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2 years ago

I NEED HELP ASAPPPP!!!!!!!!!!!!!!

Mathematics

1 answer:

vovikov84 [41]2 years ago

7 0

A: 178 feet away

B: 14 feet (?)

C: 1250 away from jonathon
1222feet away from seth
1386 away from mason (?)

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A fair die is rolled four times. Find the probability that all four rolls show different numbers

tensa zangetsu [6.8K]

Number of such outcomes, in which each number is no less than the preceding number, will be equal to number of ways of selecting 4 numbers from 6, with replacement; without considering order!

Why?

Lets choose any such set of size 4, say {2, 1, 1, 3}. we can sort it to get a sequence {1,1,2,3} which is one of our desired outcomes. So each such sorted sequence corresponds to one selection of size 4, with replacement and without considering order.

Number of such selections will be equal to number of solutions of following equation:

x1 + x2 + x3 + x4 + x5 + x6 = 4

where:

x1: number of 1 in the selections

x2: number of 2s in the selection

x6: number of 6s in the selection

Number unique solutions of such equation = 9 choose 5 = 126

(For more details on this see : Unordered sampling with replacement )

Number of possible outcomes = 6^4 = 1296

Probability of favorable outcomes = 126/1296=7/72

4 0

2 years ago

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Crystal is writing a coordinate proof to show that the diagonals of a parallelogram bisect each other. She starts by assigning c

belka [17]

c,c/2,a+b/2,AE,DE are the answers

4 0

3 years ago

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Can someone help me I’m so confuse on this question help me

Stels [109]

I don’t know I’m lost use a calculator

7 0

2 years ago

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Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

case A) $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

case B) $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

case C) $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

case D) $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

the answer is

$f(x)=(x-2)(x+2)$

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3 years ago

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Please help me!! Solve for 'a'. b=(4*a)-3 ---------- 4

lilavasa [31]

here u go! i hope this is correct :)

8 0

2 years ago

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