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Alla [95]
3 years ago
6

Question

Mathematics
1 answer:
Sonbull [250]3 years ago
3 0

Answer:

1

Step-by-step explanation:

f(x) = a(2) + 6 = 2a + 6

Given f(2) = 8

so 2a + 6 = 8

2a = 8 - 6 = 2

so a = 1

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5+3w+3-w combine like terms to simplify
andreev551 [17]

Answer:

8 + 2w

Step-by-step explanation:

⇒ Change all signs to addition (subtraction is the same as adding a negative):

5 + 3w + 3 + -w

⇒ Since the expression is now all addition, you can rearrange them in any order. So group the terms together by their type:

5 + 3 + 3w + -w

⇒ Simplify:

8 + 2w

<u>Answer:</u> 8 + 2w

<em />

<em>Hope this helps!</em> :)

8 0
3 years ago
Read 2 more answers
I really need the answer to this
bezimeni [28]

Answer:

$350

Step-by-step explanation:

9550-9200=350

7 0
3 years ago
1/2 to exponent of 4
Neporo4naja [7]

Answer:

0.0625 is the answer.

Hope it helps you!!!

4 0
3 years ago
Si una persona aprovecha un descuento del 20% y paga 60.000 pesos por una camisa,¿Cuál era el precio de la camisa sin el descuen
oksano4ka [1.4K]

Answer:

the price of the shirt without discount is 75,000 pesos

Step-by-step explanation:

The computation of the price of the shirt without discount as follows:

There is a discount of 20%

And the person pays 60,000 pesos after considering the discount

So, the price without discount is

= 60,000 pesos ÷ 0.80

= 75,000 pesos

hence, the price of the shirt without discount is 75,000 pesos

6 0
3 years ago
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
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