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Alja [10]
4 years ago
13

What's The Domain And Range

Mathematics
1 answer:
Ostrovityanka [42]4 years ago
6 0
The domains are the first ones like 5,2,-3,-1, 7 and range are the second ones
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What you would do is 24÷3 to get the answer of 8:1 as your unit rate
8 0
3 years ago
Which equation represents the relationship between s and t? A. s=1/7t B. s=7t C. s=t+2 D. s=t+18
xeze [42]

Answer A

Step-by-step explanation: 2+2=22-55= -12-5= -7 then take away the negative then look outside and u should see a tree. Your final answer should be s=1/7t

4 0
3 years ago
Find the distance between C and D on the number line c=3 and d=11
777dan777 [17]

Answer:

The distance between these two is 8.

Step-by-step explanation:

To find this, simply subtract the smaller from the larger.

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4 0
4 years ago
How do you multiply and then use distributive property?
morpeh [17]
It depends on the question given, for example

you can do distributive property will multiplying.

Example 1:

there is two ways on doing distributive property.

I can do 4(8+3)

add 8+3 since it is inside the parenthesis which equals = 11 then multiply the outside on what you added. 4x11= 44

example 2: I can also do it another way. 4(8+3) distribute the 4 by multiplying the 4 with both numbers. 4x8 and 4x3 which equals 32, and 12. add 32+12=44

so either way you will get the same answer. hope I helped.

6 0
3 years ago
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
3 years ago
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