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Alenkinab [10]
3 years ago
15

Given the parallelogram

Mathematics
1 answer:
Mashutka [201]3 years ago
4 0

Answer: 55.

Step-by-step explanation:R = 55. You put all the angles equal to 180 and then get the variable by itself.

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Find the geometric mean of the pair of numbers
Svetach [21]

Answer: a. 55

Step-by-step explanation:

The formula to find the geometeric mean between two numbers a and b is given by :-

G.M. =\sqrt{ab}

The given numbers are : 275 and 11

The geometric mean of 275 and 11 is given by :-

G.M.=\sqrt{275\times11}\\\\\Rightarrow\ G.M. =\sqrt{3025}\\\\\Rightarrow\ G.M.=\sqrt{55\times55}\\\\\Rightarrow\ G.M.=\sqrt{55^2}\\\\\Rightarrow\ G.M.=55

Hence, the geometric mean of 275 and 11 is 55.

So , the correct option is a. 55 .

4 0
3 years ago
What is a coefficient? Give me an example.
aksik [14]

Answer:

A coefficient is the number next to a letter. It's used to multiply a variable. If there is just a variable then the coefficient is 1. For example 6x.

Step-by-step explanation:

7 0
3 years ago
3A + 4N - 5A + 3 - 8N + 11 =
Anit [1.1K]

Answer:

-2a-4n+14

Step-by-step explanation:

7 0
3 years ago
roberto has 12 tiles. each tile is 1 square inch. he will arrange them into a rectangle and glue 1-inch stones around the edge.
LUCKY_DIMON [66]
To answer this you will need to determine which area made with the 12 square tiles would give the shortest perimeter.

Think of all the combinations to get 12 square inches
1x12, 2x6, and 3x4. The perimeters of these are 26 in, 16 in, and 14 in.

You should arrange your tiles using the 3 x 4 option. You would need 14 stones.
3 0
3 years ago
Prove that root 7 is irrational by the method of contradiction
Alchen [17]

Let assume that \sqrt7 is a rational number. Therefore it can be expressed as a fraction \dfrac{a}{b} wherea,b\in\mathbb{Z} and \text{gcd}(a,b)=1.

\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2

This means that a^2 is divisible by 7, and therefore also a is divisible by 7.

So, a=7k where k\in\mathbb{Z}

(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2

Analogically to a^2=7b^2 ------- b^2 is divisible by 7 and therefore so is b.

But if both numbers a and b are divisible by 7, then \text{gcd}(a,b)=7 which contradicts our earlier assumption that \text{gcd}(a,b)=1.

Therefore \sqrt7 is an irrational number.

8 0
3 years ago
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