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mars1129 [50]
3 years ago
9

-7(2-3x) help solve please

Mathematics
2 answers:
sp2606 [1]3 years ago
6 0

Answer:

<u>T</u><u>h</u><u>e</u><u> </u><u>answer</u><u> </u><u>i</u><u>s</u><u> </u><u>-14 + 21x</u>

Step-by-step explanation:

Distribute -7 to 2 and -3x

-7 × 2 = -14

-7 × - 3x = 21x

= -14 + 21x

nadezda [96]3 years ago
4 0

Answer:

-14+21x

Step-by-step explanation:

Distrubute

-7 times 2= -14

-7 times 3x= -21x

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In a competition, a school awarded medals in different categiories.40 medals in sport 25 medals in danceand 212 medals in music,
aleksley [76]

Answer:

210

Step-by-step explanation:

Given:

Medals in sports = 40

Medals in dance = 25

Medals in music = 212

Total students that received medals = 55

Total students that received medals in all three categories = 6

Required:

How many students get medals in exactly two of these categories?

Take the following:

A = set of persons who got medals in sports.

B = set of persons who got medals in dance

C = set of persons who got medals in music.

Therefore,

n(A) = 40

n(B) = 25

n(C) = 212

n(A∪B∪C)= 55

n(A∩B∩C)= 6

To find how many students get medals in exactly two of these categories, we have:

n(A∩B) + n(B∩C) + n(A∩C) −3*n(A∩B∩C)

=n(A∩B) + n(B∩C) + n(A∩C) −3*6 ……............... (1)

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)

Thus, n(A∩B)+n(B∩C)+n(A∩C)=n(A)+n(B)+n(C)+n(A∩B∩C)−n(A∪B∪C)

Using equation 1:

=n(A)+n(B)+n(C)+n(A∩B∩C)−n(A∪B∪C)−18

Substitute values in the equation:

= 40 + 25 + 212 + 6 − 55 − 18

= 283 - 73

= 210

Number of students that get medals in exactly two of these categories are 210

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If f(x) = -5x - 1 and g(x) = x - 6 what is (f g) (7)
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You have

(f\circ g)(x) = f(g(x)) \implies (f\circ g)(7) = f(g(7))

So, we have to compute g(7) and feed it as input to f(x). We have

g(7)=7-6=1 \implies f(g(7))=f(1)=-5-1=-6

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