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3 years ago

Can someone please help me with this. is it A B C. OR D

Mathematics

1 answer:

creativ13 [48]3 years ago

6 0

A because mean is average.

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How does 8 4/3 = 9 1/3?

irga5000 [103]

You're question asks why the given mixed numbers are equal.

In fact, they're equal.

Lets make it easier for you to understand:

8 4/3 has a improper fraction "4/3", that fraction is over 1 whole.

You could look at 8 4/3 as:

8 + 3/3 + 1/3

The 3/3 turns into 1:

8 + 1 + 1/3

9 + 1/3

9 1/3

You would see that it turns into 9 1/3, therefore they are equal.

7 0

3 years ago

Help dued in 10 mintues

melamori03 [73]

You can find the improper fractions of both.
14 1/2=29/2
1 3/5=8/5
Divide the two numbers
(29/2)/(8/5) and you will get 145/16 or 9.0625.
Therefore, you can fit 9 books on the bookshelf.

4 0

4 years ago

Which equation is in point-slope form and depicts the equation of this line?

crimeas [40]

Answer:

i believe the answer is A

Step-by-step explanation:

(y + 4) = -(1)/(3)(x + 1)

distribute -1/3 to (x+1)

(y+4)= -1/3x + -1/3

subtract 4 from each side

y= -1/3x - 13/3

4 0

3 years ago

X+21.4=47.36<br> x=25.96

mixer [17]

Answer:

x=25.96

25x=649

Step-by-step explanation:

6 0

3 years ago

Jamison graphs the function ƒ(x) = x4 − x3 − 19x2 − x − 20 and sees two zeros: −4 and 5. Since this is a polynomial of degree 4

Art [367]

We are given polynomial function f(x)= $x^4-x^3-19x^2-x-20$

We are given two zeros -4 and 5.

Therefore, x=-4 and x=5 is given.

So, the two factors of the polynomial will be (x+4)(x-5).

Let us write some steps to factor the given polynomial.

Let us take above factor (x+4) first.

On factoring (x+4) we get factors

$x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x^3-5x^2+x-5\right)$

$\mathrm{:Let \ us \ factor}\:x^3-5x^2+x-5 \ now.$

Grouping

$=\left(x^3-5x^2\right)+\left(x-5\right)$

Factoring out gcf from each group, we get

$=x^2\left(x-5\right)+\left(x-5\right)$

$=\left(x-5\right)\left(x^2+1\right)$

So, the final factored form of given polynomial will be

$x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x-5\right)\left(x^2+1\right)$

For first to factors (x+4) and (x-5) we have given roots: -4 and 5.

Let us find the root of third factor we got.

$x^2+1=0$

Subtracting both sides by 1.

x^2 = - 1.

On taking square root on both sides, we get a square root(-1)

$x=\sqrt{-1}=$ + i and -i.

Those are imaginary solutions.

Therefore, correct option is B) No, there are two imaginary solutions.

5 0

4 years ago