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3 years ago

For the expression 6 - y + 3, determine the coefficient for the variable

Mathematics

1 answer:

natulia [17]3 years ago

3 0

Answer:

Step-by-step explanation:

The coefficient of a variable is always the number before the variable, if there is no number before a variable the coefficient is always 1.

In this case, since the subtraction sign is before the variable, you will need to make the variable positive by making the coefficient of variable y, -1.

In short terms, your answer is -1. :)

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5tens and 13ones is equal?

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A coin is flipped eight times where each flip comes up either heads or tails. How many possible outcomes a) are there in total?

attashe74 [19]

Answer:

There are 256 ways in total.

There are 56 possible outcomes contain exactly three heads.

The possible outcomes contain at least three heads is 219.

The possible outcomes contain the same number of heads and tails are 70.

Step-by-step explanation:

Consider the provided information.

A coin is flipped eight times where each flip comes up either heads or tails.

Part (a) How many possible outcomes are there in total?

Each time we flip a coin it comes up either heads or tail.

Therefore the total number of ways are:

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^8=256$

Hence, there are 256 ways in total.

Part (b) contain exactly three heads?

We want exactly 3 heads, therefore,

n=8 and r=3

According to the definition of combination: $\binom{n}{r}=\frac{n!}{r!(n-r)!}$

$\binom{8}{3}=\frac{8!}{3!(5)!}=56$

Hence, there are 56 possible outcomes contain exactly three heads.

Part (c) contain at least three heads?

For 3 heads: $\binom{8}{3}=\frac{8!}{3!(5)!}=56$

For 4 heads: $\binom{8}{4}=\frac{8!}{4!(4)!}=70$

For 5 heads: $\binom{8}{5}=\frac{8!}{5!(3)!}=56$

For 6 heads: $\binom{8}{6}=\frac{8!}{6!(2)!}=28$

For 7 heads: $\binom{8}{7}=\frac{8!}{7!(1)!}=8$

For 8 heads: $\binom{8}{8}=1$

Now add them as shown:

56+70+56+28+8+1=219

Hence, the possible outcomes contain at least three heads is 219.

Part (d) contain the same number of heads and tails?

Same number of heads and tails means that the value of r=4.

Therefore,

$\binom{8}{4}=\frac{8!}{4!(4)!}=70$

Hence, the possible outcomes contain the same number of heads and tails are 70.

6 0

3 years ago

Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ell

Tanzania [10]

Answer:

Length (parallel to the x-axis): $2 \sqrt{2}$ ;

Height (parallel to the y-axis): $4\sqrt{2}$ .

Step-by-step explanation:

Let the top-right vertice of this rectangle $(x,y)$ . $x, y >0$ . The opposite vertice will be at $(-x, -y)$ . The length the rectangle will be $2x$ while its height will be $2y$ .

Function that needs to be maximized: $f(x, y) = (2x)(2y) = 4xy$ .

The rectangle is inscribed in the ellipse. As a result, all its vertices shall be on the ellipse. In other words, they should satisfy the equation for the ellipse. Hence that equation will be the equation for the constraint on x and y.

For Lagrange's Multipliers to work, the constraint shall be in the form: $g(x, y) =k$ . In this case

$\displaystyle g(x, y) = \frac{x^{2}}{4} + \frac{y^{2}}{16}$ .

Start by finding the first derivatives of $f(x, y)$ and $g(x, y)$ with respect to x and y, respectively:

$f_x = y$ ,
$f_y = x$ .

$\displaystyle g_x = \frac{x}{2}$ ,
$\displaystyle g_y = \frac{y}{8}$ .

This method asks for a non-zero constant, $\lambda$ , to satisfy the equations:

$f_x = \lambda g_x$ , and

$f_y = \lambda g_y$ .

(Note that this method still applies even if there are more than two variables.)

That's two equations for three variables. Don't panic. The constraint itself acts as the third equation of this system:

$g(x, y) = k$ .

$\displaystyle \left\{ \begin{aligned} &y = \frac{\lambda x}{2} && (a)\\ &x = \frac{\lambda y}{8} && (b)\\ & \frac{x^{2}}{4} + \frac{y^{2}}{16} = 1 && (c)\end{aligned}\right.$ .

Replace the $y$ in equation $(b)$ with the right-hand side of equation $(b)$ .

$\displaystyle x = \lambda \frac{\lambda \cdot \dfrac{x}{2}}{8} = \frac{\lambda^{2} x}{16}$ .

Before dividing both sides by $x$ , make sure whether $x = 0$ .

If $x = 0$ , the area of the rectangle will equal to zero. That's likely not a solution.

If $x \neq 0$ , divide both sides by $x$ , $\lambda = \pm 4$ . Hence by equation $(b)$ , $y = 2x$ . Replace the $y$ in equation $(c)$ with this expression to obtain (given that $x, y >0$ ) $x = \sqrt{2}$ . Hence $y = 2x = 2\sqrt{2}$ . The length of the rectangle will be $2x = 2\sqrt{2}$ while the height will be $2y = 4\sqrt{2}$ . If there's more than one possible solutions, evaluate the function that needs to be maximized at each point. Choose the point that gives the maximum value.