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3 years ago

Plz help. Will give brainliest

Mathematics

2 answers:

diamong [38]3 years ago

8 0

Answer:

17.1

Step-by-step explanation:

Kazeer [188]3 years ago

3 0

Answer 17. 1

Step-by-step explanation:

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I don’t know how to do this lol this is a test

Artemon [7]

Answer:

D) $x+y=-3$

Step-by-step explanation:

Basically, if you plug in the x and y-coordinates into an equation, both sides should work out to be equal to each other:

Since $-5+2\neq 3$ , we can eliminate choice A

Since $-5-2\neq -3$ , we can eliminate choice B

Since $-5-2\neq 3$ , we can eliminate choice C

Therefore, since $-5+2=-3$ , then D is correct. Also, $3+(-6)=-3$

3 0

2 years ago

Read 2 more answers

HELPPP ILL DO ANYTHING HELPPPP

katovenus [111]

Answer:

-2

Step-by-step explanation:

hi! to find f(-3) on this graph, go to the x-value -3 and find the y-value there. when we look at -3 on the x-axis, we go down and see that there is a point at (-3,-2). therefore, f(-3)=-2.

3 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

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

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago

What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?

Pavel [41]

$\boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}$

<h2>Explanation:</h2>

Using the quadratic formula:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ Here: \\ \\ f(x) = x^2 + 5x + 5 \\ \\ \\ So: \\ \\ a=1 \\ \\ b=5 \\ \\ c=5 \\ \\ \\ x=\frac{-5 \pm \sqrt{5^2-4(1)(5)}}{2(1)} \\ \\ x=\frac{-5 \pm \sqrt{25-20}}{2} \\ \\ x=\frac{-5 \pm \sqrt{5}}{2} \\ \\ \\ Two \ solutions: \\ \\ \boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}$

<h2>Learn more:</h2>

Quadratic functions: brainly.com/question/12164750

#LearnWithBrainly

8 0

2 years ago

NEED HELP!!!!!!!!!!!!!!!!!!!!!

kondor19780726 [428]

The quadratic formula is -b plus minus the square root of b^2-4ac all over 2a.
Here, a=1, b=13, and c=30.
The only option that fills in the values correctly is D

6 0

3 years ago