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4 years ago

What is the slope of the line between (-3,6) and (-5,9)?

Mathematics

2 answers:

Margaret [11]4 years ago

3 0

the answer would be 3/-2.

Anettt [7]4 years ago

3 0

Answer:

3/-2

Step-by-step explanation:

You do the slop intercept form which is y2-y1/x2-x1

So it would be

y2=9

y1=6

9-6=3

x2=-5

x1=-3

-5+3=-2

so you get 3/-2

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. The figure below shows a circular driveway. What is the area of the driveway?

Elanso [62]

Answer:

Step-by-step explanation:

A= πr squared

D=r*2

r=24

π=3.14

A= π(24 * 24) = 576

A= π576

A=10808.64

A=10808.6

4 0

3 years ago

Jackson deposited $5,000 at 3.8% interest, compounded continuously, when he was 18 years old. How much will be in the account wh

Ivahew [28]

5,000 × 3.8%= 19,000×18= 342,000 × 40 = 13,680,000

6 0

4 years ago

Solve the given equation for X. X/3 + X/7 = 5/3

enyata [817]

The answer is 7/2
In decimal form it’s 3.5

4 0

3 years ago

Read 2 more answers

For which equation is x = 1 a solution?

boyakko [2]

Answer:

The answer is x+3=4

x+3=4

x=4-3

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5 0

3 years ago

Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1

leva [86]

Answer:

$S_{1893} =5632.98$

Step-by-step explanation:

The correct form of the question is:

$S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1$

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

$S_n = \frac{a(1 - r^n)}{1 - r}$

But first, we need to get the number of terms in the sequence using:

$T_n = ar^{n-1}$

Where

$a = First\ Term$

$a = 1.001^{1892}$

$r = common\ ratio$

$r = \frac{1}{1.001}$

$T_n = Last\ Term$

$T_n = 1$

So, we have:

$T_n = ar^{n-1}$

$1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}$

Apply law of indices:

$1 = 1.001^{1892} * (1.001^{-1})^{n-1}$

$1 = 1.001^{1892} * (1.001)^{-n+1}$

Apply law of indices:

$1 = 1.001^{1892-n+1}$

$1 = 1.001^{1892+1-n}$

$1 = 1.001^{1893-n}$

Represent 1 as $1.001^0$

$1.001^0 = 1.001^{1893-n}$

They have the same base:

So, we have

$0 = 1893-n$

Solve for n

$n = 1893$

So, there are 1893 terms in the sequence given.

Solving further:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Where

$a = 1.001^{1892}$

$r = \frac{1}{1.001}$

$n = 1893$

So, we have:

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }$

Simplify the numerator

$S_{1893} =\frac{1.001^{1892} -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{1892-1893}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{-1}}{\frac{0.001}{1.001} }$

$S_{1893} =(1.001^{1892} -1.001^{-1})/({\frac{0.001}{1.001} })$

$S_{1893} =(1.001^{1892} -1.001^{-1})*{\frac{1.001}{0.001}}$

$S_{1893} =\frac{(1.001^{1892} -1.001^{-1}) * 1.001}{0.001}$

Open Bracket

$S_{1893} =\frac{1.001^{1892}* 1.001 -1.001^{-1}* 1.001 }{0.001}$

$S_{1893} =\frac{1.001^{1892+1} -1.001^{-1+1}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1.001^{0}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1}{0.001}$

$S_{1893} =5632.97970294$

Hence, the sum of the sequence is:

$S_{1893} =5632.98$ ----- approximated

4 0

3 years ago