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MArishka [77]
3 years ago
9

Part A picture above

Mathematics
1 answer:
-BARSIC- [3]3 years ago
8 0

Answer:

<em>Part A </em>C = (10,5)<em>  Part B </em>C. D'(0,10)

Step-by-step explanation:

<em>Part A</em>

Since c is at the point (2,1) in relation to the origin, we can multiply those distances by our scale factor of 5

(2,1) * 5 = (10,5)

The new point C is going to be (10,5)

<em>Part B</em>

If you dilate with a factor of 5 -- relative to the origin -- you have to multiply the distance from <em>the origin</em> by 5.

In this case, point D is already on the y axis, so it's x value wouldn't be affected. Point D is currently 2 units away from (0,0), so we can multiply 2*5 to get 10 -- our ending point is (0,10)

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(3n^2+13n^3+5n)-(7n+4n^3)
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Step-by-step explanation:

4 0
3 years ago
You buy a book of 5 basketball tickets for $62.50. What is the rate of cost per ticket?
Kisachek [45]
12.50 as 12.50 x 5 is 62.50 or as we know 62.50 for 5 we do 62.50/5 to get 12.50
4 0
3 years ago
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When you find the surface area of a cube, you ________ the length of the cube and multiply by 6.
garik1379 [7]

Answer:

A-square

Step-by-step explanation:

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3 years ago
Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks​
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Let

P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

P(1):\ 1 = \dfrac{1\cdot 2}{2}=1

So, the base case is ok. Now, we need to assume P(n) and prove P(n+1).

P(n+1) states that

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Since we're assuming P(n), we can substitute the sum of the first n terms with their expression:

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Which terminates the proof, since we showed that

P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}

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4 0
3 years ago
The sum of two integers is 6 and the difference between the numbers is 40. Find the numbers
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Set up a system of equations.

X+y=6
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Substitution method.

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Plug it back into the equation.
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(-17,33) is one of the possible answers.







8 0
4 years ago
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