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2 years ago

What are the zeros of the polynomial function f(x) = x3 − 2x2 − 24x?

Mathematics

1 answer:

Morgarella [4.7K]2 years ago

4 0

Answer:

x = 0, x = -4, and x = 6

Step-by-step explanation:

To find the zeros of this polynomial, we can begin by factoring out a common factor of each term. 'x' is a common factor. We can distribute this variable out, giving us:

f(x) = x(x²- 2x- 24)

Now, factor the polynomial inside of the parenthesis into its simplest form. Factors of -24 that add up to -2 are -4 and 6.

f(x) = x( x + 4) (x - 6)

From this, we can derive the zeros x = 0, x = -4 and x = 6.

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Question: Find The Sum Of The Integers From -6 To 58 Of Ss. O 1,500 O 1,734 O 1,690 O 1,621

REY [17]

Answer:

The Sum Of The Integers From -6 To 58 is 1690.

Step-by-step explanation:

Given,

$a=-6$

$T_n=58$

We have to find out the sum of integers from -6 To 58.

Firstly we will find out the total number of terms that is 'n'.

Here $a_1=-6\ and\ a_2=-5$

$\therefore d=a_2-a_1=-5-(-6)=-5+6=1$

Now we use the formula of A.P.

$T_n=a+(n-1)d$

On substituting the values, we get;

$58=-6+(n-1)1\\\\n-1=58+6\\\\n-1=64\\\\n=64+1=65$

So there are 65 terms in between -6 To 58.

That means we have to find the sum of 65 terms in between -6 To 58.

Now we use the formula of Sum of n_terms.

$S_n=\frac{n}{2}(2a-(n-1)d)$

On substituting the values, we get;

$S_{65}=\frac{65}{2}(2\times-6+(65-1)1)\\\\S_{65}=\frac{65}{2}(-12+64)\\\\S_{65}=\frac{65}{2}\times52\\\\S_{65}=65\times26=1690$

Hence The Sum Of The Integers From -6 To 58 is 1690.

5 0

3 years ago

Please help with number 10

Flura [38]

Answer:

see below for a drawing
the area of one of the trapezoids is 20 units²

Step-by-step explanation:

No direction or other information about the desired parallelogram is given here, so we drew one arbitrarily. Likewise for the segment cutting it in half. It is convenient to have the bases of the trapezoids be the sides of the parallelogram that are 5 units apart.

The area of one trapezoid is ...

A = (1/2)(b1 +b2)h = (1/2)(3+5)·5 = 20 . . . . square units

The sum of the trapezoid base lengths is necessarily the length of the base of the parallelogram, so the area of the trapezoid is necessarily 1/2 the area of the parallelogram. (The area is necessarily half the area of the parallelogram also because the problem has us divide the parallelogram into two identical parts.)

3 0

2 years ago

Two and one-third times two and three-fourths.

vesna_86 [32]

Answer:

$6\frac{5}{12}$

Step-by-step explanation:

$2\frac{1}{3}* 2\frac{3}{4}\\\rule{150}{0.5}\\\frac{7}{3} * \frac{11}{4}\\\\\frac{7*11}{3*4}\\\\\frac{77}{12}\\\\\text{77 goes into 12 6 times with 5 left over.}\\\\\boxed{6\frac{5}{12}}$