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AlexFokin [52]
3 years ago
9

How do i find x in #21

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
8 0
In quadrilaterals inscribed in circles, the sum of opposite angles=180 degrees. so, x+134=180
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Pleas I really need help with this all I need is the answer
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It’s 40% i hope i answered it in time for you
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2 years ago
Ben is 121212 years older than Ishaan. Ben and Ishaan first met two years ago. Three years ago, Ben was 444 times as old as Isha
Maksim231197 [3]
Ishaan is 53816799 (121212-3(444)+3)
5 0
3 years ago
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Ricky the rabbit eats 1 on Sunday 2carrots on Monday 4 carrot on Tuesday and so on how many carrots does he eat every week?
Anna11 [10]

Answer:

7

Step-by-step explanation:

Because I added up all of the numbers by using ow many carrots he eats a day and then that gets me my answer of 7 hope this helps!!

7 0
3 years ago
Please help me right away with these problems.
viktelen [127]

Q.34

\sum\limits_{k=1}^{\infty}420\left(1.002\right)^{k-1}

The infinite geometric series is converges if |r| < 1.

We have r =1.002 > 1, therefore our infinite geometric series is Diverges

Answer: c. Diverges, sum not exist.

Q.35

\sum\limits_{k=1}^{\infty}-5\left(\dfrac{4}{5}\right)^{k-1}

The infinite geometric series is converges if |r| < 1.

We have r = 4/5 < 1, therefore our infinite geometric series is converges.

The sum S of an infinite geometric series with |r| < 1 is given by the formula :

S=\dfrac{a_1}{1-r}

We have:

a_1=-5\left(\dfrac{4}{5}\right)^{1-1}=-5\left(\dfrac{4}{5}\right)^0=-5\\\\r=\dfrac{4}{5}

substitute:

S=\dfrac{-5}{1-\frac{4}{5}}=-\dfrac{5}{\frac{1}{5}}=-5\cdot\dfrac{5}{1}=-25

Answer: c. Converges, -25.

4 0
3 years ago
Which number produces a rational number when added to 1/5?
dimaraw [331]

Answer:

1/5 is a rational number

Step-by-step explanation:

"In mathematics, a rational number is a number such as -3/7 that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Every integer is a rational number: for example, 5 = 5/1."

6 0
3 years ago
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