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2 years ago

What is the product of (2p + 7)(3p2 + 4p – 3)?

Mathematics

1 answer:

Nady [450]2 years ago

3 0

Answer:

(2p+7)(3p²+4p-3)

2p(3p²+4p-3)+7(3p²+4p+3)

(6p³+8p²-6p)+ (21p²+28p+21)

6p³+29p²-6p+21

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WILL GIVE BRAINLIEST!!!

Mnenie [13.5K]

The answer to your question is 34.13%

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3 years ago

HELP !!! Use the elimination method to solve the system of equations.

Dafna1 [17]

Answer:

C. (5, -3)

Step-by-step explanation:

4x + 3y = 11

x - y = 8 (multiply by 3) --> 3x - 3y = 24

4x + 3y = 11

3x - 3y = 24

---------------------add

7x = 35

x = 5

5 - y = 8

y = 5 - 8

y = -3

(5, -3)

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2 years ago

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HELP ASAP PLEASE.

ValentinkaMS [17]

Answer:

pls give brainlist

Step-by-step explanation:

a. $\frac{50 blocks (L)}{9 blocks(W)\\}$ 50 x 9 = 450. A=450

b. I can scale the unit rate of the area given above and largen the area of the map. Ex. $\frac{7}{3}$ = $\frac{21}{6}$ <---(scaled by 2)

Then we can enlarge it however we want so,

$\frac{21}{6}$ = $\frac{42}{12}$ <--(scaled by 2 once more)

If the statement above is true then we can scale times 4.

$\frac{7}{3}$ = $\frac{42}{12}\\$ <---(scaled by 4)

A= $\frac{42}{12\\}$

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2 years ago

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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

2 years ago

Solve for x! Please try and show work. ANSWER ASAP SCAM ANSWERS WILL BE REPORTED. Also will award brainliest to CORRECT answer

Tamiku [17]

Answer:

x = 12 degrees

Step-by-step explanation:

m < LNM = (70+136)/2

= 206 / 2

= 103 degrees

So 7x + 19 = 103

7x = 103 - 19 = 84

x = 84/7

x = 12.

8 0

3 years ago