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3 years ago

Which is the best answer??

Mathematics

1 answer:

GuDViN [60]3 years ago

5 0

Answer:

$\large\boxed{C.\ (-4x+3)(x-2)}$

Step-by-step explanation:

$-4x^2+11x-6=-(4x^2-11x+6)=-(4x^2-8x-3x+6)\\\\=-\left[4x(x-2)-3(x-2)]=-(x-2)(4x-3)=(-4x+3)(x-2)$

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Lilly has 25/2 josh has 36/23 how much is there all together

mamaluj [8]

Answer:

14.0652173913

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2 years ago

(Inscribed Angles)

Alik [6]

Answer:

(D) 30 degrees

Step-by-step explanation:

An inscribed angle is an angle formed by two chords in a circle which have a common endpoint.

Theorem(Inscribed Angles Conjecture I): In a circle, the measure of an inscribed angle is half the measure of the central angle with the same intercepted arc.

By the theorem above and from the attached diagram,

Inscribed Angle p =Half of 60 degrees

$\angle p=0.5 X 60^\circ\\p=30^\circ$

The correct option is D.

4 0

2 years ago

A over b is in parentheses with the exponent -4 on the outside then there's another set of parentheses and on the outside of tho

JulijaS [17]

Answer:

Step-by-step explanation:

(((a/b)^(-4))^0)=

((b/a)^4)^0)=

((b^4)/(a^4))^0=

(ANYTHING to the power of 0 is equal to 1)

7 0

3 years ago

Which equation shows the correct use of the Distributive Property? −4(32x−12)=−15

KengaRu [80]

Answer:

-128x + 48 = -15

Step-by-step explanation:

I don't any choices to pick from but using the distributive property you will multiply the -4 by everything in the parenthesis.

-4(32x - 12) = -15

=-128x + 48 = -15.

6 0

3 years ago

Read 2 more answers

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago