I can't get this it keeps saying its wrong.

A student is asked to find a polynomial f(x) of degree four with the zeros 2, -3, 0, and -7. What factors would the student mult

lord [1]

Answer:

(x-2), (x+3),x and (x + 7)

Step-by-step explanation:

Here, the zeros of the polynomial are 2, -3, 0 and -7.

Let’s consider a case of a quadratic equation where x = 5 is a solution. This means that x + 5 is a factor of that quadratic equation.

Now, let’s apply same approach to this polynomial. If x = 2 is a solution, then x -2 is a factor. If x = -3, then, x + 3 is a factor. If x = 0, this means x itself is a solution and lastly, if x = -7 is a solution, that means x + 7 is a factor.

Thus, the factors to multiply together are; (x-2), (x+3),x and (x + 7)

4 0

3 years ago

The sum of 3x2 + x – 7 and x² + 10 can be expressed as

Andrei [34K]

The sum of all that stuff is a number from here to there

6 0

3 years ago

Arya and Cody each manage a landscaping business.

Viktor [21]

And? Congrats? Please write the full question

3 0

2 years ago

What is the domain of the ordered pairs?

BigorU [14]

Answer:

b. {-3, -1, 2}

Step-by-step explanation:

An ordered pair is a pair of elements written as (x, y) where the first element is the input value and the second element is the output value.

The domain is the set of input values (x-values)

The range is the set of output values (y-values)

Therefore, for the given ordered pairs (-1, 0), (2, 4) and (-3, 6)

Domain: {-3, -1, 2}

Range: {0, 4, 6}

3 0

2 years ago

(I've been trying to figure this out for 3 days and I really need help)

liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

1)

$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7$

2)

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

$\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55$

3)

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

4)

pretty much the same thing, we get the volume of the cone and its top, add them up.

$\bf \stackrel{\textit{cone's volume}}{\cfrac{\pi (3)^2(8)}{3}}~~~~+~~~~\stackrel{\stackrel{\textit{half a sphere}}{\textit{top's volume}}}{\cfrac{4\pi 3^3}{3}\div 2}\implies 24\pi +18\pi \implies 42\pi ~~\approx~~131.95~in^$

8 0

3 years ago