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Llana [10]
3 years ago
6

In the diagram, AB = AC and ∠DCE = 75°, BCD and ACE are straight lines. ∠BAC =

Mathematics
1 answer:
RoseWind [281]3 years ago
7 0

Again, what is the shape?

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G(t) = -9t-4 <br> g(. )=23
Mila [183]

Answer:

Is this high school work?

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Whats 126/399 in its simplest form
marusya05 [52]
A common factor of both 126 and 399 is 3 so we can divide both numbers by 3.

126 / 3 = 42
399 / 3 = 133

A common factor for both 42 and 133 is 7 so we can divide both numbers by 7.

42 / 7 = 6
133 / 7 = 19

19 is a prime number so we can't reduce it anymore, therefore 126/399 in its simplest form is 6/19
4 0
3 years ago
I need to solve this equasion. Could you please help me ?<img src="https://tex.z-dn.net/?f=y%20%3D%204%20x%5E%7B2%7D%20%2B3x%2B5
SOVA2 [1]
Hope I helped you. You didn't even asked any question.

3 0
3 years ago
Michael has 3 quarters, 2 dimes, and 3 nickels in his pocket. He randomly draws two coins from his pocket, one at a time, and th
salantis [7]

Answer:

No. Choosing two dimes are dependent events. The probability of choosing the first dime is 1/4 and the probability of choosing the second dime is 1/7 . The probability that both coins are dimes is

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please help me. these problems<br>​
jeyben [28]

Answer:

1st problem:

Converges to 6

2nd problem:

Converges to 504

Step-by-step explanation:

You are comparing to \sum_{k=1}^{\infty} a_1(r)^{k-1}

You want the ratio r to be between -1 and 1.

Both of these problem are so that means they both have a sum and the series converges to that sum.

The formula for computing a geometric series in our form is \frac{a_1}{1-r} where a_1 is the first term.

The first term of your first series is 3 so your answer will be given by:

\frac{a_1}{1-r}=\frac{3}{1-\frac{1}{2}}=\frac{3}{\frac{1}{2}=6

The second series has r=1/6 and a_1=420 giving me:

\frac{420}{1-\frac{1}{6}}=\frac{420}{\frac{5}{6}}=420(\frac{6}{5})=504.

3 0
3 years ago
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