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Find the equation for the plane through the points upper p 0 left parenthesis negative 2 comma negative 5 comma negative 4 right parenthesis, upper q 0 left parenthesis 5 comma 1 comma negative 4 right parenthesis, and upper r 0 left parenthesis negative 1 comma 2 comma 5 right parenthesis.

2 months ago

Solution 1

Guest #10085

2 months ago

We can suppose the equation of the plane is in the form ...
ax +by +cz = 1
By using the given point coordinates for x, y, and z, we end up with three equations in 3 unknowns. These can be described by the augmented matrix ...
$\left[\begin{array}{ccc|c}-2&-5&-4&1\\5&1&-4&1\\-1&2&5&1 \end{array}\right]$
Putting this in reduced row-echelon form, we find
a = 54/35
b = -63/35
c = 43/35

The equation of the plane through $P_{0}(-2,-5,-4), Q_{0}(5,1,4), R_{0}(-1,2,5)$ is ...
54x -63y +43z = 35

📚 Related Questions

Question

In a family with 8 children, excluding multiple births, what is the probability of having 7 boys and 1 girl, in any order? assume that a boy is as likely as a girl at each birth

Solution 1

There is a 7/8 chance there will be boys and a 1/8 chance there will be a girl

Question

For the particular problem raised in the introduction, assume that the total bill is $44. to answer the question "how should the bill be split?" we will create a linear equation. the unknown is how much money a single person (besides anika) must pay, so call that x. although four people (you plus three friends) went to dinner, only three are paying the unknown amount x for a total of 3x. since anika is paying $2, the total amount paid is 3x+2 dollars, which must equal the amount of the bill, $44. thus, the equation to find x is 3x+2=44. the steps for solving a linear equation are as follows: move all of the constants to the right side. move all of the variable terms (terms containing x) to the left side. divide both sides by the coefficient of the variable to isolate the variable. you will go through these steps one at a time to solve the equation and determine how much each person should pay.

Solution 1

3x+2=44
3x+2(-2)=44-2
3x=42
3x/3=42/3
x=14
So each person would have to pay $14, except Anika

Question

A pair of two distinct dice are rolled six times. suppose none of the ordered pairs of values (1, 5), (2, 6), (3, 4), (5, 5), (5, 3), (6, 1), (6, 2) occur. what is the probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice?

Solution 1

The probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice is 1/6.

What is probability?

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

It is given that:

A pair of two distinct dice are rolled six times. suppose none of the ordered pairs of values (1, 5), (2, 6), (3, 4), (5, 5), (5, 3), (6, 1), (6, 2) occur.

Total number of outcomes = 6x6

Total number of outcomes = 36

Total favourable outcome = 6

The probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice:

P = 6/36 = 1/6

Thus, the probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice is 1/6.

Learn more about the probability here:

brainly.com/question/11234923

#SPJ2

Solution 2

But them in order of add them all up i guess

Question

If the numerator of a fraction is increased by 3, the fraction becomes 3/4. If the denominator is decreased by 7, the fraction becomes 1. Determine the original fraction. Which of the following equations represents "If the numerator of a fraction is increased by 3, the fraction becomes 3/4"?

Solution 1

For this case, the original fraction is:
x / y
Where,
x = numerator
y = denominator:
If the numerator of a fraction is increased by 3, the fraction becomes 3/4:
(x + 3) / y = 3/4
If the denominator is decreased by 7, the fraction becomes 1:
x / (y-7) = 1
Solving the system of equations we have:
x = 9
y = 16
The original fraction is:
9/16
Answer:
the original fraction is:
9/16
"If the numerator of a fraction is increased by 3, the fraction becomes 3/4" is:
(x + 3) / y = 3/4

Solution 2

Let n = numerator
d = denominator

n+3/d = 3/4

n/(d-7) = 1

Which of the following equations represents "If the numerator of a fraction is increased by 3, the fraction becomes 3/4"?

The equation is (n+3)/d = 3/4

Question

A single fair die is tossed. find the probability of rolling a number greater than 55.

Solution 1

0% no # greater than 6

Question

A jar contains 8 large red marbles, 5 small red marbles, 7 large blue marbles, and 4 small blue marbles. if a marble is chosen at random, which of the conditional probabilities is larger, p(red|large) or p(large|red)?

Solution 1

8 large red, 5 small red, 7 large blue, 4 small blue:

totally = 24 marbles

P(red)= 13/24

let A red and B large

P(A/B)= P(A∩B)/P(B)

P(B)= 15/24 =.625

P(A∩B)= 8/24 = .333

P(A/B) =.333/.625 =.533

P(B/A) =.625/.333= 1.875

so the probability of p(large/red) is larger.

Question

"a fair coin is tossed 5 times. what is the probability of exactly 4 heads?"

Solution 1

5 flips render40 % 4 will turn up heads

Question

Evaluate the surface integral. s xyz ds, s is the cone with parametric equations x = u cos(v), y = u sin(v), z = u, 0 ≤ u ≤ 3, 0 ≤ v ≤ π 2

Solution 1

Answer:

$\frac{243\sqrt{2}}{10}$

Step-by-step explanation:

Let $r(u,v)=\left \langle u\cos v,u\sin v,u \right \rangle$

Differentiate partially with respect to u, v.

