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xeze [42]
3 years ago
9

Graph the inverse function of f(x) = 2x + 4.​

Mathematics
1 answer:
motikmotik3 years ago
7 0

Answer:

Step-by-step explanation:

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How many zero pairs must be added to the function
lakkis [162]

Answer:

In short, Your Answer would be Option D) 25

Hope this helps!

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
How do you solve this
fiasKO [112]
(((x + y) / 3) + (1 / x)) / (5 + (15 / x))
The best way is to make it one fraction.
Multiply by ((3x/3x) / (3x/3x)) to remove the other fractions.
((x(x + y)) + 3(1)) / (5(3x) + 3(15))
(x^2 + xy + 3) / (15x + 45)
Then factor to simplify
(x^2 + xy + 3) / (3(x + 15))



4 0
3 years ago
For number 2 I know the answer but I just need work so if you could please explain the work I’d be thankful
const2013 [10]

Answer:

Step-by-step explanation: when you take 521 and divide it by 4 you get 130.25, which is about 130, making you choose answer choice B. (lol i like your profile picture too!!)

4 0
3 years ago
the diameter of a piston for an automobile is 3 11/16in with tolerence of 1/64in. find the upper and lower limits of the pistons
padilas [110]

Upper Tolerance

Remark

The 11/16 is the only thing that will be affected. The three won't go up or down when we add 1/64 so we should just work with the 11/16. We need only add 11/16 and 1/64 together to see what the upper range is. Later on we can add 3 into the mix.

Solution

<u>Upper Limit</u>

\dfrac{11}{16} +  \dfrac{1}{64}

Now change the 11/16 into 64. Multiply numerator and denominator or 11/16 by 4

\dfrac{11*4}{16*4} +  \dfrac{1}{64}

\dfrac{11*4}{16*4} +  \dfrac{1}{64} Which results in

\dfrac{44}{64} +  \dfrac{1}{64}

With a final result for the fractions of 45/64

So the upper tolerance = 3 45/64

<u>Lower Tolerance</u>

Just follow the same steps as you did for the upper tolerance except you subtract 1/64 like this.

\dfrac{11}{16} - \dfrac{1}{64}

Your answer should be 3 and 43/64


4 0
3 years ago
Which expression is equivalent to b^-2/ab^-5?<br>a) a/b^5<br>b)1/ab^5<br>c)a^3b/1<br>d)b/a
Snezhnost [94]
\bf a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad&#10;\cfrac{1}{a^{-n}}\implies a^{{ n}}\\\\&#10;-------------------------------\\\\&#10;\cfrac{b^{-2}}{ab^{-5}}\implies \cfrac{b^{-2}\cdot b^{+5}}{a}\implies \cfrac{b^{-2+5}}{a}\implies \cfrac{b^3}{a}
4 0
3 years ago
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