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sineoko [7]
3 years ago
11

How to solve 5h+7=17

Mathematics
1 answer:
Alex777 [14]3 years ago
5 0
If you are trying to find h, then its very simple.
So lets see 5+7=12
So 17-12=5
So lets check to see if h=5
5+5+7
5+5=10+7=17 so h is 5. :)
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1.05

Step-by-step explanation:

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Write an expression with an exponent that has a value between zero and one
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4 0
2 years ago
3. Which relation is a function?
Pani-rosa [81]

Answer:

B

Step-by-step explanation:

Recall that functions are defined only if for each value in the domain produces one and only one value in the range.

If we view the relations in the questions as x-y coordinates, this means that for every x-value, you can only have one y-value

Lets evaluate the options:

A) we can see that for x = -3, this gives 2 possible values for y i.e (-3,4) and (-3,8) (hence this is not a function)

C) we can see that for x = 3, this gives 2 possible values for y i.e (3,-8) and (3,8) (hence this is not a function)

D) we can see that for x = -3, this gives 2 possible values for y i.e (-3,4) and (-3,8) (hence this is not a function)

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6 0
3 years ago
Please help
Harman [31]

Answer:

we get \mathbf{gf(2)=-\frac{1}{3}}

Step-by-step explanation:

We are given:

f(x)=1-2x\\g(x)= \frac{1}{x}\\h(x)=x^3+1

We need to find value of gf(2)

gf(2)=g(f(2))

First putting f(x) inside of g(x) and then putting x=2

gf(x)=g(f(x))\\gf(x)=\frac{1}{1-2x}

Now putting x=2 to find gf(2)

gf(x)=\frac{1}{1-2x}\\gf(2)=\frac{1}{1-2(2)}\\gf(2)=\frac{1}{1-4}\\gf(2)=\frac{1}{-3}\\gf(2)=-\frac{1}{3}

So, we get \mathbf{gf(2)=-\frac{1}{3}}

3 0
2 years ago
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