Can you do multiplication.

Add the following polynomials, then place the answer in the proper location on the grid.

aivan3 [116]

Answer:

$x^3 - 6x + 2$

Step-by-step explanation:

The given polynomials can be added together by adding like terms together as shown below:

=> Add together, the coefficient of all terms that have the power of 3.

$(2x^3) + (x^3) + (-3x^3)$

$2x^3 + x^3 - 3x^3$

$3x^3 - 3x^3$

$x^3$

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

$-4x^2 + 8x^2$

$4x^2$

=> $(6x) + (-8x) + (-4x)$

$6x - 8x - 4x$

$-6x$

=> $(-3) + (12) +(-7)$

$-3 + 12 - 7$

$2$

The answer = $x^3 - 6x + 2$

5 0

3 years ago

If f(x) = 2x and g(x) = 5x + 2, what is (f ÷ g)(x) when x = 3

anygoal [31]

Answer:

$= \frac{2x}{5x + 2} \\ = \frac{2 \times 3}{5 \times 3 + 2} \\ = \frac{3}{17}$

3 0

3 years ago

An astronaut weighs 685 N on earth. What would the astronaut's weight (in FORCE) be on Jupiter where gravity= 25.9 m/s2? (*Tip:

VladimirAG [237]

Answer:

Step-by-step explanation:

I'm sure by now you have learned the difference between mass and weight. Mass will never change regardless of where something is while weight changes depending upon the pull of gravity. If we want the mass, then we have to take the weight on Earth and divide by its pull of gravity. The equation for that will be

W = mg where W is the weight in Newtons, m is mass and g is gravity.

685 = m(9.8) so

m = 7.0 × 10¹ kg

Now that we know that mass, and also because we know that the mass is constant no matter where the astronaut is, we can find his weight on Jupiter.

W = (7.0 × 10¹)(25.9) so

W = 1800 N

3 0

2 years ago

Find the first Taylor polynomial T1(x) for f(x)=exp(x) based at b=0:

avanturin [10]

Answer:

f(x) = 1 + x + (x²/2!) + (x³/3!) + ....... = Σ (xⁿ/n!) (Summation from n = 0 to n = ∞)

Step-by-step explanation:

f(x) = eˣ

Expand using first Taylor Polynomial based around b = 0

The Taylor's expansion based around any point b, is given by the infinite series

f(x) = f(b) + xf'(b) + (x²/2!)f"(b) + (x³/3!)f'''(b) + ....= Σ (xⁿfⁿ(b)/n!) (Summation from n = 0 to n = ∞)

Note: f'(x) = (df/dx)

So, expanding f(x) = eˣ based at b=0

f'(x) = eˣ

f"(x) = eˣ

fⁿ(x) = eˣ

And e⁰ = 1

f(x) = 1 + x + (x²/2!) + (x³/3!) + ....... = Σ (xⁿ/n!) (Summation from n = 0 to n = ∞)

6 0

3 years ago

I need help with my homework

maxonik [38]

Answer:

The measure of the arc RQ is 160°

Step-by-step explanation:

<em>In any circle, the measure of the subtended arc of an inscribed angle is twice the measure of this inscribed angle</em>

<em>The measure of any circle is 360°</em>

Let us use these rules to solve the question

In the given figure

∵ ∠RPQ is an inscribed angle subtended by arc RQ

∴ m of arc RQ = 2 × m∠RPQ

∴ m∠RPQ = 5x + 15

∴ m of arc RQ = 2 × (5x + 15)

→ Multiply the bracket by 2

∴ m of arc RQ = 2(5x) + 2(15)

∴ m of arc RQ = 10x + 30

∵ The measure of the circle = 360°

∵ The sum of the measures of arcs RQ, QP, and PR equals the measure

of the circle

∵ m of arc QP = 6x - 12

∵ m of arc PR = 9x + 17

→ Add them and equate the sum by 360°

∴ 10x + 30 + 6x - 12 + 9x + 17 = 360

→ Add the like terms in the left side

∵ (10x + 6x + 9x) + (30 - 12 + 17) = 360

∴ 25x + 35 = 360

→ Subtract 35 from both sides

∴ 25x = 325

→ Divide both sides by 25 to find x

∴ x = 13

→ To find the measure of arc RQ substitute x by 13 in its measure

∵ m of arc RQ = 10(13) + 30 = 130 + 30

∴ m of arc RQ = 160°

∴ The measure of the arc RQ is 160°

4 0

2 years ago