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horrorfan [7]
3 years ago
15

Rewrite this exponential equation as a logarithmic equation.

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
3 0

Answer:

Step-by-step explanation:

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We want to find word problems that gives rise to the following systems of algebraic equations.

a)

x+y=13

x-y=1.

Answer: The sum of the ages of two students is 13. The difference between their ages is 1. Find the ages of the two students.

b)

3x-30y=15

2x+10y=40

Answer:

The difference between 3 times Dan's age and 30 times Mark's age is 15. If the sum of 2 times Dan's age and 10 times Mark's age is 40. Find the ages of Dan and Mark.

c)

8x+3y=37

8x-3y=50

Answer:

The sum of 8 times an eagle's distance above sea level in feet and a herring's distance below sea level is 37 feet. The difference between 8 times an eagle's distance in feet above sea level and 3 times the herring's distance below sea level is 50. Find the distance of the eagle and the herring relative to the surface of the sea.

d) x-5y=4

3x+5y=32

The difference between a pig's age and 5 times the age of a piglet is 4 years. If the sum of 3 times pigs and 5 times the piglet's age is 32 years, find the ages of the pig and its piglet.

7 0
3 years ago
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DaniilM [7]

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Step-by-step explanation:

7 0
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100 Points! Will give brainliest!
disa [49]

-3 ≤ 6x - 6 < 39 (seperate into two inequalities)

6x - 9 ≥ -3

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6 0
3 years ago
Read 2 more answers
Consider the function f(x) = x^3 - 7x^2 - 2x + 14
mars1129 [50]

Answer:

There are 3 zeros, which are:

x=7,\:x=-\sqrt{2},\:x=\sqrt{2}

Step-by-step explanation:

Given the function

f\left(x\right)\:=\:x^3\:-\:7x^2\:-\:2x\:+\:14

To get the zeros of f(x), set y or f(x) =0

so

0=x^3-7x^2-2x+14

as

x^3\:-\:7x^2\:-\:2x\:+\:14=\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)

so

\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0

Using the zero factor principle:

\mathrm{ \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

so

x-7=0\quad \mathrm{or}\quad \:x+\sqrt{2}=0\quad \mathrm{or}\quad \:x-\sqrt{2}=0

solving

x-7=0

  • x=7

x+\sqrt{2}=0

  • x=-\sqrt{2}

x-\sqrt{2}=0

  • x=\sqrt{2}

Therefore, there are 3 zeros, which are:

x=7,\:x=-\sqrt{2},\:x=\sqrt{2}

8 0
3 years ago
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