There were 50 dogs and 30 of them died how many is left

Mathematics

1 answer:

Elis [28]3 years ago

3 0

If 50 dogs are alive and 30 die there are 20 left

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If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability of getting two queens and one king

NeX [460]

There are 4 choose 3 ways to hold 3 kings and 4 possible queens to make up the fourth card of the hand out of a total of 54 choose 4 possible four-card hands, which gives us the following expression for the probability of being dealt 3 kings and a queen out of 4 cards: Or roughly 1 in every 16920 four card hands.

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3 years ago

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A bathtub contains 20 gallons of water. After the drain is opened, there are 14 gallons of water remaining after 30 seconds. Usi

leva [86]

Answer

a decrease of 0.2 gallons per second

Explanation

The drain was open for 30 seconds.

The water that ran out for 30 seconds = 20 - 14

= 6 gallons

Rate of decrease = (change in gallons)/(time taken)

= 6/30

= 0.2

The rate of change in the amount of water remaining in the tub per second will be the same as the one that ran out since the outlet is the same.

So the correct answer from the choices is "a decrease of 0.2 gallons per second".

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4 years ago

If u(x)=-2x²+3 and v(x)=1/x, what is the range of (u ° v)(x)?

Ludmilka [50]

$u(x)=-2x^2+3\\\\v(x)=\dfrac{1}{x}\\\\(u\ \circ\ v)(x)=-2\left(\dfrac{1}{x}\right)^2+3=-2\left(\dfrac{1}{x^2}\right)+3=-\dfrac{1}{x^2}+3\\\\\text{The range of}\ y=\dfrac{1}{x^2}\ \text{is all positive real numbers.}\\\\\text{The range of}\ y=-\dfrac{1}{x^2}\ \text{is all negative real numbers.}\\\\\text{The range of }\ (u\ \circ\ v)(x)=-\dfrac{1}{x^2}+3\ is\ (-\infty,\ 3)$

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4 years ago

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Find the equation of the tangent of x^2- xy + y^2 =7 at (-1, 2)

Citrus2011 [14]

Answer:

It is 3y = 4x + 10

Step-by-step explanation:

Let's first get the slope of the curve.

[ slope is the derivative of the equation ]

${x}^{2} - xy + {y}^{2} = 7$

introduce dy/dx :

$\frac{d}{dx} ( {x}^{2} - xy + {y}^{2} ) = \frac{d}{dx} (7) \\ \\ 2x - (y + \frac{dy}{dx} ) + 2y \frac{dy}{dx} = 0$

make dy/dx the subject:

$2x - y - \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \\ \\ 2y \frac{dy}{dx} - \frac{dy}{dx} = y - 2x \\ \\ \frac{dy}{dx} (2y - 1) = y - 2x \\ \\ \frac{dy}{dx} = \frac{y - 2x}{2y - 1}$