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3 years ago

HELP THIS IS THE 5TH TIME I POSTED THIS BRUH

Mathematics

1 answer:

Delvig [45]3 years ago

4 0

Answer:

it's blurry all i see is pixels...

i'll help if you send me a picture...

Step-by-step explanation:

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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

An amount of money b divided by 20 is $7. What is the amount?

olga2289 [7]

The amount of money, b, is $140.

4 0

2 years ago

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Can someone help me with this quickly????

iren2701 [21]

Answer:

Step-by-step explanation:

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3 years ago

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(Right angle Trigonometry) help me solve for X please!

Vinvika [58]

Answer:

x = 52.0

Step-by-step explanation:

Reference angle = 68°

Length of side Opposite to reference angle = x

Adjacent length = 21

Apply the trigonometric function, TOA, which is given as:

Tan 68° = Opp/Adj

Substitute

Tan 68° = x/21

21 × Tan 68° = x

51.9768238 = x

x = 51.9768238 ≈ 52.0 (approximated to nearest tenth)

4 0

3 years ago

Simplify the following <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%20%5Cfrac%7B1%7D%7Bx%7D%20%20%2B%201%20%7D%20%7B%20%5Cfrac%2

Anon25 [30]

Good evening, Riley!

Used properties:

$\star~\boxed{\mathsf{a = \dfrac{ax}{x}, \ \ x\neq 0}}\\ \\ \\ \star~\boxed{\mathsf{\dfrac{\frac{a}{b}}{\frac{c}{d}} = \dfrac{ad}{bc}}}$

Now, note that:

$\mathsf{1 = \dfrac{x}{x},\ x\neq 0}$

So:

$\mathsf{\dfrac{\frac 1 x + \frac x x }{\frac 1 x} = \dfrac{\frac{x+1}{x}}{\frac 1 x}}=} \mathsf{\dfrac{x}{1}\dfrac{x+1}{x} = \boxed{x + 1}}$

8 0

3 years ago