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3 years ago

1/3 x 1/2 PLS PSL eellelele

Mathematics

2 answers:

Darina [25.2K]3 years ago

6 0

1/6
explanation:
1 x 1 = 1
2 x 3 = 6
1/6
Hope it helps!

Vikki [24]3 years ago

4 0

Step-by-step explanation:

1/3 x 1/2

3 x 2=6

so ∴1/3 x 1/2=1/6

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A researcher wishes to conduct a study of the color preferences of new car buyers. suppose that 50% of this population prefers t

Amanda [17]

We will use the Binomial probability formula to find the answer

The formula is given by ⁿCₓ (p)ˣ (1-p)ⁿ-ˣ

We have
n, the number of trial = 16
x, the sample we aim to try = 7
p, the probability of success = 0.5

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P(7) = 0.1746 (rounded to four decimal places)

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3 years ago

How are the semicircle and the diameter of a circle related?

k0ka [10]

Answer:

rThe degree of measure of the simicircle are the same.

Step-by-step explanation:

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2 years ago

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PLEASE HELP!! DUE REALLY SOON (LINK)

andre [41]

Answer:

youngest=115

middle=150

elder=230

Step-by-step explanation:

let the youngest be x

the middle will be x+35

the oldest will be 2x

x+x+35+2x=495

4x+35=495

4x=495-35

4x=460

x=115

since the youngest is x,the money he will receive is $115.

the middle will be x+35 which is equal to 115+35=$150.

lastly the oldest will be 2x which is equal to 2×115=$230.

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babunello [35]

It would be a rectangle

5 0

2 years ago

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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