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lord [1]
3 years ago
8

The graph of the function f(x) is shown below.

Mathematics
1 answer:
marishachu [46]3 years ago
6 0

Answer:-1.8 is going to be the answer do good on the rest of your test

Step-by-step explanation:

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I only need number 11 in the picture please help and show work ​
IgorLugansk [536]

Answer:

1 = 115°

2 = 115°

3 = 65°

Step-by-step explanation:

P=65, so S (3) =65

360-65-65=230

230/2=115

Q (1) = 115

R (2) = 115

3 0
2 years ago
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What type of triangle is shown below check all that apply
sineoko [7]

Answer:

The first image is an Isosceles Triangle and Acute

Step-by-step explanation:

3 0
3 years ago
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PLEASE HELP ME WITH THIS QUESTION!!!!!
prohojiy [21]

Answer:

we have, 1953125=5⁹, so it cannot be a perfect square. If the last digit of a given number is 5, then the last three digits must be perfect squares, 025 or 225 or 625. Otherwise, that number cannot be a perfect square. And as 125 is not a perfect square, so no number ending with 125 can be a perfect square

4 0
2 years ago
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3. Simplity: 1-7721<br>a. -772<br>b. - 7<br>c. 772<br>d. 72​
Tems11 [23]

Answer:

1 - 7721 = -7220  so i don't really see the answer in the choices you gave.

Step-by-step explanation:

3 0
3 years ago
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Use mathematical induction to show that 4^n ≡ 3n+1 (mod 9) for all n equal to or greater than 0
cestrela7 [59]
When n=0, you have

4^0=1\equiv3(0)+1=1\mod9

Now assume this is true for n=k, i.e.

4^k\equiv3k+1\mod9

and under this hypothesis show that it's also true for n=k+1. You have

4^k\equiv3k+1\mod9
4\equiv4\mod9
\implies 4\times4^k\equiv4(3k+1)\mod9
\implies 4^{k+1}\equiv12k+4\mod9

In other words, there exists M such that

4^{k+1}=9M+12k+4

Rewriting, you have

4^{k+1}=9M+9k+3k+4
4^{k+1}=9(M+k)+3k+3+1
4^{k+1}=9(M+k)+3(k+1)+1

and this is equivalent to 3(k+1)+1 modulo 9, as desired.

3 0
3 years ago
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