Ask question

Ask question

All categories

4 years ago

14

Calculate the sum of the infinite series

1 answer:

Rina8888 [55]4 years ago

4 0

Answer:

$S_\infty=250$ .

Step-by-step explanation:

The given infinite series is $50+40+32+\frac{128}{5}+...$ .

The first term of this series is $a_1=50$ .

The common ratio is $r=\frac{40}{50}$ .

$\Rightarrow r=\frac{4}{5}$ .

The sum of this infinite series is given by the formula,

$S_\infty=\frac{a_1}{1-r}$

We now substitute all the above values in to this formula obtain,

$S_\infty=\frac{50}{1-\frac{4}{5}}$

This implies that,

$S_\infty=\frac{50}{\frac{1}{5}}$

This simplifies to,

$S_\infty=50\times 5$

$S_\infty=250$ .

The correct answer is C.

You might be interested in

$5.99 for 18 lb this is the question

-Dominant- [34]

Wait- what’s the question.

7 0

3 years ago

100 POINTS HELP PLEASEEEE !!!HURRY!!!

SIZIF [17.4K]

<em>Question # 1 Solution</em>

<em>Answer:</em>

$(h-f)(-8)=-229$

<em>Step-by-step Solution:</em>

Given

$h(r)=-4r^{2}+4$

$f(r)=2r-7$

As

$(h-f)(r)=h(r)-f(r)$

$(h-f)(r)=-4r^{2}+4-(2r-7)$

$(h-f)(r)=-4r^{2}+4-2r+7)$

We have to find $(h-f)(-8)$

So,

$(h-f)(-8)=-4(-8)^{2}+4-2(-8)+7)$

$(h-f)(-8)=-4(64)+4+16+7$

$(h-f)(-8)=-4(64)+4+16+7$

$(h-f)(-8)=-256+4+16+7$

$(h-f)(-8)=-229$

∴ $(h-f)(-8)=-229$

<em> Question # 2 Solution</em>

Answer:

∴ $(f.p)(9)=38979$

Step-by-step Solution:

Given

$f(s)=7s-2$

$p(s)=s^{3}-10s$

As

$(f.p)(s)=f(s).p(s)$

$(f.p)(s)=(7s-2)(s^{3}-10s)$

$(f.p)(s)=7s^4-2s^3-70s^2+20s$

We have to find $(f.p)(9)$

So,

$(f.p)(9)=7\left(9\right)^4-2\left(9\right)^3-70\left(9\right)^2+20\left(9\right)$

