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3 years ago

5

The five starters on Raylene's basketball team lined up in order of height from shortest to tallest? Which statistic must descri

be the height of the player standing in the middle? A. mean B. median C. mode D. maximum

2 answers:

UkoKoshka [18]3 years ago

4 0

It should be median

melisa1 [442]3 years ago

3 0

It would be B.median

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Write in factored form:

Katyanochek1 [597]

Answer:

( 3x + 5 ) ( x - 4 )

Step-by-step explanation:

1.) Factors of 20 = (1,20) (2,10) and (4,5)

2.) ( 3x + 5 ) ( x - 4 )

3.) 3x - 12x + 5x - 20 = 0

4.) 3x^2 + 7x - 20 = 0

4 0

3 years ago

Express the edge length of a cube as a function of the cube's diagonal length d. Then express the surface area & volume of t

Lorico [155]

D = sqrt(3s^2) where s is the length of the side. Solving for s,

<span>3s^2 = d^2 iff </span>
<span>s^2 = d^2 / 3 iff </span>
<span>s = sqrt(d^2 / 3) </span>
<span>= d / sqrt(3) or d sqrt(3) / 3 </span>

<span>Surface area of the cube = 6 s^2. Thus, </span>
<span>A = 6 (d / sqrt(3))^2 </span>
<span>= 6d^2 / 3 </span>
<span>= 2d^2 </span>

<span>Volume = s^3. Thus, </span>
<span>V = (d / sqrt(3))^3 </span>
<span>= d^3 / 3sqrt(3) </span>
<span>= d^3 sqrt(3) / 9</span>

8 0

3 years ago

Read 2 more answers

If \triangle △RST is isosceles. find x then find m∠S.

rusak2 [61]

Answer:

s=32

Step-by-step explanation:

15x-31 = 9x+11

6x=42

x=7

2(63+11)= 2(74)= 148

180-148=32

3 0

3 years ago

Read 2 more answers

Evaluate v(x) = 12 - 2x - 5 when x = -2,0. and 5.

EleoNora [17]

Answer:

11,7,-3 respectively

Step-by-step explanation:

when x is -2 ,0 and 5 the result will be 11, 7 and -3 respectively

4 0

3 years ago

Which of the following is not a possible number of solutions when solving a system of equations containing a quadratic and a lin

Kobotan [32]

Answer:

3

Step-by-step explanation:

We have a system with two equations, one equation is a quadratic function and the other equation is a linear function.

To solve this system we have to clear "y" in both equations, and then equal both equations, then we will have a quadratic function and equal it to zero:

$ax^2+bx+c=0, a\neq 0$

Then to resolve a quadratic equation we apply Bhaskara's formula:

$x_{1}=\frac{-b+\sqrt{b^2-4ac} }{2a}$

$x_{2}=\frac{-b-\sqrt{b^2-4ac} }{2a}$

It usually has two solutions.

But it could happen that $\sqrt{b^2-4ac}$ then the equation doesn't have real solutions.

Or it could happen that there's only one solution, this happen when the linear equation touches the quadratic equation in one point.

And it's not possible to have more than 2 solutions. Then the answer ir 3.

For example:

In the three graphs the pink one is a quadratic function and the green one is a linear function.

In the first graph we can see that the linear function intersects the quadratic function in two points, then there are two solutions.

In the second graph we can see that the linear function intersects the quadratic function in only one point, then there is one solutions.

In the third graph we can see that the linear function doesn't intersect the quadratic function, then there aren't real solutions.

7 0

3 years ago