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Natasha_Volkova [10]
3 years ago
7

Are the ratios 18:9 and 7:6 the same?

Mathematics
1 answer:
Anna007 [38]3 years ago
7 0

Answer:

No there not the same

Step-by-step explanation:

There are not equivalents to each other the answer would be 2:1

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What is the area? (IXL)
Zina [86]

Answer:

45 square meters

Step-by-step explanation:

we know that

The area of trapezoid is equal to

A=\frac{1}{2}(b1+b2)h

where

b1 and b2 are the parallel bases

h is the height of trapezoid (perpendicular distance between the parallel bases)

we have

b1=6\ m\\b2=9\ m\\h=6\ m

substitute

A=\frac{1}{2}(6+9)6

A=45\ m^2

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2 3/7...1 6/7..2 5/8...2 5/10<br> Least from greatest
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Can someone answer this question please I need it right now
Lady bird [3.3K]

Answer: Choice B) -5.2 degrees

=====================

Work Shown:

Add up the given temperatures

-42 + (-17) + 14 + (-4) + 23 = -26

Then divide by 5 since there are 5 values we're given

-26/5 = -5.2

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F(x)=1+log4^x the 4 is low to the log
SVEN [57.7K]
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What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

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Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
3 years ago
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