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3 years ago

A^3b^2 divided by a^-1b^-3

Mathematics

1 answer:

Naya [18.7K]3 years ago

3 0

Answer:

$\frac{a^3 b^2}{\frac{1}{a} \frac{1}{b^3}}$

And simplifying we got:

$a^3 b^2 a b^3$

$a^3 a b^2 b^3 = a^{3+1} b^{2+3} = a^4 b^5$

Step-by-step explanation:

We want to simplify the following expression:

$\frac{a^3 b^2}{a^{-1} b^{-3}}$

And we can rewrite this expression using this property for any number a:

$a^{-1}= \frac{1}{a}$

And using this property we have:

$\frac{a^3 b^2}{\frac{1}{a} \frac{1}{b^3}}$

And simplifying we got:

$a^3 b^2 a b^3$

$a^3 a b^2 b^3 = a^{3+1} b^{2+3} = a^4 b^5$

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Forty percent of the homes constructed in the Quail Creek area include a security system. Three homes are selected at random: Wh

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Answer: 0.064

Step-by-step explanation:

Binomial probability formula :-

$P(X)=^nC_x \ p^x\ (1-p)^{n-x}$ , where P(x) is the probability of getting success in x trials, n is total number of trials and p is the probability of getting succes in each trial.

Given : The proportion of the homes constructed in the Quail Creek area include a security system : $p=0.40$

Now, if three homes are selected at random, then the probability all three of the selected homes have a security system is given by :-

$P(3)=^3C_3 \ (0.40)^3\ (1-0.40)^{3-3}\\\\=(0.40)^3=0.064$

Hence, the probability all three of the selected homes have a security system = 0.064

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3 years ago

Simplify each expression as much as possible, and rationalize denominators when applicable. √72=?

Misha Larkins [42]

Answer:

6√2

Step-by-step explanation:

Solving the given expression step by step:

72 = 2 × 2 × 2 × 3 × 3

Now for finding the square root we will make pair of two numbers of same values.

We have one pair of 2's and one pair of 3's and left one 2

Thus, √72 = 2 × 3 × √2 = 6√2

We rationalize denominator and change it into a simpler form as soon as possible.

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Answer:

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Step-by-step explanation:

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3 years ago

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = 1.5 m/s downwards

(b) The rate the area formed by the ladder is changing is approximately -75.29 ft.²/sec

Reasons:

The given parameter are;

Length of the ladder, l = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let y represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = 1.5 m/s downwards

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ -6.86 m/s downwards

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ -75.29 ft.²/sec

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ 0.286 rad/sec

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

3 years ago