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GalinKa [24]
4 years ago
11

HELP!! please I can't figure this out​

Mathematics
1 answer:
Fofino [41]4 years ago
7 0
12.875 pounds is the answer
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6. Find the sum or difference<br> 1/3(9 - 6m) + 1/4 (12m-8)
romanna [79]

Answer:

m+1

Step-by-step explanation:

1/3(9 - 6m) + 1/4 (12m-8)

Distribute

1/3 *9  - 1/3 *6m  + 1/4 * 12m - 1/4 *8

3 -2m + 3m -2

Combine like terms

-2m+3m +3-2

m+1

6 0
3 years ago
Help I'm really stuck!!!
faust18 [17]
399.6 square meters
7•8/2=28
28•3=84

8•6.9/2=27.6

(12•8)•3=288

84+27.6+288=399.6
5 0
3 years ago
Read 2 more answers
Whats half of 30 i need a answer
Flura [38]

Answer:

15

Step-by-step explanation:

30 / 2 = 15

4 0
3 years ago
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For each function below, determine whether or not the function is injective and whether or not the function is surjective. (You
Harlamova29_29 [7]

Answer: First, injective means that for every y, there is only one x such f(x) = y, and surjective means that if f is a function that goes from the set {x} to the set {y}, for every element y in the codomain there is at least one element x in the domain such f(x) = y

(a) f:R----> + R given by f(x) = x2 (i guess you written x^{2})

The function is not injective, because f(-2) = f(2), and is surjective, because the codomain is +R, and x^{2} is always a positive real number.

(b) f:N----> + N given by f(n) = n2  (i guess you written n^{2})

As the naturals have no negative numbers, in this case the function is injective, but isn't surjective, because there is no number that when squared is equal to 5, for example.

(c) f: Zx Z → Z given by f(n, k) = n +k:

here the domain is of the form (n,k) and the function is n +k, so the numbers (z,w) and (w,z) return the same value when evaluated in f, then f is not injective. And is easy to see that f is surjective, because in the sum you can reach al the integers on the codomain.

3 0
3 years ago
What is 2 1/4 as a fraction
arlik [135]
I don’t know if that 2 and 1/4 or 21/4 so I’ll do both.
2 and 1/4 is = 9/4
21/4 is = 5 and 1/4
6 0
3 years ago
Read 2 more answers
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