answer:

GenaCL600 [577]3 years ago

7 0

The answer is I because S is at -2 and T is at 4.

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Which of the following represents the prime factortzation of 72?

rjkz [21]

Answer:

2³ x 3²

Step-by-step explanation:

The prime factorization of a number is the process of breaking down a number to a set of prime numbers that their product will give the original number:

2 x 36

2 x 2 x 18

2 x 2 x 2 x 9

2 x 2 x 2 x 3 x 3

So;

The prime factorization is 2³ x 3²

8 0

3 years ago

Triangle RST is shown. What is the y-coordinate of the final image of vertex T after the triangle is reflected over the x-axis f

gavmur [86]

Answer:

R¹(-7, -3) , S¹(-2, -1), T¹(-5, 4)

Step-by-step explanation:

Step(i):-

From the graph, the points are R(-4,5), S(1,3),T(-2,-2)

First, the points are reflected over the x-axis then transform in to

(x,y)→(x,-y)

R(-4,5)→R¹(-4,-5)

S(1,3)→S¹(1,-3)

T(-2,-2)→T¹(-2,2)

Step(ii):-

Next, The points are shifted to 3 units left then the transform

(x,y)→ (x -3 ,y)

R¹(-4,-5)→ R¹(-4-3, -5) → R¹(-7, -5)

S¹(1,-3) → S¹(1-3, -3) → S¹(-2, -3)

T¹(-2,2) → T¹(-2-3, 2)→ T¹(-5, 2)

Step(iii):-

Now again the points are shifted to '2' units up then transform to

(x,y) )→ (x ,y+2)

R¹(-7, -5+2) → R¹(-7, -3)

S¹(-2, -3+2) → S¹(-2, -1)

T¹(-5, 2+2) → T¹(-5, 4)

4 0

3 years ago

Act scene where Macbeth and last Macbeth plan to kill king Duncan

inessss [21]

Answer:

Step-by-step explanation:

inside the castle

5 0

3 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

Someone help me asap?

Rzqust [24]

1. 66

2. -36

3. 10

4. 1

8 0

3 years ago

Read 2 more answers