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2 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Mathematics

1 answer:

Rudik [331]2 years ago

7 0

Answer:

a=1/2b, 2a=b

Step-by-step explanation:

Since angle a is 1/2 the size of angle b, a=1/2b and b=2a. In order to get a numeral answer, you need the number of one of the angles.

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IM REALLY CONFUSED!??? giving away 20pts!!

Olenka [21]

Answer:

A. y = -12.5 + 28000

B. $700

Step-by-step explanation:

7 0

3 years ago

What is 5.092 rounded to the nearest hundredth?

goblinko [34]

5.09 because 2 is less then 5 so you round down

6 0

3 years ago

Read 2 more answers

Given: AB ∩ CD ∩ GF = O, E ∈ interior of ∠GOB, H ∈ interior of ∠AOF. Without changing the picture indicate: Vertical angle to ∠G

lys-0071 [83]

Answer:

The verticle angle to GOB is AOF.

Step-by-step explanation:

Look at the picture, it's pretty clear

5 0

3 years ago

What is an equation of the line perpendicular to 5x-4y=10 that passes through the point (-1,-6)?

Jobisdone [24]

Answer:

$y = - \frac{4}{5} x - \frac{34}{5}$

Step-by-step explanation:

Attached above

3 0

3 years ago

The math question is on the image<br>find the nth term of the sequence

Rus_ich [418]

Notice how Pattern 2 is Pattern 1 with 4 balls added in the bottom row.

Pattern 3 is Pattern 2 with 5 more balls.

Pattern 4 is Pattern 3 with 6 more balls.

Generalizing the trend, we expect Pattern $n$ to be identical to Pattern $n-1$ with $n+2$ more balls.

If $b_n$ is the number of balls in the $n$ -th pattern, then we have the recursive relation

$\begin{cases} b_1 = 6 \\ b_n = b_{n-1} + n + 2 & \text{for } n>1 \end{cases}$

We can solve this recurrence by substitution. Using the definition of $b_n$ , we have

$b_{n-1} = b_{n-2} + (n-1) + 2 \\\\ \implies b_n = (b_{n-2} + (n-1) + 2) + n + 2 \\\\ \implies b_n = b_{n-2} + 2\times2 + \bigg(n + (n-1)\bigg)$

$b_{n-2} = b_{n-3} + (n-2) + 2 \\\\ \implies b_n = (b_{n-3} + (n-2) + 2) + 2\times 2 + \bigg(n + (n-1)\bigg) \\\\ \implies b_n = b_{n-3} + 3\times2 + \bigg(n + (n-1) + (n-2)\bigg)$

and so on, down to

$b_n = b_1 + (n-1)\times2 + \bigg(n + (n-1) + (n-2) + \cdots + 2\bigg)$

Recall that

$\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

Then we find

$\displaystyle b_n = 6 + 2(n-1) + \sum_{i=2}^n i$

$\displaystyle b_n = 2n + 4 + \left(\sum_{i=1}^n i - 1\right)$

$\displaystyle b_n = 2n + 3 + \frac{n(n+1)}2$

$\displaystyle \boxed{b_n = \frac{n^2+5n+6}2}$

8 0

1 year ago