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3 years ago

A lot of points for this...

Mathematics

2 answers:

andriy [413]3 years ago

8 0

Rate of change, slope, constant are the answers

Tom [10]3 years ago

3 0

Answer:

Rate of change, slope, constant are the answers

Step-by-step explanation:

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Which is an equation in point-slope form for the given point and slope? Point: (5, 9); Slope: 2

Ray Of Light [21]

$(\stackrel{x_1}{5}~,~\stackrel{y_1}{9})~\hfill \stackrel{slope}{m}\implies 2 ~\hfill \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{9}=\stackrel{m}{2}(x-\stackrel{x_1}{5})$

7 0

2 years ago

What is the smallest power of 10 that exceeds this number 235,452?

taurus [48]

${10}^{6}$
10 to the power of 6 equals 1,000,000.

8 0

3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

3 years ago

I don't know how to solve it.

Nataly [62]

Step-by-step explanation:

Take the first derivative

$\frac{d}{dx} ( {x}^{3} - 3x)$

$3 {x}^{2} - 3$

Set the derivative equal to 0.

$3 {x}^{2} - 3 = 0$

$3 {x}^{2} = 3$

${x}^{2} = 1$

$x = 1$

$x = - 1$

For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).

For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.

Since we are going to Positve to negative slope, we have a local max at x=-1

Plug in -1 for x into the original function

$( - 1) {}^{3} - 3( - 1) = 2$

So the local max is 2 and occurs at x=-1,

For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.

Since we are going to decreasing to increasing, we have minimum at x=1,

Plug in 1 for x into original function

${1}^{3} - 3(1)$

$1 - 3 = - 2$

So the local min occurs at -2, at x=1

8 0

2 years ago

Beaker A contains 1 liter which is 30 percent oil and the rest is vinegar, thoroughly mixed up. Beaker B contains 2 liters which

alekssr [168]

Answer:

1.25 liters of oil

Step-by-step explanation:

Volume in Beaker A = 1 L

Volume of Oil in Beaker A = 1*0.3 = 0.3 L

Volume of Vinegar in Beaker A = 1*0.7 = 0.7 L

Volume in Beaker B = 2 L

Volume of Oil in Beaker B = 2*0.4 = 0.8 L

Volume of Vinegar in Beaker B = 1*0.6 = 1.2 L

If half of the contents of B are poured into A and assuming a homogeneous mixture, the new volumes of oil (Voa) and vinegar (Vva) in beaker A are:

$V_{oa} = 0.3+\frac{0.8}{2} \\V_{oa} = 0.7 \\V_{va} = 0.7+\frac{1.2}{2} \\V_{va} = 1.3$

The amount of oil needed to be added to beaker A in order to produce a mixture which is 60 percent oil (Vomix) is given by:

$0.6*V_{total} = V_{oa} +V_{omix}\\0.6*(V_{va}+V_{oa} +V_{omix}) = V_{oa} +V_{omix}\\0.6*(1.3+0.7+V_{omix})=0.7+V_{omix}\\V_{omix}=\frac{0.5}{0.4} \\V_{omix}=1.25 \ L$

1.25 liters of oil are needed.

6 0

3 years ago