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2 years ago

Le help wanted for variables part two

Mathematics

2 answers:

Lerok [7]2 years ago

4 0

The coefficient is -9

disa [49]2 years ago

3 0

The coefficient for the first question is c and the coefficient for the second question is gh

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(3x^4-x^3-4x^2-2x-20)÷(x^2+2) using long division

rosijanka [135]

Answer is below..............

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3 years ago

The parallel boxplots below display the quiz grades (out of 30) for two of Mrs. Smith’s statistics classes.

stich3 [128]

Answer:

The quiz grades for class 1 have more variability than the quiz grades for class 2.

Step-by-step explanation:

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3 years ago

The product of x and its opposite is always 1.<br> O True<br> O False

kari74 [83]

Answer:

False

Step-by-step explanation:

The opposite of x is -x

x* -x = -x^2

This is not always 1

3 0

2 years ago

Read 2 more answers

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

Homework help please I'll give you 10 points and brainliest

DaniilM [7]

Answer:

Step-by-step explanation:

Use 49 instead of 14 for 7²

Reason: 7² = 7 * 7 = 49

Not 14 as 7*2 = 14

6 0

2 years ago