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3 years ago

How do you solve this

Mathematics

1 answer:

Vlad1618 [11]3 years ago

8 0

Try unplugging your computer all the way and them leave it unplugged about 3 min and then re plug it in

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a population of insects increases at a rate of 1.5% per day about how long will it take to double A)2.5 days b) 5.0 days c) 46.6

ANTONII [103]

We can use population formula

$P(t)=P_0(1+r)^{t}$

we are given

interest rate is 1.5% per day

so,

$r=0.015$

now, we can plug values

$P(t)=P_0(1+0.015)^{t}$

$P(t)=P_0(1.015)^{t}$

we are given

P(t)=2Po

we can plug it

$2P_0=P_0(1.015)^{t}$

we can solve for t

$2=(1.015)^{t}$

$t=46.55$

so,

number of days is 46.5..........Answer

6 0

3 years ago

Read 2 more answers

PLS HELP ASPA I WILL GIVE BRAINIEST FOR FIRST PERSON WHO ANSWERS!!!!!!!

ANEK [815]

Answer:

8/15 or 0.53

Step-by-step explanation:

Please mark brainliest ty :)

7 0

3 years ago

Read 2 more answers

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.