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Firlakuza [10]
3 years ago
7

Express 80 miles per hour in kilometers per hour

Mathematics
2 answers:
Anna [14]3 years ago
8 0
80 miles per hour is 128.74752 kilometers per hour
nataly862011 [7]3 years ago
5 0
It is 128.748 kilometers per hour
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Solve for x in the equation x^2-14 x + 31 = 63.
nadya68 [22]

Answer:

x=11.2426

or

x=2.7574

Step-by-step explanation:

3 0
3 years ago
Uhh hiii... i won't be back uhh coz I'm really sad, I'm really tired.... and i....i need a hug... I'm not taking ur attention i
jarptica [38.1K]

Answer:

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Step-by-step explanation:

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3 years ago
Which expressions are polynomials? select each correct answer
Zielflug [23.3K]
Note:
A polynomial of order n is of the form
f(x)=a_{n}x^{n}+a_{n-1}x^{n-1} + \, ..., \, +a_{1}x+a_{0}
where the a's are constants.

Consider the given expressions.
x/5 + 2                  Not a polynomial
35x²                      It is a polynomial of order 2
7x⁷ - x⁵                  It is a polynomial of order 7
4x + √(x-1)             It is not a polynomial

Answer:
The polynomials are
35x²
7x⁷ - x⁵

8 0
3 years ago
81 POINTS
Jobisdone [24]

Base Case: plug in n = 1 (the smallest positive integer)

If n = 1, then 3n-2 = 3*1-2 = 1. Square this and we see that (3n-2)^2 = 1^2 = 1

On the right hand side, plugging in n = 1 leads to...

n*(6n^2-3n-1)/2 = 1*(6*1^2-3*1-1)/2 = 1

Both sides are 1. So that confirms the base case.

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

Inductive Step: Assume that

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

is a true statement for some positive integer k. If we can show the statement leads to the (k+1)th case being true as well, then we will have sufficiently proven the overall statement to be true by induction.

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 + (3(k+1)-2)^2 = (k+1)*(6(k+1)^2-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3(k+1)-2)^2 = (k+1)*(6(k^2+2k+1)-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3k+3-2)^2 = (k+1)*(6k^2+12k+6-3k-3-1)/2

k*(6k^2-3k-1)/2 + (3k+1)^2 = (k+1)*(6k^2+9k+2)/2

k*(6k^2-3k-1)/2 + 9k^2+6k+1 = (k+1)*(6k^2+9k+2)/2

(6k^3-3k^2-k)/2 + 2(9k^2+6k+1)/2 = (k*(6k^2+9k+2)+1(6k^2+9k+2))/2

(6k^3-3k^2-k + 2(9k^2+6k+1))/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3-3k^2-k + 18k^2+12k+2)/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3+15k^2+11k+2)/2 = (6k^3+15k^2+11k+2)/2

Both sides simplify to the same expression, so that proves the (k+1)th case immediately follows from the kth case

That wraps up the inductive step. The full induction proof is done at this point.

7 0
3 years ago
To decrease an amount by 19% multiply by
yawa3891 [41]
Think of the original amount as 100%
If we want 100% of something we multiply by 1
To get from 100% to 1, we divide the percentage by 100
Decrease 100% by 19%
100% - 19% = 81%
Then to get the multiplier, divide by 100
81% / 100 = 0.81
To decrease by 19%, multiply by 0.81
6 0
3 years ago
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