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4 years ago

Im not afraid to fight for love

Mathematics

2 answers:

EastWind [94]4 years ago

7 0

I don’t think this is a question but this means the character is willing to do anything for love

svetoff [14.1K]4 years ago

4 0

Answer:

I DONT KNOW IF THIS IS A REAL QUESTION

Step-by-step explanation:

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Solve the following equation for y: −8x+ 4y = 20

MrRissso [65]

Answer:

y=5+2x

Normally I would explain it, but im in a bit of a hurry today. I apologize for the inconvenience!

8 0

3 years ago

What is the completely factored form of d4 -8d2 + 16

irinina [24]

According to Vieta's Formulas, if $x_1,x_2$ are solutions of a given quadratic equation:

$ax^2+bx+c=0$

Then:

$a(x-x_1)(x-x_2)$ is the completely factored form of $ax^2+bx+c$ .

If choose $x=d^2$ , then:

$\displaystyle x^2-8x+16=0\\\\x_{1,2}= \frac{8\pm \sqrt{64-64} }{2}=4$

So, according to Vieta's formula, we can get:

$x^2-8x+16=(x-4)(x-4)= (x-4)^2$

But $x=d^2$ :

$d^4-8d^2+16=(d^2-4)^2=[(d+2)(d-2)]^2=(d+2)^2(d-2)^2$

8 0

3 years ago

What is the side length of a square with the area of 20cm2

rodikova [14]

Answer:

S = a² => 20 = a² => a = √20 ≈ 4.5 cm

Step-by-step explanation:

5 0

4 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago

What percent of 48 is 18

arsen [322]

To find the percent you take 18 and divide it by 48, so you get 9/24 or 3/8, by using calculator you get .375 which is 37.5% then you multiply by 100 to convert a decimal to a percentage.

6 0

3 years ago