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Irina-Kira [14]
3 years ago
9

What is the slope of the line shown?

Mathematics
1 answer:
zlopas [31]3 years ago
8 0

Step-by-step explanation:

slope = rise/run or apply slope formula (you can find online or on my insta page)

Topic: coordinate geometry

If you like to venture further, feel free to check out my insta <em><u>(learntionary)</u></em>. I'll be constantly posting math tips and notes! Thanks!

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Quick answer pleaseee
Nikolay [14]
D is the correct answer
3 0
2 years ago
What is the solution to the equation -5(s-30)=-10
melomori [17]

Answer: S=28

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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
2 years ago
Please help this was due yesterday
lys-0071 [83]

Answer:

2) 2^8

3)7^9

4)  1/3^6

5)x^10

6) y^8

7) x^10

Step-by-step explanation:

5 0
1 year ago
4 2/3 - (-1 4/5) (by the way 4/5 is a fraction)
Free_Kalibri [48]

Answer: 6

Step-by-step explanation:

Eliminate the negative and change the operation.

4 + 1

Rewrite equation with parts separated

4 + + 1 +

Add whole numbers

4 + 1 = 5

Solve fraction

+ = ??

Find LCD of and . LCD = 15

+ =

Simplify fraction

= 1

Combine whole number and fraction

5 + 1 + = 6

4 0
2 years ago
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