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kozerog [31]
3 years ago
5

What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

Mathematics
1 answer:
vivado [14]3 years ago
7 0
Successive terms have a common factor of -\dfrac23, which means the nth partial sum of the series can be written as

S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}
S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}

\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)

\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)
\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)
S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)

As n\to\infty, the geometric term vanishes, leaving you with

S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5
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\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}

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6 0
3 years ago
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For every boy in an advanced math course there are 3 girls who are taking the course.
Nutka1998 [239]

Answer:

If 32 students are enrolled in the course, there would be 24 girls.

Step-by-step explanation:

Given that For every boy in an advanced math course there are 3 girls who are taking the course.

It means if there are four students in the class. There would be 3 girls and 1 boy. Thus, 75% of the students are girls.

Given that there are 32 students enrolled in the course.

so

75% of 32 = 75/100 × 32

                 = 24

Thus, if 32 students are enrolled in the course, there would be 24 girls.

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2 years ago
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In the circle below, AD is a diameter and AB is tangent at A. suppose mADC=228. Find the measures of mCAB and mCAD. Type your nu
Tju [1.3M]

Answer:

m∠CAB = 66°

m∠CAD = 24°

Step-by-step explanation:

<em>m∠CAB</em>

The given parameters are;

The measure of arc m\widehat{ADC} = 228°

The diameter of the given circle = \overline{AD}

The tangent to the circle = \underset{AB}{\leftrightarrow}

The measure of m∠CAB and m∠CAD = Required

By the tangent and chord circle theorem, we have;

m∠CAB = (1/2) × m\widehat{AC}

However, we have;

m\widehat{AC} + m\widehat{ADC} = 360° the sum of angles at the center of a circle is 360°

∴ m\widehat{AC} = 360° - m\widehat{ADC}

Which gives;

m\widehat{AC} = 360° - 228° = 132°

m\widehat{AC} = 132°

Therefore;

m∠CAB = (1/2) × 132° = 66°

m∠CAB = 66°

<em>m∠CAD</em>

Given that  \overline{AD} is the diameter of the given circle, we have

The tangent, \underset{AB}{\leftrightarrow}, is perpendicular to the radius of the circle, and therefore \underset{AB}{\leftrightarrow} is also perpendicular to the diameter of the circle

∴ m∠DAB = 90° which is the measure of the angle formed by two perpendicular lines

By angle addition property, we have;

m∠DAB = m∠CAB + m∠CAD

∴ m∠CAD =  m∠DAB - m∠CAB

By substitution, we have;

m∠CAD = 90° - 66° = 24°

m∠CAD = 24°

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3 years ago
Rewrite 16 in exponent form using 2 as the base.
VashaNatasha [74]

Answer:

16=24

Step-by-step explanation:

The first step to write any number in exponential form is to do the prime factorization of the given number. As we can see that 2 is multiplied 4 times so by using exponent we can write 16 as 24

8 0
2 years ago
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