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Naily [24]
3 years ago
6

Which of the following are true statements about any regular polygon? Check all that apply

Mathematics
2 answers:
arlik [135]3 years ago
6 0

Is it this question?

bekas [8.4K]3 years ago
5 0
I need you to give me the answer choices so I can help you. Share a picture of the problem and I can help you.
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The temperature of the oven/cooking thingy. The weight and time always change. But the temp doesn't.
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3 years ago
Can you solve this equation?​
Leya [2.2K]

Answer:

10 + 10 + 10 = 30 \\ 10 + 5 + 5 = 20 \\ 5 + 4 + 4 = 13 \\ 10 + 5 + 2 = 17

5 0
3 years ago
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Find the quotient 5 68
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8 0
3 years ago
While bowling with friends, Brandy rolls a strike in 57 out of 72 frames. what is the experimental probability that brandy will
gogolik [260]

Answer:

79.17% probability that brandy will roll a strike the first frame of the next game

Step-by-step explanation:

The experimental probability is the probability based on previous outcomes, that is, the number of desired outcomes divided by the number of total outcomes.

While bowling with friends, Brandy rolls a strike in 57 out of 72 frames. What is the experimental probability that brandy will roll a strike the first frame of the next game?  

Desired outcomes:

57 strikes

Total outcomes:

72 frames

Probability:

p = 57/72 = 0.7917

79.17% probability that brandy will roll a strike the first frame of the next game

3 0
3 years ago
Find the zeros of the function f(x) = x2 + 5x + 6. A) y = 6 because the graph crosses the y-axis at 6. B) y = -0.25 because that
Fofino [41]

Answer:

D.)

Step-by-step explanation:

The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts.  You can graph the function and see when the graph crosses the x-axis or solve for the x-values.  I will solve it via factoring and so:

f(x)=x^2+5x+6

Multiply the outer coefficients, in this case 1 and 6, and 1×6=6.  Now let's think about all the factors of 6 we have: 6×1 and 2×3.  Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5?  Actually we can say 6-1=5 and 2+3=5.  Let's try both.

First let's use 6 and -1 and so:

x^2+5x+6\\\\x^2+6x-x+6\\\\x(x+6)-1(x-6)

Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.

Now let's try 2 and 3 and so:

x^2+5x+6\\\\x^2+3x+2x+6\\\\x(x+3)+2(x+3)\\\\(x+2)(x+3)

Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:

Case 1:

f(x)=x+2\\\\0=x+2\\\\x=-2

Case 2:

f(x)=x+3\\\\0=x+3\\\\x=-3

So your zero's are when x=-2 and x=-3.

D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.


~~~Brainliest Appreciated~~~

8 0
3 years ago
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