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3 years ago

Log 15 (2x − 2) = log 15 (x+9)

Mathematics

1 answer:

notka56 [123]3 years ago

5 0

Answer:

$x=11$

Step-by-step explanation:

Given the equation

$\log_{15}(2x-2)=\log_{15}(x+9)$

Note that

$2x-2>0\\ \\2x>2\\ \\x>1$

and

$x+9>0\\ \\x>-9$

Combined, $x>1$

Solve the equation:

$2x-2= x+9\\ \\2x-x=9+2\\ \\x=11$

Since $11>1,$ this is the solution to the equation.

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Counting back from 5 what number follow 4

qwelly [4]

Answer:

5 4 3 2 1

Step-by-step explanation:

is this what ur asking

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Answer ASAP 15 points

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A or 5/7, they’re both odd numbers so it cannot be simplified any further

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3 years ago

PLEASE HELP LAST QUESTION

igomit [66]

ANSWER:

a) y = 15x + 75

b) 50 weeks

EXPLANATION:

P A R T A

2 weeks, 105 dollars --> (2, 105)

5 weeks, 150 dollars --> (5, 150)

if we use slope intercept form (y=mx+b), we must:

1) find the slope

y2 - y1 -->> 150 - 105 = 45

x2 - x1 -->> 5 - 2 = 3

45/3 = 15

2) find the y intercept

y = 15x + b -->> 150 = 15 (5) + b -->> 150 = 75 + b -->> 75 = b

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P A R T B

y = 15x + 75 (substitute the y with 825 because that is the goal)

825 = 15x + 75 (solve normally)

750 = 15x

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5 0

3 years ago

Pyramid A has a triangular base where each side measures 4 units and a volume of 36 cubic units. Pyramid B has the same height,

omeli [17]

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

Pyramid A

$s = 4$ -- base sides

$V = 36$ -- Volume

Pyramid B

$s = 6$ --- base sides

Required

Determine the volume of pyramid B [Missing from the question]

From the question, we understand that both pyramids are equilateral triangular pyramids.

The volume is calculated as:

$V = \frac{1}{3} * B * h$

Where B represents the area of the base equilateral triangle, and it is calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

$B = \frac{1}{2} * s^2 * sin(60)$

$B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}$

$B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}$

$B = 4\sqrt 3$

The height is calculated from:

$V = \frac{1}{3} * B * h$

This gives:

$36 = \frac{1}{3} * 4\sqrt 3 * h$

Make h the subject

$h = \frac{3 * 36}{4\sqrt 3}$

$h = \frac{3 * 9}{\sqrt 3}$

$h = \frac{27}{\sqrt 3}$

To calculate the volume of pyramid B, we make use of:

$V = \frac{1}{3} * B * h$

Since the heights of both pyramids are the same, we can make use of:

$h = \frac{27}{\sqrt 3}$

The base area B, is then calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where

$s = 6$

So:

$B = \frac{1}{2} * 6^2 * sin(60)$

$B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}$

$B = 9\sqrt 3$

So:

$V = \frac{1}{3} * B * h$

Where

$B = 9\sqrt 3$ and $h = \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9 * 27$

$V = 81$

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3 years ago

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Soloha48 [4]

Answer:

rod waves can give u the answer for that just play a some him

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