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3 years ago

12

3^sqrt -64w^12

qrt[3]{ - 64w^{12}} " alt=" \sqrt[3]{ - 64w^{12}} " align="absmiddle" class="latex-formula">

1 answer:

Agata [3.3K]3 years ago

7 0

-64=(-4)³, w^12=(w^4)³
the result is -4w^4

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Diego is solving the equation x^2-12x = 21

uysha [10]

Answer:

The solutions to the quadratic equations will be:

$x=\sqrt{57}+6,\:x=-\sqrt{57}+6$

Step-by-step explanation:

Given the expression

$x^2-12x\:=\:21$

Let us solve the equation by completing the square

$x^2-12x\:=\:21$

Add (-6)² to both sides

$x^2-12x+\left(-6\right)^2=21+\left(-6\right)^2$

simplify

$x^2-12x+\left(-6\right)^2=57$

Apply perfect square formula: (a-b)² = a²-2ab+b²

i.e.

$x^2-12x+\left(-6\right)^2=\left(x-6\right)^2$

so the expression becomes

$\left(x-6\right)^2=57$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x-6=\sqrt{57}$

add 6 to both sides

$x-6+6=\sqrt{57}+6$

Simplify

$x=\sqrt{57}+6$

also solving

$x-6=-\sqrt{57}$

add 6 to both sides

$x-6+6=-\sqrt{57}+6$

Simplify

$x=-\sqrt{57}+6$

Therefore, the solutions to the quadratic equation will be:

$x=\sqrt{57}+6,\:x=-\sqrt{57}+6$

5 0

3 years ago

An envelope contains three cards: a black card that is black on both sides, a white card that is white on both sides, and a mixe

Over [174]

Answer:

There is a 2/3 probability that the other side is also black.

Step-by-step explanation:

Here let B1: Event of picking a card that has a black side

B2: Event of picking a card that has BOTH black side.

Now, by the CONDITIONAL PROBABILITY:

$P(B_2/B_1 ) = \frac{P(B_1\cap B_2)}{P(B_1)}$

Now, as EXACTLY ONE CARD has both sides BLACK in three cards.

⇒ P (B1 ∩ B2) = 1 /3

Also, Out if total 6 sides of cards, 3 are BLACK from one side.

⇒ P (B1 ) = 3 /6 = 1/2

Putting these values in the formula, we get:

$P(B_2/B_1 ) = \frac{P(B_1\cap B_2)}{P(B_1)} = \frac{1}{3} \times\frac{2}{1} = \frac{2}{3}$

⇒ P (B2 / B1) = 2/3

Hence, there is a 2/3 probability that the other side is also black.

5 0

3 years ago

PLEASE HELP ASAP! need answer now!

BARSIC [14]

Answer:

i have the same question

Step-by-step explanation:

7 0

3 years ago

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Give the properties for the equation -2x 2 - y + 10x - 7 = 0.

Sladkaya [172]

Given:

The quadratic equation is

$-2x^2-y+10x-7=0$

To find:

The vertex of the given quadratic equation.

Solution:

If a quadratic function is $f(x)=ax^2+bx+c$ , then

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

We have,

$-2x^2-y+10x-7=0$

It can be written as

$-2x^2+10x-7=y$

$y=-2x^2+10x-7$ ...(i)

Here, $a=-2,b=10,c=-7$ .

$\dfrac{-b}{2a}=\dfrac{-10}{2(-2)}$

$\dfrac{-b}{2a}=\dfrac{-10}{-4}$

$\dfrac{-b}{2a}=\dfrac{5}{2}$

Putting $x=\dfrac{5}{2}$ in (i), we get

$y=-2(\dfrac{5}{2})^2+10(\dfrac{5}{2})-7$

$y=-2(\dfrac{25}{4})+\dfrac{50}{2}-7$

$y=\dfrac{-50}{4}+25-7$

$y=\dfrac{-25}{2}+18$

On further simplification, we get

$y=\dfrac{-25+36}{2}$

$y=\dfrac{11}{2}$

So, the vertex of the given quadratic equation is $\left(\dfrac{5}{2},\dfrac{11}{2}\right)$ .

Therefore, the correct option is A.

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3 years ago

Is 4.1 greater than 4

EleoNora [17]

Answer: yes it is.

let me know if you need anything else

Step-by-step explanation:

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3 years ago

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