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aleksley [76]
3 years ago
8

What is the range of the function graphed below? 1 -3 -2 -3

Mathematics
1 answer:
Jet001 [13]3 years ago
4 0

Answer:

1-(-3)=4

Step-by-step explanation:

To find the range you have to look at the list of numbers:

1, -2, -3, -3

Then you take the highest number, 1, and the lowest, -3, and then subtract.

1-(-3)

And once you do the math, that'll leave you with 4.

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Which equation represents the graph?
frosja888 [35]

Answer:

The answer is A) y+4 =3(x+3)

Step-by-step explanation:

The slope is the rate of change, which you can see from the graph is m = 3/1 = 3.

Your options are given in point-slope form:

y-y1=m(x-x1)

Pick a point from the graph. I used;

(-3, -4). Now plug this into your point slope form equation.

y-(-4) = 3(x-(-3))

y+4=3(x+3)

8 0
3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
Answer correctly please , will mark brainliest answer !!!!
Afina-wow [57]

Answer:

x-2

Step-by-step explanation:

(7x^2-14x)/7x

Distribute the 7

x-2

7 0
3 years ago
Last year, 11 students tried out for the
umka2103 [35]
136% increase I think
5 0
3 years ago
Pls help with my geometry
Georgia [21]
AB, CA, BC hope this helps
7 0
3 years ago
Read 2 more answers
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