$r_u=\left \langle \cos v,\sin v,1 \right \rangle\\r_v=\left \langle -u\sin v,u\cos v,0 \right \rangle$

$r_u\times r_v=\left | \begin{matrix} i&j&k\\ \cos v&\sin v&1\\-u\sin v&u\cos v&0 \end{matrix} \right |=\left \langle -u\cos v,-u\sin v,u \right \rangle\\\left \|r_u\times r_v \right \|=\sqrt{u^2\cos ^2v+u^2\sin ^2v+u^2}\\=\sqrt{u^2\left (\cos ^2v+\sin ^2v \right )+u^2}\\=\sqrt{u^2+u^2}\\=\sqrt{2u^2}=\sqrt{2}u$

$dS=\left \| r_u\times r_v \right \|\,du\,\,dv$

$\int \int f\,dS=\int_v \int_u f\,\left \| r_u\times r_v \right \|\,du\,\,dv$

To find f, put $x = u \cos(v)\,,\, y = u \sin(v)\,,\, z = u$ in xyz.

$f=xyz\\=(u \cos(v))( u \sin(v))(u)\\=u^3\sin v\,\cos v\\=u^3\left ( \frac{2\sin v\,\cos v}{2} \right )\\=\frac{u^3}{2}\sin 2v\,\,\left \{ \because 2\sin v\,\cos v=\sin 2v \right \}$

So,

$\int \int f\,dS=\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \frac{u^3}{2}\sin 2v\,\left ( \sqrt{2}u \right )\,du\,\,dv\\=\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}} \int_{0}^{3}u^4\sin 2v\,\,du\,\,dv$

Integrate with respect to u.

$\int \int f\,dS=\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}}\sin 2v\left [ \frac{u^5}{5} \right ]_{0}^{3}\,\,dv\,\,\left \{ \because \int u^n\,du=\frac{u^{n+1}}{n+1} \right \}\\=\frac{\sqrt{2}}{10}(243)\int_{0}^{\frac{\pi}{2}}\sin 2v\,\,dv$

Integrate with respect to v.

$\int \int f\,dS=\frac{243\sqrt{2}}{10}\left [ \frac{-\cos 2v}{2} \right ]_{0}^{\frac{\pi}{2}}\,\,\left \{ \because \int \sin v=-\cos v \right \}\\=\frac{243\sqrt{2}}{20}\left ( -1-1 \right )\\=\frac{243\sqrt{2}}{10}$

Solution 2

Let's capture the parameterization of the surface $\mathcal S$ by the vector function

$\mathbf s(u,v)=\langle u\cos v,u\sin v,u\rangle$

Then the surface element is given by

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv$
$\implies\mathrm dS=\sqrt 2u\,\mathrm du\,\mathrm dv$

So the surface integral is equivalent to

$\displaystyle\iint_{\mathcal S}xyz\,\mathrm dS=\sqrt2\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=3}u^4\cos v\sin v\,\mathrm du\,\mathrm dv=\frac{243}{5\sqrt2}$

Question

Determine the values of the constants b and c so that the function given below is differentiable. f(x)={2xbx2+cxx≤1x>1

Solution 1

Assuming the function is

$f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}$

For $f(x)$ to be differentiable, it necessarily has to be continuous. For this condition to be met, we need

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$
$\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)$
$\iff2=b+c$

For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have

$f'(x)=\begin{cases}2&\text{for }x1\end{cases}$

$\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)$
$\iff2=2b+c$

Now we solve for $b$ and $c$ :

$\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2$

Question

A tank with a capacity of 100 gallons initially contains 50 gallons of water with 10 pounds of salt in solution. fresh water enters at a rate of 2 gallons per minute and a well-stirred mixture is pumped out at the rate of 1 gallon per minute. compute the amount of salt (in pounds) in the tank at the first moment when the tank is filled.

Solution 1

Let $A(t)$ be the amount of salt (in pounds) in the tank at time $t$ . We're given that $A(0)=10\text{ lb}$ .

The rate at which the amount of salt in the tank changes is given by the ODE

$A'(t)=\dfrac{2\text{ gal}}{1\text{ min}}\cdot\dfrac{0\text{ lb}}{1\text{ gal}}-\dfrac{1\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{50+(2-1)t\text{ gal}}$

$A'(t)+\dfrac{A(t)}{50+t}=0$

$(50+t)A'(t)+A(t)=0$

$\bigg((50+t)A(t)\bigg)'=0$

$(50+t)A(t)=C$

$A(t)=\dfrac C{50+t}$

Given that $A(0)=10$ , we find that

$10=\dfrac C{50+0}\implies C=500$

so that the amount of salt in the tank is described by

$A(t)=\dfrac{500}{50+t}$

The tank will be filled when $50+t=100$ , or after $t=50$ minutes. At this time, the amount of salt in the tank is

$A(50)=\dfrac{500}{50+50}=5\text{ lb}$

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Find the equation for the plane through the points upper p 0 left parenthesis negative 2 comma negative 5 comma negative 4 right parenthesis​, upper q 0 left parenthesis 5 comma 1 comma negative 4 right parenthesis​, and upper r 0 left parenthesis negative 1 comma 2 comma 5 right parenthesis.

Solution 1

📚 Related Questions

What is probability?

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Find the equation for the plane through the points upper p 0 left parenthesis negative 2 comma negative 5 comma negative 4 right parenthesis, upper q 0 left parenthesis 5 comma 1 comma negative 4 right parenthesis, and upper r 0 left parenthesis negative 1 comma 2 comma 5 right parenthesis.