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3 years ago

Help just look at the photo please

Mathematics

1 answer:

Katarina [22]3 years ago

6 0

Answer:

is this Baldy's basics question 2 wt#

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What is (0.55x - 0.8) - (0.6x - 0.55)?

storchak [24]

The answer is 2054 yes ruega

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3 years ago

Write an equation for the line that passes through (-2,4) and (0,7)

xz_007 [3.2K]

Answer:

y = 3/2x + 7

Step-by-step explanation:

The slops is rise over run, so in this case the slope is 3/2 because it goes up 3 units while going 2 units to the right.

The points (0,7) is on the y axis. This means that 7 is the y intercept

With this information, we can follow the format of the slope intercept equation: y = mx + b

m stands for the slope and b stands for the y intercept. Plugging the information in, our equation is:

y = 3/2x + 7

8 0

3 years ago

What type of

ololo11 [35]

undefined

because zero slope is side ways

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3 years ago

Which statement is true about the net and the solid it can form?

Sever21 [200]

The length of side c will be 2 m.

3 0

2 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