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3 years ago

You roll a number cube.find each probability p(2)

Mathematics

1 answer:

hram777 [196]3 years ago

7 0

Answer:

they are 6 possibilities in one roll

Step-by-step explanation:

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Please answer correctly !!!!! Will mark brainliest answer !!!!!!!!!!

AnnZ [28]

Answer:

type 2 in the first box,

13/4 in the second box, and

-9/8 in the third one

Step-by-step explanation:Notice that you are asked to write the following quadratic expression in vertex form, so you need to find the "x" value of the vertex, and then the "y" value of the vertex:

$x_{vertex}= -b/2a$

Which in our case is: -13/4

and the value of the y for the vertex is obtained using the functional expression when x equals -13/4:

$f(-13/4)= -9/8$

Then your expression for this quadratic should be:

$f(x)=2\,(x+\frac{13}{4} )^2+(-\frac{9}{8})$

Then type 2 in the first box, 13/4 in the second box, and -9/8 in the third one

7 0

3 years ago

I need your help guys help me differentiate this question

Elden [556K]

Answer:

$6((4 {x}^{3} - 5x + 1) ^{ - 4} + 4 {x}^{10} + 4) ^{5} \times (- 4(4x ^{3} - 5x + 1)^{ - 5} \times 12 {x}^{2} - 5 ) + 40 {x}^{9}$

3 0

3 years ago

Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>

$x^ax^b = x^{(a+b)}$
$\dfrac{x^a}{x^b} = x^{(a-b)}$
$(a^a)^b =x^{(a\times b)}$
$(xy)^a = x^ay^a$

$x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4$

The solution of the expression is $\dfrac{4x^4}{y^6}$ .

<h2>Explanation</h2>

Given to us

$(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$ ,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$ ,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$ ,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

$=2^2x^4y^{-6}$

Using identity $\dfrac{x^a}{x^b} = x^{(a-b)}$ ,

$=\dfrac{2^2x^4}{y^6}$

$=\dfrac{4x^4}{y^6}$

Hence, the solution of the expression is $\dfrac{4x^4}{y^6}$ .

Learn more about Exponentiation:

brainly.com/question/2193820

8 0

2 years ago

Use the substitution method to solve.

lord [1]

Answer:

x = 7, y = 6

Step-by-step explanation:

solve for y for the first equation

2x + y = 20

-2x -2x , eliminate 2x

y = 20-2x

now that we have found y, substitute the y in for the second equation

-5y = -6x + 12

-5(20-2x) = -6x + 12, we just changed y into 20-2x. remember that we are multiplying all of 20 - 2x by -5

-100+10x = -6x + 12, we multiplied everything from the parathesis by -5

+100 + 100, eliminate -100

10x = -6x + 112

+6x +6x , eliminate 6x

16x = 112 , solve for x

x = 7

then y = 20 - 2x = 20 - 2*7 = 6

check:

2 * 7 + 6 = 20

20 = 20

-5 * 6 = -6 * 7 + 12

-30 = -42 + 12

-30 = -30

4 0

3 years ago

The graph of y = f(x) is shown below. Determine the value of x when f(x) = -4? ? y 10 9 8 7 6 5 ur 4 3 NO C 2 1 X -10 -9 -8 -7 -

Brilliant_brown [7]

Answer:

Step-by-step explanation:

the function f(x) is a straight line. The "-4" means that the graph moves down 4 units. Point A is at y=-4, so this is the point. You can graph the equation and you'll see that it hits point A.

4 0

3 years ago