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2 years ago

Two planes intersect in exactly one point

Mathematics

1 answer:

adell [148]2 years ago

7 0

Answer:

If two planes intersect, then their intersection is a line. A line contains at least two points. If two lines intersect, then exactly one plane contains both lines. If a point lies outside a line, then exactly one plane contains both the line and the point.

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Use graph to answer question

slamgirl [31]

Answer:

cos(θ) = 3/5

Step-by-step explanation:

We can think of this situation as a triangle rectangle (you can see it in the image below).

Here, we have a triangle rectangle with an angle θ, such that the adjacent cathetus to θ is 3 units long, and the cathetus opposite to θ is 4 units long.

Here we want to find cos(θ).

You should remember:

cos(θ) = (adjacent cathetus)/(hypotenuse)

We already know that the adjacent cathetus is equal to 3.

And for the hypotenuse, we can use the Pythagorean's theorem, which says that the sum of the squares of the cathetus is equal to the square of the hypotenuse, this is:

3^2 + 4^2 = H^2

We can solve this for H, to get:

H = √( 3^2 + 4^2) = √(9 + 16) = √25 = 5

The hypotenuse is 5 units long.

Then we have:

cos(θ) = (adjacent cathetus)/(hypotenuse)

cos(θ) = 3/5

5 0

2 years ago

Find the solution of 3 times the square root of the quantity of x plus 6 equals negative 12, and determine if it is an extraneou

Anna35 [415]

For this case we have the following equation:
$3 \sqrt{x+6}=-12$
Rewriting we have:
$\sqrt{x+6}= \frac{-12}{3}$
$\sqrt{x+6}=-4$
We raise both members of the equation to the square:
$(\sqrt{x+6})^2=(-4)^2$
Rewriting we have:
$x+6=16$
$x=16-6$
$x=10$
It's a extraneous solution because equality is not met by substituting x = 10 in the original equation:
$3 \sqrt{10+6}=-12$
$3 \sqrt{16}=-12$
$3(4)=-12$
$12=-12$
Answer:
The solution is:
$x=10$
It's a extraneous solution

6 0

3 years ago

How do i divide an equation over a negitive number?

mina [271]

Multiply both sides by negative four to get rid of the fraction

j+18=-32

subtract 18

j=-50

5 0

2 years ago

Henge

stiks02 [169]

Answer:

$356.78

Step-by-step explanation:

$178.39 * 2

3 0

2 years ago

Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$

c) cumulative distribution function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

6 0

3 years ago