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zvonat [6]
2 years ago
15

Find the measure of the arc or angle indicated.

Mathematics
1 answer:
Ksenya-84 [330]2 years ago
5 0

Answer: The correct is 114

Step-by-step explanation: Hope this helps

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Fill in the blanks for this geometry problem. Translation, reflection, etc type of problem. 70 POINTS!
Shkiper50 [21]

The answer is translation of 2 units left and a reflection over the x-axis.

The triangle is translated 2 units left because the triangle moved two units to the left and it is reflected over the x-axis because the bottom triangle is like a reflection of the original triangle.

7 0
2 years ago
Is the expression 6(1 + m) equal to 6 + 6m? Why or why not? How do you know?​
lesya692 [45]

Answer:

Yes, they are equal because of distributive property

Step-by-step explanation:

6(1 + m) = 6 + 6m

3 0
2 years ago
Show all four symbols of multiplication using the variables a/b
OlgaM077 [116]

Answer:

Multiplication Symbols:

<u>1.</u><u> Times</u> ×

<u>2.</u><u> Dot</u>  ⋅

<u>3.</u><u> Parentheses</u> ()

<u>4.</u><u> Variables next to each other</u> <em>ab</em>

Step-by-step explanation:

The symbols can be used like this...

Times: <em>a</em> × <em>b</em> = <em>c</em>

Dot: <em>a</em> ⋅ <em>b</em> = <em>c</em>

Parentheses:<em> </em>(<em>a</em>)(<em>b</em>) = <em>c</em>

Variables next to each other: <em>ab </em>= <em>c</em>

5 0
2 years ago
Which equation represents a linear function?
Blababa [14]

Answer:

The third one, y=x-3

Step-by-step explanation:

A linear equation is any equation that can be written in the form. ax+b=0. Meaning it will not have a radical or be squared.

4 0
3 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
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