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DerKrebs [107]
4 years ago
5

For the geometric sequence of a1=2 and r=2 find a5

Mathematics
1 answer:
Brrunno [24]4 years ago
3 0

Answer:

a_5 = 32

Step-by-step explanation:

The nth term for the geometric sequence is given by:

a_n = a_1 \cdot r^{n-1}

where,

a_1 is the first term

r is the common ratio

n is the number of terms.

As per the statement:

For the geometric sequence of a_1=2 and r=2

We have to find a_5

for n = 5;

a_5=a_1 \cdot r^{n-1}

Substitute the given values we have;

a_5 = 2 \cdot 2^4 = 2 \cdot 16

⇒a_5 = 32

Therefore, the value of a_5 is, 32

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Answer: Length = 4√5 units.
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A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions
faust18 [17]

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3><u>Solution:</u></h3>

Given that  

Area of rectangular parking lot = 15000 square feet

Length is 20 feet more than the width.

Need to find the dimensions of rectangular parking lot.

Let assume width of the rectangular parking lot in feet be represented by variable "x"

As Length is 20 feet more than the width,

so length of rectangular parking plot = 20 + width of the rectangular parking plot

=> length of rectangular parking plot = 20 + x = x + 20

<em><u>The area of rectangle is given as:</u></em>

\text {Area of rectangle }=length \times width

Area of rectangular parking lot = length of rectangular parking plot \times width of the rectangular parking

\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}

But it is given that Area of rectangular parking lot = 15000 square feet

\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}

Solving the above quadratic equation using quadratic formula

<em><u>General form of quadratic equation is  </u></em>

{ax^{2}+\mathrm{b} x+\mathrm{c}=0

And quadratic formula for getting roots of quadratic equation is

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}

\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}

As variable x represents width of the rectangular parking lot, it cannot be negative.

=> Width of the rectangular parking lot "x" = 112.882 feet  

=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882

Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.

3 0
3 years ago
Approximate the square root value of 5 to the nearest Hundredth (50 points and will give most brainiest answer)
Softa [21]

Answer:The approximate square root is 2.23

Step-by-step explanation:√ 5  = 2.23

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One positive number exceeds three times another positive number by 5. The product of the two numbers is 68. Find the number
kati45 [8]

Answer:12

Step-by-step explanation:

Let the numbers be a and b

a-3b=5 ........equation (1)

There product is :

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From equation (1)

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Make a the subject of the formula

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Substitute equation (3) into equation (2)

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So the number is 12

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4 years ago
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