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3 years ago

Someone pls answer this correctly pls ! IXL Geometry area between two shapes.

Mathematics

2 answers:

SVETLANKA909090 [29]3 years ago

5 0

Answer:

240 square centimeters

1. Find the area of the green figure, which is 529. Then find the area of the white figure, which is 289.

2. Subtract 289 from 529.

529 - 289 = 240

Anika [276]3 years ago

3 0

Answer:

$240\ cm^{2}$

Step-by-step explanation:

$A = 23^{2} - 17^{2} = 529 - 289 = 240\ cm^{2}$

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Find a polynomial function f(x) of degree 3 with real coefficients that satisfies the following conditions.

frosja888 [35]

Answer:

The polynomial will be $f(x) = \frac{4}{9}x(x - 2)^{2}$ .

Step-by-step explanation:

The three degree polynomial function f(x) has zeros at x = 0 and at x = 4 having multiplicity 2.

Therefore, (x - 0) and (x - 4)² will be factors of the polynomial f(x).

Hence, the polynomial will be f(x) = Ax(x - 4)² .............. (1), where A is any constant that we have to evaluate.

Now, given that f(5) = 20

So, from the equation (1) we have

20 = A(5)(5 - 2)²

⇒ $A = \frac{4}{9}$

Therefore, the polynomial will be $f(x) = \frac{4}{9}x(x - 2)^{2}$ (Answer)

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3 years ago

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3 years ago

Read 2 more answers

M is the midpoint of QR<br><br> M = (-2,6)<br> Q= (8,-10)<br> R = ????

Zolol [24]

Answer:

R = (-12, 22)

Step-by-step explanation:

Unfortunately it's not a straight line, so gonna need to put in some extra work. Basically think of it like breaking it into it's horizontal and vertical components then doubling them.

From Q to M you move 10 spaces to the left, so to go from M to R you will go another 10 to the left. Similarly, you start at -10 for the y value and go up 16 to 6, and you will need to go up another 16 to get to R.

So for x, 8 -10 = -2 then -2 - 10 = -12, so R's x value is -12

y we do the same thing. Gonna do it in one step though. -10 + 16 + 16 = 22

So R is (-12, 22)

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3 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

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3 years ago