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leva [86]
3 years ago
12

John has 6 pairs of pants and 3 shirts. How many possible outfits consisting of one shirt and

Mathematics
1 answer:
kogti [31]3 years ago
3 0

Answer:

4*5*7=140 ways

Step-by-step explanation:

First he has to decide on a pair of pants and he has 4 different choices.

For each of those choices he has a choice of 7 different shirts, so that gives him 4 times 7 = 28 different pant/shirt combinations. (I hope none of the pants are striped and none of the shirts have polka-dots).

For each those 28 possibilities, he has 5 different ties he could select, so altogether he has 28 times 5 = 140 different outfits, and some of them may be acceptable to wear in public.

It is simply , 4*5*7=140 ways.

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8 or 3

Step-by-step explanation:

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X^2 -9X = 0. Factor and use the Zero property to solve
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The first step of factoring is to try to factor out a common factor.
The terms x^2 and -9x have the factor x in common.
Factor out x from both terms.

x^2 - 9x = 0

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Now you have a product of fully factored terms equaling zero, so you can apply the zero product property to solve.

x = 0   or   x - 9 = 0

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3 years ago
If 100 percent is 480 what is 10 percent?
LenaWriter [7]

Answer:

48

Step-by-step explanation:

10% of 480 = 10/100 × 480

= 48

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3 years ago
What is 4278649 rounded to the nearest hundred
satela [25.4K]

Answer:

4278600

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You go to the number that would be in the 100th place (6) look to the number to the right (4), greater than 5 round up less than 5 round down.

3 0
3 years ago
Which are equivalent expressions?
bonufazy [111]

Answer:

B) 4x + 8y = 2(2x + 4y)

C) 4x + 8y = 4(x + 2y)

D) 4x + 8y + 12z = 4(x + 2y + 3z)

E) 5x + 10y = 5(x + 2y)

Step-by-step explanation:

We would assume a value for the variables a, b, c, x, y and z.

Let x, a = 1

Let y, b = 2

Let z, c = 3

A) a + b + c = 3abc

When a = 1, b = 2 and c = 3

Substituting into the above equation, we have;

1 + 2 + 3 = 3(1*2*3)

6 ≠ 3(6)

6 ≠ 18 (not equivalent)

B) 4x + 8y = 2(2x + 4y)

When x = 1 and y = 2

Substituting into the above equation, we have;

4(1) + 8(2) = 2(2*1 + 4*2)

4 + 16 = 2(2 + 8)

20 = 2(10)

20 = 20 (equivalent expression)

C) 4x + 8y = 4(x + 2y)

When x = 1 and y = 2

Substituting into the above equation, we have;

4(1) + 8(2) = 4(1 + 2*2)

4 + 16 = 4(1 + 4)

20 = 4(5)

20 = 20 (equivalent expression).

D) 4x + 8y + 12z = 4(x + 2y + 3z)

When x = 1, y = 2 and z = 3

Substituting into the above equation, we have;

4(1) + 8(2) + 12(3) = 4(1 + 2*2 + 3*3)

4 + 16 + 36 = 4(1 + 4 + 9)

56 = 4(14)

56 = 56 (equivalent expression).

E) 5x + 10y = 5(x + 2y)

When x = 1 and y = 2

Substituting into the above equation, we have;

5(1) + 10(2) = 5(1 + 2*2)

5 + 20 = 5(1 + 4)

25 = 5(5)

25 = 25 (equivalent expression).

6 0
3 years ago
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