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3 years ago

HELP PLEASEEEEEE simplify cos^2 θ(1+tan^2 θ)

Mathematics

1 answer:

expeople1 [14]3 years ago

7 0

Answer:

Step-by-step explanation:

cos²Ф(1 + tan²Ф) Simplify 1 + tan²Ф

cos²Ф(1/cos²Ф) Multiply

If this answer is correct, please make me Brainliest!

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The temperature readings from a thermocouple in a furnace fluctuate according to a cumulative distribution function,

Oliga [24]

Answer:

$P(x$

$P(x>808.40) = 0.58$

$P(x\le805.40\ or\ x\ge 808.40) = 0.85$

Step-by-step explanation:

Given:

$F(x) = \left\{\begin{array}{ll}{0} &; {x$

Solving (a): P(x < 810)

To solve this, we make use of:

$P(x$

Because

$800^\circ C \le 810 \le 820^\circ C$

So:

$F(x) = 0.05 * 810 -40$

$F(x) = 40.5 -40$

$F(x) = 0.5$

Hence:

$P(x$

Solving (b): P(x < 800)

To solve this, we make use of:

$P(x$

Because:

$x < 800^\circ C$

So:

$P(x$

Solving (c): P(x > 808.40)

Here, we make use of complement rule:

$P(x>808.40) = 1 - P(x \le 808.40)$

Where:

$P(x\le808.40) = 0.05x - 40$

$P(x\le808.40) = 0.05*808.40 - 40$

$P(x\le808.40) = 0.42$

So:

$P(x>808.40) = 1 - 0.42$

$P(x>808.40) = 0.58$

Solving (d) The probability that the temperature lies outside 805.60 and 808.40

First, we calculate the probability that it lies within.

This is represented as:

$P(805.40 < x < 808.40)$

This is calculated using:

$P(805.40 < x < 808.40) = F(808.40) - F(805.40)$

$P(805.40 < x < 808.40) = 0.05*808.40-40 - (0.05*805.40-40)$

$P(805.40 < x < 808.40) = 0.42 - (0.27)$

$P(805.40 < x < 808.40) = 0.42 - 0.27$

$P(805.40 < x < 808.40) = 0.15$

Using complement rule, the required probability is:

$P(x\le805.40\ or\ x\ge 808.40) = 1 - 0.15$

$P(x\le805.40\ or\ x\ge 808.40) = 0.85$

5 0

3 years ago

\WILL GIVE BRAINLY, PLEASE HELP

MissTica

Answer:

....

Step-by-step explanation:

3 0

3 years ago

Consider the function f(x, y) = x2 + xy + y2 defined on the unit disc, namely, d = {(x, y)| x2 + y2 ≤ 1}. use the method of lagr

Pani-rosa [81]

First we note that the partial derivatives vanish simultaneously at one point:

$\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)$

so there is one critical point within the region $D$ .

The Lagrangian is

$L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)$

and has partial derivatives

$L_x=2x+y+2\lambda x$
$L_y=x+2y+2\lambda y$
$L_\lambda=x^2+y^2-1$

Set each partial derivative to 0 to find the possible critical points within the disk $D$ . Then we notice that

$yL_x=2xy+y^2+2\lambda xy=0$
$xL_y=x^2+2xy+2\lambda xy=0$
$L_\lambda=0\implies x^2+y^2=1$

$\implies xL_y-yL_x=x^2-y^2=0$

Since $x^2+y^2=1$ , we have

$x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}$

And since $x^2-y^2=0$ , or $x^2=y^2$ , we also have

$x=\pm\dfrac1{\sqrt2}$

So we have four possible additional critical points to consider:

$f(0,0)=0$
$f\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac32$
$f\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac32$

It should be clear enough which of these correspond to the absolute extrema of $f$ over $D$ .

8 0

3 years ago

6*10^-2 is how many times as 2•10-8

andreev551 [17]

Answer:

200

Step-by-step explanation:

6*10^-2=0.06

2*10-8=12

12 divided by 0.06 is 200

8 0

3 years ago

Explain the steps you would take to find the area of the following composite shape. (Use 3.14 for Tr.)

USPshnik [31]

The first composite shape we would need to find is the rectangle. We would use the formula length multiplied by width. The second composite shape is a hemisphere. The area of a hemisphere is Pi multiplied by the radius squared and then divide it by 2. To find the area of both of these figures, add up the area for each one and then you will get the
area of both.

4 0

2 years ago