$(f.p)(9)=7\cdot \:9^4-2\cdot \:9^3-70\cdot \:9^2+20\cdot \:9$

$(f.p)(9)=9^4\cdot \:7-9^3\cdot \:2-9^2\cdot \:70+180$

$(f.p)(9)=45927-1458-5670+180$

$(f.p)(9)=38979$

∴ $(f.p)(9)=38979$

<em> Question # 3 Solution</em>

<em>Answer:</em>

$(\frac{f}{p})(r)=\frac{11r+8}{r\left(r^2+6\right)}$

<em>Step-by-step Solution:</em>

Given

$p(r)=r^{3}+6r$

$f(r)=11r+8$

We have to find $(\frac{f}{p})(r)$

Using the formula

$(\frac{f}{p})(r)=\frac{f(r)}{p(r)}$

As

$p(r)=r^{3}+6r$

$f(r)=11r+8$

So

$(\frac{f}{p})(r)=\frac{11r+8}{r^{3}+6r}$

$(\frac{f}{p})(r)=\frac{11r+8}{r\left(r^2+6\right)}$

∴ $(\frac{f}{p})(r)=\frac{11r+8}{r\left(r^2+6\right)}$

<em> Question # 4 Solution</em>

<em>Answer:</em>

$(h-p)(k)=5k^2-3-k^3-8k$

<em>Step-by-step Solution:</em>

Given

$h(k)=5k^{2}-3$

$p(k)=k^{3}+8k$

We have to find $(h-p)(k)$

Using the formula

$(h-p)(k)=h(k)-p(k)$

As

$h(k)=5k^{2}-3$

$p(k)=k^{3}+8k$

So

$(h-p)(k)=5k^{2}-3-(k^{3}+8k)$

$(h-p)(k)=5k^2-3-k^3-8k$

∴ $(h-p)(k)=5k^2-3-k^3-8k$

<em> Question # 5 Solution</em>

<em>Answer:</em>

$(\frac{p}{g})(11)=\frac{1287}{155}$

<em>Step-by-step Solution:</em>

Given

$p(b)=b^{3}-4b$

$g(b)=b^{2}+4b-10$

We have to find $(\frac{p}{g})(11)$

As

$(\frac{p}{g})(b)=\frac{p(b)}{g(b)}$

$(\frac{p}{g})(b)=\frac{b^{3}-4b}{b^{2}+4b-10}$

So

$(\frac{p}{g})(11)=\frac{11^{3}-4(11)}{11^{2}+4(11)-10}$

$(\frac{p}{g})(11)=\frac{1287}{155}$

∴ $(\frac{p}{g})(11)=\frac{1287}{155}$

<em> Question # 6 Solution</em>

<em>Answer:</em>

$(f+g)(x)=x^2+20x-18$

<em>Step-by-step Solution:</em>

Given

$g(x)=x^{2}+11x-7$

$f(x)=9x-11$

We have to find $(f+g)(x)$

As

$(f+g)(x)=f(x)+g(x)$

$(f+g)(x)=9x-11+(x^{2}+11x-7)$

$(f+g)(x)=9x-11+x^{2}+11x-7$

$(f+g)(x)=x^2+20x-18$

∴ $(f+g)(x)=x^2+20x-18$

<em> Question # 7 Solution</em>

<em>Answer:</em>

$h(10)+g(10)=-983$

<em>Step-by-step Solution:</em>

Given

$h(w)=-11w^{2}-7$

$g(w)=w^{2}+3w-6$

We have to find $h(10)+g(10)$

So,

$h(10)=-11(10)^{2}-7$

$h(10)=-1107$ .....[1]

and

$g(10)=10^{2}+3(10)-6$

$g(10)=124$ .....[2]

Adding Equation [1] and Equation [2]

$h(10)+g(10)=-1107+124$

$h(10)+g(10)=-983$

∴ $h(10)+g(10)=-983$

<em> Question # 8 Solution</em>

<em>Answer:</em>

$(f+g)(-3)=-6$

<em>Step-by-step Solution:</em>

Given

$g(b)=b^{2}+9b+10$

$f(b)=3b+11$

We have to find $(f+g)(-3)$

As

$(f+g)(b)=f(b)+g(b)$

So

$(f+g)(b)=3b+11+b^{2}+9b+10$

$(f+g)(-3)=3(-3)+11+(-3)^{2}+9(-3)+10$

$(f+g)(-3)=-6$

∴ $(f+g)(-3)=-6$

<em> </em>

<em> Question # 9 Solution</em>

<em>Answer:</em>

$(f.h)(k)=33k^3+22k-12k^2-8$

<em>Step-by-step Solution:</em>

Given

$f(k)=-11k+4$

$h(k)=-3k^{2}-2$

We have to find $(f.h)(k)$

As

$(f.h)(k)=f(k).h(k)$

So,

$(f.h)(k)=(-11k+4).(-3k^{2}-2)$

$\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$

∵ FOIL means (First, Outer, Inner, Last)

$(f.h)(k)=\left(-11k\right)\left(-3k^2\right)+\left(-11k\right)\left(-2\right)+4\left(-3k^2\right)+4\left(-2\right)$

$(f.h)(k)=33k^3+22k-12k^2-8$

∴ $(f.h)(k)=33k^3+22k-12k^2-8$

<em> Question # 10 Solution</em>

<em>Answer:</em>

$p(-8)-f(-8)=-581$

<em>Step-by-step Solution:</em>

Given

$p(s)=s^{3}+6s$

$f(s)=-2s+5$

We have to find $p(-8)-f(-8)$

So,

$p(-8)=(-8)^{3}+6(-8)$

$p(-8)=-560.....[1]$

and

$f(-8)=-2(-8)+5$

$f(-8)=21.....[2]$

Subtracting Equation [2] from Equation [1]

$p(-8)-f(-8)=-560-21$

$p(-8)-f(-8)=-581$

∴ $p(-8)-f(-8)=-581$

<em>Keywords: function operation</em>

<em>Learn more about function operations from brainly.com/question/3725682</em>

<em>#learnwithBrainly</em>

6 0

3 years ago

Read 2 more answers

What is the circumference of the circle? (use 3.14 for )

wariber [46]

C=pi*d
C=pi*12
C=37.699...cm

Answer= 37.7cm (to 1 dp)

4 0

3 years ago

Read 2 more answers

The interior angles of a triangle have measures kº, 27°, and 10°<br> What is the value of k?

S_A_V [24]

Answer:

k = 143

Step-by-step explanation:

The sum of the 3 interior angles of a triangle = 180°, thus

k = 180 - (27 + 10) = 180 - 37 = 143

4 0

3 years ago

-please help this is the last question i have for tonight and im struggling big time D:

Mrrafil [7]

Answer:

108 or 72

Step-by-step explanation:

:) cdhhyvfhjyvfyhgfvhghghhhjbh

do you have S s-n-a-p-c-h-a-t

8 0

3 years ago