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3 years ago

For y = 3x - 2x^2 +5x^3 show how to find the value for y, when x = 3. Show your work

2 answers:

8 0

$y \: = \: f(x) \: = \: 3x \: - \: 2 {x}^{2} \: + \: 5 {x}^{3}$
Here, y is a cubic function of x.
When x = 3,
$y \: = \: f(3) \: = \: 3(3) \: - \: 2 {(3)}^{2} \: + \: 5 {(3)}^{3}$
$y \: = \: 9 \: - \: 18 \: + \: 135$
$y \: = \: f(3) \: = \: 126$
Hence, when x = 3, y = 126.

liq [111]3 years ago

7 0

Y=3(3)-2(3)^2+5(3)^3
y=9-2(3)^2+5(3)^3
y=9-6^2+5(3)^3
y=9-36+5(3)^3
y=9-36+15^3
y=9-36+3,375
y=-27+3,375
y=3,348

Just do the steps as I showed you. Just go little by little for each thing so that you don't mess up our get super confused.

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Question 5 (2 points) <br> Find x, y, and z.

Verizon [17]

Answer:

Its actually really easy.

Y is = to 76 degree (Vertically Opposite Angle)

The same way x = z (Vertically Opposite Angles)

So as y = 76 and given angle also 76 we'll add them. You should get 152. Now we know x = z. And we also know angles around a point add up to 360. So 360 - 152 gives 208. Since there are two angles we'll divided them by 2. So you now have 104

Given angle : 76

Y : 76

X : 104

Z : 104

8 0

3 years ago

Read 2 more answers

If the radius of circle A is 6 units, calculate the area of the shaded region...

meriva

The formula of an area of a triangle:

$A_{\triangle}=\dfrac{base\cdot height}{2}$

We have:

$\triangle_{FDE}:\ base=height=1\\\triangle_{GDH}:\ base=height=9$

substitute:

$A_{\triangle_{FDE}}=\dfrac{1\cdot1}{2}=0.5\\\\A_{\triangle_{GDH}}=\dfrac{9\cdot9}{2}=40.5$

The formula of an area of a circle:

$A_O=\pi r^2$

We have $r=6$

Substitute:

$A_O=\pi\cdot6^2=36\pi\approx36\cdot3.14=113.04$

The area of the shaded region

$A=A_O-(A_{\triangle_{FDE}}+A_{\triangle_{GDH}})\\\\A=113.04-(0.5+40.5)=113.04-41=72.04\ u^2$

4 0

3 years ago

Question 1 Part B (3 points): Using the GCF that you chose in part A, factor that GCF out of the polynomial:

Aleks [24]

The factor of the polynomial 18x³ + 6x²y - 9x² - 3xy will be 3x(6x² - 2xy - 3x - y). Then the correct option is C.

<h3>What is a factorization?</h3>

It is a method for dividing a polynomial into pieces that will be multiplied together. At this moment, the polynomial's value will be zero.

The polynomial is given below.

⇒ 18x³ + 6x²y - 9x² - 3xy

Then the factor of the polynomial will be

⇒ 18x³ + 6x²y - 9x² - 3xy

⇒ 3x(6x² - 2xy - 3x - y)

More about the factorization link is given below.

brainly.com/question/6810544

#SPJ1

5 0

2 years ago

Evaluate the surface integral ∫sf⋅ ds where f=⟨2x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with

skad [1K]

Parameterize S by the vector function

$\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

$\vec n = \dfrac{\partial\vec s}{\partial v} \times \dfrac{\partial \vec s}{\partial u} = \left\langle 16 \cos(u) \sin^2(v), 16 \sin(u) \sin^2(v), 16 \cos(v) \sin(v) \right\rangle$

The integral of the field over S is then

$\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \vec f(\vec s) \cdot \vec n \, du \, dv$

$\displaystyle = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv$