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Alja [10]
2 years ago
8

Do this work If one do I will make brainliest her or him please

Mathematics
1 answer:
denpristay [2]2 years ago
6 0

Answer:

If you need the explanation, just ask.

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If f(x)=4x-7 and g(x)=2x+4 evalvate f(x)+g(x) for x=-3
galben [10]

Answer:

-21

Step-by-step explanation:

We are told to find f(x) + g(x) for x= -3. Therefore, we must evaluate f(-3) and g(-3), then add them together.

First, evaluate f(-3).

f(x)=4x-7

To find f(-3), we need to substitute -3 in for x.

f(-3)= 4(-3)-7

Solve according to PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction First, multiply 4 and -3.

f(-3)= -12-7

Next, subtract 7 from -12

f(-3)= -19

Next, find g(-3).

g(x)=2x+4

To find g(-3), substitute -3 in for x.

g(-3)= 2(-3)+4

Solve according to PEMDAS. First, multiply 2 and -3.

g(-3)= -6+4

Next, add -6 and 4

g(-3)= -2

Now, we can add f(-3) and g(-3) together.

f(-3) + g(-3)

f(-3)= -19

g(-3)= -2

-19 + -2

Add

-21

7 0
3 years ago
Read 2 more answers
Solve for c: a=7b+8c+9d-10
polet [3.4K]
<span>a=7b+8c+9d-10
a = 8c + 16d - 10
solve for c then
8c = a - 16d + 10
  c = (</span>a - 16d + 10) / 8
or
  c = a/8 - 2d + 5/4
7 0
3 years ago
I'm confused how you split the fraction
slava [35]
1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 = 6/3 
8 0
3 years ago
Read 2 more answers
A manufacturer produces cherry, peppermint, cinnamon, and orange-flavored gum. Each flavor is available with sugar or sugar-free
Semenov [28]

Answer: The total number of different types of labels will the manufacturers have to produce = 24.

Step-by-step explanation:

Given: Choices for flavours = 4

Choices for sugar = 2  {either sugar or sugar free}

Choices for the qunatity = 3

By Fundamental counting principle,

Total number of different types of labels will have to produce = (Choices for flavors) x (Choices for sugar)x (Choices for the quantity )

= 4 x  2x 3

=24

Hence, the total number of different types of labels will the manufacturers have to produce = 24.

4 0
3 years ago
The cost of membership in the school kayaking club is a $10.00 fee and $3.00
Luden [163]

Answer:

igwif

Step-by-step explanation:

392e6×0273=2048

3 0
3 years ago
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