Ask question

Ask question

All categories

2 years ago

8

HELP PLEASE!! BRAINLIEST AND 10 POINTS!! if the sixth term of a geometric sequence is 224, and the eleventh term is 7168, what i

s the first term?

1 answer:

Klio2033 [76]2 years ago

8 0

The first term should be 7

a1 = (224) / (2)6 - 1 = 7

You might be interested in

I NEED HELP!! 60 POINTS!! QUESTION IN IMAGE BELOW

Lelu [443]

Option 1 is the Answer.

For explanation, then please check the attached image.

<em>By </em><em>Benjemin</em>

7 0

2 years ago

WILL GIVE BRAINLIEST!!

MA_775_DIABLO [31]

given a polynomial with factor (x - a)

Then x = a is a root of the polynomial and f(a) = 0

(1)

(x+ 2) is a factor , hence x = - 2 is a root

Evaluate the polynomial for x = - 2

f(- 2) = 5(- 2)³ + 8(- 2)² - 7(- 2) - 6 = - 40 + 32 + 14 - 6 = 0

Hence (x + 2 ) is a factor

(2)

If (x + 1) is a factor then x = - 1 is a root

5(- 1)³ + 8(-1)² - 7(- 1) - 6 = -5 + 8 + 7 - 6 = 4

Since f( - 1) ≠ 0

Then (x + 1) is not a factor

5 0

3 years ago

Let A be the area of a circle with radius r. If dr/dt=3, find dA/dt when r=2.

astraxan [27]

$\bf \textit{area of a circle}=A=\pi r^2
\\\\\\
\cfrac{dA}{dt}=\pi \cdot 2r^1\cdot \cfrac{dr}{dt}\implies \cfrac{dA}{dt}=\pi \cdot 2r\cdot \cfrac{dr}{dt}\quad 
\begin{cases}
r=2\\
\frac{dr}{dt}=3
\end{cases}
\\\\\\
\cfrac{dA}{dt}=\pi \cdot 2(2)(3)$

3 0

3 years ago

Read 2 more answers

How do u graph 2y=-4x+6

gladu [14]

Answer:

3 would be your y intercept

then from the point 0,0 put 2 in as slope

Step-by-step explanation:

First turn the equation into y=mx+b (slope-intercept form)

Which turns into y= -2x+3

in y=mx+b

m is slope and b is y intercept

3 would be your y intercept

then from the point 0,0 put 2 in as slope

Hope this helps! have a nice day! Please consider making me Brainliest!

4 0

2 years ago

Find the complex fourth roots \[-\sqrt{3}+\iota \] in polar form.

____ [38]

Let $z=-\sqrt3+i$ . Then

$|z|=\sqrt{(-\sqrt3)^2+1^2}=2$

$z$ lies in the second quadrant, so

$\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6$

So we have

$z=2e^{i5\pi/6}$

and the fourth roots of $z$ are

$2^{1/4}e^{i(5\pi/6+k\pi)/4}$

where $k\in\{0,1,2,3\}$ . In particular, they are

$2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}$

$2^{1/4}e^{i(5\pi/6+2\pi)/4}=2^{1/4}e^{i17\pi/24}$

$2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}$

$2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}$

7 0

2 years ago