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gavmur [86]
2 years ago
10

3. Use each number line to show a different way to count from $580 to $994

Mathematics
1 answer:
Amiraneli [1.4K]2 years ago
6 0

What are the number lines though?

Give more info.

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Find the volume of 22 ft
Mazyrski [523]
V=4πr^3V=4π(11)^3<span>
V=5575.28
22 is the diameter and the radius is half of the diameter </span>
5 0
3 years ago
What divisor is represented by the synthetic division below? A vertical line and horizontal line combine to make a L shape. Ther
g100num [7]

One of the ways of dividing polynomials is by synthetic division

The divisor of the synthetic division is x + 5

<h3>How to determine the divisor of the division</h3>

From the question, we understand that:

Entry -5 is on the outside to the left of the shape

The entry represents the zero of the divisor.

Next, we set this entry to the variable (assume the variable is x)

So, we have:

x = -5

Add 5 to both sides

x + 5= 0

Hence, the divisor of the synthetic division is x + 5

Read more about synthetic division at:

brainly.com/question/24662212

8 0
2 years ago
Read 2 more answers
China issued the largest stamp ever, 210mm Long and 65mm wide. Find the ratio of length to width
lora16 [44]
210 : 65 
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42 : 13 is the ratio of 210mm long and 65mm wide
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<span>divide both numbers by 5</span>
7 0
3 years ago
Help me... I did it again! I skipped the practice and study guides
Katyanochek1 [597]

Answer:

X = 1 3/4

Step-by-step explanation:

Hope this helps!

4 0
2 years ago
Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks​
Iteru [2.4K]

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

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

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

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

Which terminates the proof, since we showed that

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

as required

4 0
3 years ago
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