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2 years ago

7

The expression 3r - 6 can be written as _____. 3(r - 2) 3(r - 6) 3(3r - 6) user: 18 - 2[x (x - 5)] user: simplify 18 - 2[x (x -

5)]. user: simplify 18 - 2[x (x - 5)]. 8 - 4x 28 - 4x 28 - 2x

1 answer:

Vlada [557]2 years ago

4 0

The expression 3r - 6 can be written as 3(r-2), because when you distribute 3(r-2) you get 3r -6. So they are equivalent.

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Peter shot 40 free throws. He made 80% of his shots. Which fraction is equivalent to

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2 years ago

Pls pls help I’m really confused could someone explain to me pls

kenny6666 [7]

Answer: 11.6

--------------------------------------

Explanation:

See the attached image for a visual reference

The distance from point D to E is 21 units. Point C is the midpoint, so CD is 10.5 units long (21/2 = 10.5)

We have a right triangle ACD. The legs are
AC = 5
CD = 10.5
The hypotenuse is
AD = x
Because AD is another radius of the same circle

Use the pythagorean theorem to find x
a^2 + b^2 = c^2
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2 years ago

There are 5 red balls, 4 blue balls, 6 yellow balls and 10 green balls in a box,

lys-0071 [83]

<h2 /><h2>Here we go ~ </h2>

According to given information there are :

5 red balls

4 blue balls

6 yellow balls

10 green balls

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-

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{total \: red \: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{5}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{5}{25}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{1}{5}$

<h3>2. what is the probability that the ball chosen is blue ?</h3>

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{total \: blue \: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{4}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{4}{25}$

<h3>3. what is the probability that the ball chosen is yellow ?</h3>

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{total \: yellow\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{6}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{6}{25}$

<h3>4. what is the probability that the ball chosen is green ?</h3>

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{total \: green\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{10}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{10}{25}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{2}{5}$

<h3>5. what is the probability that the ball chosen is not green ?</h3>

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{total \: non \: green\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{5 + 4 + 6}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{15}{25}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{3}{5}$

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