All categories

2 years ago

Please please do a step by step of how to solve the equation and the answer! please and thank you!

Mathematics

1 answer:

Yuliya22 [10]2 years ago

6 0

Answer:

Value of x using quadratic formula (x = -4) and (x = -5)

Step-by-step explanation:

Given:

Equation

x² + 9x + 20 = 0

Find:

Value of x using quadratic formula

Computation:

Quadratic formula = [-b±√b²-4ac] / 2a

Given equation;

x² + 9x + 20 = 0

a = 1 , b = 9 , c = 20

By putting value

[-9±√9²-4(1)(20)] / 2(1)

[-9±√81-80] / 2

[-9±√1] / 2

(-9 + 1) / 2 , (-9 - 1) / 2

-8 / 2 , 10 / 2

-4 , - 5

Value of x using quadratic formula (x = -4) and (x = -5)

You might be interested in

Find the surface area no links no pdfs just type it into the answer section will give brainliest

rewona [7]

Check the picture below.

let's firstly convert the mixed fractions to improper fractions.

$\stackrel{mixed}{7\frac{1}{2}}\implies \cfrac{7\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{15}{2}} ~\hfill \stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{\Large Areas}}{\stackrel{two~triangles}{2\left[ \cfrac{1}{2}\left(\cfrac{15}{2} \right)(10) \right]}~~ + ~~\stackrel{\textit{three rectangles}}{(10)(15)~~ + ~~\left( \cfrac{15}{2} \right)(15)~~ + ~~\left( \cfrac{25}{2} \right)(15)}} \\\\\\ 75~~ + ~~150~~ + ~~112.5~~ + ~~187.5\implies \boxed{525}$

8 0

2 years ago

Solve the equation using square roots. x2 + 10 = 9

Oksana_A [137]

$x^2 + 10 = 9$

Subtract 10 from both sides.

$x^2 = -1$

Since the square of a number, x, equals a negative number, -1, the answer cannot be a real number because the square of a real number is always non-negative. Therefore, there is no real number solution to this equation.

Answer: B. no real number solutions

5 0

3 years ago

Brainliest is given to the correct answer.

Svet_ta [14]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Simplify the expression state any excluded values 2a^2-4a+2 --------------- 3a^2-3

chubhunter [2.5K]

Answer:

The simplified form is $\dfrac{2(x-1)}{3(x+1)}$ .

$x =1$ is the excluded value for the given expression.

Step-by-step explanation:

Given:

The expression given is:

$\dfrac{2a^2-4a+2}{3a^2-3}$

Let us simplify the numerator and denominator separately.

The numerator is given as $2a^2-4a+2$

2 is a common factor in all the three terms. So, we factor it out. This gives,

$=2(a^2-2a+1)$

Now, $a^2-2a+1=(a-1)(a-1)$

Therefore, the numerator becomes $2(a-1)(a-1)$

The denominator is given as: $3a^2-3$

Factoring out 3, we get

$3(a^2-1)$

Now, $a^2-1$ is of the form $a^2-b^2=(a-b)(a+b)$

So, $a^2-1=(a-1)(a+1)$

Therefore, the denominator becomes $3(a-1)(a+1)$

Now, the given expression is simplified to:

$\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}$

There is $(x-1)$ in the numerator and denominator. We can cancel them only if $x\ne1$ as for $x=1$ , the given expression is undefined.

Now, cancelling the like terms considering $x\ne1$ , we get:

$\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}$

Therefore, the simplified form is $\dfrac{2(x-1)}{3(x+1)}$

The simplification is true only if $x\ne1$ . So, $x =1$ is the excluded value for the given expression.

8 0

3 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago