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1 year ago

Question 10 of 15 What is the solution to the system of equations graphed below?

Mathematics

2 answers:

Natali [406]1 year ago

8 0

Answer: (1, 5)

Step-by-step explanation: Over to the right one, up five; I hope this helps!

Key: (xGraph, yGraph) If a number is positive on the x axis, go right, if negative, go left. The same applies to the y graph, if the number is positve, go up, if negative go down. Your solution is where the two lines meet.

storchak [24]1 year ago

3 0

Answer:

(1,5)

Step-by-step explanation:

The solution to the system of equations is where the two graphs intersect.

The two lines intersect at x=1 and y =5

The point of intersection is (1,5)

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Solve the equation -3 + 1/2n = 1/2 (-n + 14) in two different ways.

Lemur [1.5K]

Answer:

Step-by-step explanation:

-3 + 1/2n = 1/2 (-n + 14)

distribute

-3 + 1/2n = -1/2n + 7

-1/2n

cancel out -1/2n

-3 = n + 7

-10 = n

subtract 7

n = -10

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2 years ago

5. Five-eighth of a number is 15. What is the number?<br> Your answer

anygoal [31]

Answer:

I think it is 24

Step-by-step explanation:

15÷5=3

3×8=24

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2 years ago

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She runs 10 laps around the track I hope this helped ^^

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3 years ago

Solve for x:<br> please help me

solniwko [45]

Answer:

x = 10

Step-by-step explanation:

We know that is a right angle which is 90 degrees.

we know one angle is 30 degrees.

So take 90- 30 = 60

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x = 10

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2 years ago

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Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

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3 years ago