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2 years ago

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Mathematics

1 answer:

Shkiper50 [21]2 years ago

4 0

Answer:

Step-by-step explanation:

$(2^{3})^{3} = 2^{3*3}=2^{9}\\\\\\(\dfrac{4^{2}}{5^{3}})^{3} = \dfrac{(4^{2})^{3}}{(5^{3})^{3}}=\dfrac{4^{6}}{5^{9}}\\\\\dfrac{3^{6}}{3^{3}}=3^{6-3}=3^{3}\\\\\\a^{m}*a^{n}=a^{m+n}\\\\(a^{m})^{n}=a^{m*n}\\\\\dfrac{a^{m}}{a^{n}}=a^{m-n}$

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The Jones family has saved a maximum of $750 for their family vacation to the beach. While planning the trip, they determine tha

stiks02 [169]

Answer:

Step-by-step explanation:

The Jones family has saved a maximum of $750 for their family vacation to the beach. It means that whatever they would do during the vacation, they cannot spend more than $750.

The inequality can be used to determine the number of nights the Jones family could spend at the hotel is

750 ≥ 75 + 125h

This can also be written as

75 + 125h ≤ 750

We would solve the inequality for the value of h.

125h ≤ 750 - 75

125h ≤ 675

h ≤ 675/125

h ≤ 5.4

For any value of h greater than 5.4 hours, the inequality will not be true.

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2 years ago

How does the range of g(x)=6/x compare with the range of the parent function (f)=1/x?

kogti [31]

Since the only exclusion from the range is y=0, vertically scaling the graph by a factor of 6 doesn't change the range. (0 is still 0)

The appropriate choice is ...
<span>B: The range of both f(x) and g(x) is all nonzero real numbers</span>

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3 years ago

What is the solution to the equation e^3x=12 Round your answer to the nearest hundredth

Lunna [17]

Keywords:

<em>equation, variable, clear, round, centesima, neperian logarithm, exponential </em>

For this case we have the following equation $e ^ {3x} = 12$ , from which we must clear the value of the variable "x" and round to the nearest hundredth. To do this, we must apply properties of neperian and exponential logarithms. By definition:

$ln (e ^ x) = x$

So:

$e ^ {3x} = 12$ We apply Neperian logarithm to both sides:

$ln (e^{3x}) = ln (12)\\3x = ln (12)$

We divide between "3" both sides of the equation:

$\frac {3x} {3} = \frac {ln (12)} {3}\\x = \frac {ln (12)} {3}\\x = 0.828302217$

Rounding out the nearest hundredth we have:

$x = 0.83$

Answer:

$x = 0.83$

3 0

3 years ago

Read 2 more answers

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago

During the summer, the height of the water in the pool decreased by 5 inches each week due to evaporation. What is the change in

rjkz [21]

Answer:

40 inches

Step-by-step explanation:

We know that the height of water in a pool decreased 5 inches per week.

This means that we can form the equation $y=-5x$ , since the amount of water remaining is proportionate to the amount of weeks that have passed.

We know $x=8$ , since we need to find the change for 8 weeks.

$y = -5\cdot 8$

$-5 \cdot 8 = -40$

$y=-40$

The change will be the absolute value of -40, which just makes it positive, so 40.

Hope this helped!

8 0

3 years ago

Read 2 more answers