How many points are used to define a plane? O zero O one O two three ree

What is 13*14+56-31+854=

levacccp [35]

Answer:

1,061. your welcome

6 0

3 years ago

Ill give brainlist to whoever helps me w this question

spayn [35]

The answer is C because multiplication comes first

7 0

3 years ago

Find the value of k for which the line y = kx + 6 is a tangent to the curve x^2 + y^2 – 10x + 8y = 84

guajiro [1.7K]

Answer:

k = 1/2

Step-by-step explanation:

input y = kx +6 into the equation of the curve x² + y² – 10x + 8y = 84

x² + (kx + 6)² - 10x + 8(kx + 6) = 84

expand:

x² + k²x² + 12kx + 36 - 10x + 8kx + 48 = 84

simplify by collecting like terms:

x² + k²x² + 20kx - 10x + 84 = 84

subtract 84 on both sides to bring it to the left:

x² + k²x² + 20kx - 10x + 84 - 84 = 0

x² + k²x² + 20kx - 10x = 0

factorise out x:

x²(1 + k²) + x(20k - 10) = 0

using the discriminant b² - 4ac where b is 20k - 10, a is 1 + k² and c is 0, substitute them in the formula b² - 4ac:

b² - 4ac

(20k - 10)² - 4(1 + k²)(0) = 0

the part highlighted in bold is gone because it's all multiplied by 0, so we are left with (20k - 10)² = 0

(20k - 10)² is the same as

(20k - 10)(20k - 10)

equate both to 0

20k - 10 = 0 and 20k - 10 = 0

add 10 on both sides

20k = 10 and 20k = 10

divide 20 on both sides

k = 10/20 and k = 10/20 which are both the same

10/20 is simplified to 1/2

k = 1/2

5 0

2 years ago

Which of the following represents a value greater than 50,000 but less than 60,000

saul85 [17]

Answer:

Any number between them both

7 0

3 years ago

Read 2 more answers

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

So

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

So

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

3 years ago