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2 years ago

I NEED HELP GUYS PLEASE I CANT FAIL AGAIN A translation was applied to square ABCD to form

Mathematics

2 answers:

klio [65]2 years ago

8 0

B is the right answer

Pani-rosa [81]2 years ago

4 0

The correct answer is: B

You might be interested in

Sample Response: The two conditional probabilities are not equal because each has different given events. P(A|D) has event D as

Ivan

The conditional probability illustrates that's there's a 2/8 that the event A occurs.

<h3>How to illustrate the probability?</h3>

It should be noted that probability simply means the likelihood of the occurence of an event.

In this case, it can be delivered that P(AID) and P(DIA) aren't equal.

Hence, P(D|A) has event A as its given event, resulting in 2/8 for a probability.

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

7 0

2 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

A game-show spinner has these odds of stopping on particular dollar values: 40% for $5, 25% for $100, 20% for $500, and 15% for

Yakvenalex [24]

Answer:

13÷ 23

Step-by-step explanation:

The computation of the odds of player winning is shown below;

For $5 it is 40%

And, for $100 it is 25%

so for total it is

= 40% + 25%

= 65%

And, the total percentage is 100%

So, the odds in favor is

65:100

13:20

Now for winning it would be 13÷ 23

5 0

2 years ago

Simply this plz ո • ո3

anzhelika [568]

Answer:

Answer is n^4

Step-by-step explanation:

You add the exponent when they are multiplying and have the same base

4 0

2 years ago

Read 2 more answers

Find the nth degree polynomial function with real coefficients satisfying the given conditions.

RUDIKE [14]

$\bf (x+1)(x-3)(x-2-4i)(x-2+4i)
\\\\
\textit{now, let's take a peek at }(x-2-4i)(x-2+4i)\\
\textit{and recall our }\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
(x-2-4i)(x-2+4i)\implies [(x-2)-(4i)][(x-2)+(4i)]
\\\\\
[(x-2)^2-(4i)^2]\implies [(x^2-2x+4)-(4^2\boxed{i^2})]
\\\\\
[(x^2-2x+4)-(16\cdot \boxed{-1})]\implies [(x^2-2x+4)+16]
\\\\
(x^2-2x+20)\\\\
-----------------------------$

$\bf thus
\\\\

\begin{array}{llll}
(x+1)(x-3)\\(x-2-4i)(x-2+4i)
\end{array}\implies (x+1)(x-3)(x^2-2x+20)
\\\\\\
(x^2-2x-3)(x^2-2x+20)\implies 
\begin{array}{llll}
x^4-2x^3+20x^2-2x^3\\+4x^2-40x+3x^2-6x+60
\end{array}
\\\\\\
x^4-4x^3+27x^2-46x+60$

4 0

2 years ago