Find the 50th term of 0,3,6,9

1 answer:

7 0

$\text{a}_1=0 \ \text{and} \ \text{a}_2=3\\\\\text{r}=\text{a}_2-\text{a}_1=3-0=3\\\\\text{a}_{50}=\text{a}_1+49\text{r}=0+49\cdot3=0+147=147\\\\\huge\boxed{\text{a}_{50}=147}$

You might be interested in

How do I do this I need help right now

kodGreya [7K]

To answer this (specifically the first one) you have to count how many spaces each point moves. (How many spaces up/down does it go? Left/right?) then you put rise (up/down) over run (left/right) and you should get the slope!

6 0

3 years ago

(8) slope? y-intercept? (9) slope? y-intercept?

klasskru [66]

8: slope is 1/2 and y intercept is 0

9: slope is 7/4 and y-intercept is 0

8 0

3 years ago

TIMED QUESTION NEED HELP FASTWhat values of b satisfy 3(2b + 3)2 = 36?

egoroff_w [7]

Answer:

$\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}$

Step-by-step explanation:

$3(2b+3)^{2}=36\\\frac{3(2b+3)^{2}}{3}=\frac{36}{3}\\(2b+3)^{2}=12$

Taking square root both sides, we get

$\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}$

Therefore, the values of $b$ are:

$\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}$

5 0

4 years ago

Read 2 more answers

I SPENT 50 POINTS ON THIS, PLEASE HELP

Anestetic [448]

In the figure, we can consider that the base is the side that mesures 12 in and that the height is the side that measures 15 in, since that sides are perpendicular. So, we just need to use the given formula:

$A = \dfrac{bh}{2}\Longrightarrow A = \dfrac{12\cdot15}{2}\Longrightarrow A=6\cdot15\Longrightarrow \boxed{A = 90~\text{in}^2}$