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Nimfa-mama [501]
3 years ago
6

What is the range of the data Colin collected?

Mathematics
1 answer:
marin [14]3 years ago
8 0

Answer:

5

Step-by-step explanation:

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Pls answer!! giving brainliest if correct!!
guajiro [1.7K]

Answer:

78.5

Step-by-step explanation:

radius is half the diameter, so we know it is five.  Then simply plug 5 into the circle area formula.

7 0
3 years ago
Help please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Dafna11 [192]

If f(x)= 3x+2 and g(x)= 4x, then

f(x)+g(x) = 3x+2+4x

= 7x+2

For f(g(x)), you work from the inside and work your way out. Another way of thinking is to go from right to left. (Inside/out and right/left are the same simply using different wording).

So what is g(x)? 4x.

Now, you take 4x and plug it into the f(x).

** Math--> English translation:

Look at the f(x) equation. Wherever you see an x in the f(x), replace the x with 4x [because 4x is the g(x)]. Basically, plug the g(x) into f(x).

So:

f(g(x)) becomes

f(4x) becomes

3 (4x)+2 = 12x+2

** Disclaimer: Notation may be inaccurate.

That's your answer. (Usually you leave it as so, but if your instructor wants it factored, then proceed).

As for the third part, I am unsure of a clear answer, so I am unable to provide an answer at the time.

**********************

Hope that helped.

DISCLAIMER:

Always double check with a reliable source, as mistakes are a possibility. Never use this, or any, platform to cheat.

6 0
3 years ago
5. 18 m 11 m Not drawn to scale 29 m b. 445 m 7 m d. 203 m​
Luden [163]
Answer: D

Explanation: Apply the pythagorean theorem (a^2 + b^2 = c^2) to get 11^2 + b^2 = 18^2. From that you can solve for b and you get the square root of 203.

Hope this helped! :)
8 0
3 years ago
Read 2 more answers
If x 0 and y 0 where is the point (x y) located quadrant i quadrant ii quadrant iii quadrant iv
Nuetrik [128]

Answer:

true

Step-by-step explanation:

4 0
3 years ago
Find the values of x (if any) at which f(x)= x+1/x^2-1 is not continuous. If so, is the discontinuity removable?
Mkey [24]

f(x)=\dfrac{x+1}{x^2-1} has two points of discontinuity at x=\pm1, where x^2-1=0.

We have

\dfrac{x+1}{x^2-1}=\dfrac{x+1}{(x+1)(x-1)}

so as long as x\neq-1, we can cancel x+1 and be left with \dfrac1{x-1}. Then as x\to-1, we have

\displaystyle\lim_{x\to-1}f(x)=\lim_{x\to-1}\frac1{x-1}=-\frac12

The limit exists, so the discontinuity at -1 is removable (so definitely not C or D).

Meanwhile,

\displaystyle\lim_{x\to1}f(x)=\lim_{x\to1}\frac1{x-1}

does not exist, so the discontinuity at 1 is non-removable, making A the correct answer.

8 0
3 years ago
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