All categories

3 years ago

He first three terms of a geometric sequence are as follows. 10 , 30 , 90 find the next two terms of this sequence.+

1 answer:

4 0

10*3=30
30*3=90
90*3=270
270*3=810

or

10*3^1=30
10*3^2=90
10*3^3=270
10*3^4=810

the next two terms of this sequence are 270 and 810.

You might be interested in

I NEED HELP PLEASE, THANKS! :)

NISA [10]

Answer: $\bold{b)\quad \dfrac{2\pi}{3},\dfrac{4\pi}{3},0}$

<u>Step-by-step explanation:</u>

Note the following identities: tan² x = sec²x - 1

$\sec x=\dfrac{1}{\cos}$

tan² x + sec x = 1

(sec² x -1) + sec x = 1

sec² x + sec x - 2 = 0

(sec x + 2)(sec x - 1) = 0

sec x + 2 = 0 sec x - 1 = 0

sec x = -2 sec x = 1

$\dfrac{1}{\cos x}=-2\qquad \dfrac{1}{\cos x}=1\\\\\cos x=-\dfrac{1}{2}\qquad \cos x=1\\\\x=\dfrac{2\pi}{3}, \dfrac{4\pi}{3}\qquad x=0$

8 0

3 years ago

Read 2 more answers

Sheila bought several bags of balloons and boxes of candles. The candles cost $15 per bix, and the balloons cowt $8 per bag. She

snow_tiger [21]

Answer:

balloons= 9

candles= 8

Step-by-step explanation:

let the number of balloons be x

and candles be y

15x+8y=192----------1

x+y=17--------------2

from 2, x= 17-y

put x= 17-y in 1

15( 17-y)+8y=192

255-15y+8y=192

solve for y

-7y=192-255

-7y=-63

x=63/7

x=9

put x= 9 in 1

9+y=17

y= 17-9

y=8.

7 0

3 years ago

PLZ HELP QUICKLY , GET BRAINLY

Leviafan [203]

Answer:

D.) Fixed costs do not change no matter how much a business produces; variable costs do change.

Step-by-step explanation:

A variable cost varies with the amount produced, while a fixed cost remains the same no matter how much output a company produces.

I'm 100% sure that this is the answer.

3 0

3 years ago

Evaluate the indefinite integral. <br> integar x4/1 + x^10 dx

ivann1987 [24]

Answer:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c$

Step-by-step explanation:

Given

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

Required

Integrate

We have:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

Let

$u = x^5$

Differentiate

$\frac{du}{dx} = 5x^4$

Make dx the subject

$dx = \frac{du}{5x^4}$

So, we have:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

$\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}$

$\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du$

Express x^(10) as x^(5*2)

$\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du$

Rewrite as:

$\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du$

Recall that: $u = x^5$

$\frac{1}{5} \int\ {\frac{1}{1 + u^2}}} \, du$

Integrate

$\frac{1}{5} * \arctan(u) + c$

Substitute: $u = x^5$

$\frac{1}{5} * \arctan(x^5) + c$

Hence:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c$

7 0

3 years ago

Solve the following quadratic by factoring.<br> X2+14x+45

Jlenok [28]

Answer:

(x+9)(x+5)

Step-by-step explanation:

$x^{2} +14x+45\\ \\x^{2} +5x+9x+45$

x(x+5)+9(x+5)

(x+9)(x+5)

I hope this helps!

6 0

3 years ago

Read 2 more answers