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kramer
2 years ago
14

Simplify (x+4)(x^2-6x

Mathematics
1 answer:
lesya [120]2 years ago
4 0

Answer:

The answer is x^3 - 2x^2 - 24x

Step-by-step explanation:

(x+4)(x^2-6x)

= (x+4)(x^2 + -6x)

= (x)(x^2) + (x)(-6x) + (4)(x^2) + (4)(-6x)

=x^3 - 6x^2 + 4x^2 - 24x

=x^3 - 2x^2 - 24x

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∠K and ∠L are complementary angles. If m∠K = 3x + 3 and m∠L = 10x – 4, find the measures of both angles.
Oduvanchick [21]

Answer:

∠K =24

∠L =66

Step-by-step explanation:

first you need to find X in order to solve.

since they are Complementary they add together to equal 90 degrees

so we can set them equal to 90

13x-1=90

13x=91

x=7

now we can plug them in!

---

∠K = 3(7)+3 = 24

∠L =10(7)-4=66

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After 7.5 years, you will have $3,200.12
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Does anyone know how to solve this?
Ivahew [28]

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2 years ago
I need help on number 9
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Complement angles, sum is equal 90

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3 0
3 years ago
Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur
Bad White [126]

Looks like you have most of the details already, but you're missing one crucial piece.

\sigma is parameterized by

\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle

for 0\le u\le7 and 0\le v\le\frac{2\pi}3, and a normal vector to this surface is

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle

with norm

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So the integral of f(x,y,z)=x^2+y^2+z^2 is

\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}

6 0
2 years ago
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