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3 years ago

(1,4) (2,2) find the midpoint

Mathematics

1 answer:

Art [367]3 years ago

3 0

1,8 is the mid point of these two numbers

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The formula for electrical energy, P, is P = I2R, where I is current and R is resistance. The formula for I in terms of P and R

nydimaria [60]

Answer:

$i = \sqrt{ \frac{p}{r} }$

6 0

3 years ago

If a sample mean is 32, which of the following is most likely the range of

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

It is D, I did this quiz once

8 0

3 years ago

Determine the probability of a negative result for a batch of size 10 if the positivity rate

Vlad1618 [11]

The probability of a negative result for a batch will be "0.3487".

Given values are:

Positivity rate,

p = 10% or, 0.1

Batch size,

n = 10

Now,

The probability of negative result will be:

→ $P (batch \ negative) = (1-p)^n$

By substituting the values, we get

→ $=(1-0.1)^{10}$

→ $= (0.9)^{10}$

→ $= 0.3487$

Thus the answer above is right.

Learn more about the probability here:

brainly.com/question/17927863

8 0

3 years ago

Find the slope of the line passing through the points (-6,2), (0,-6).A 6B 2.C-4/3DOE-3/1

NNADVOKAT [17]

Given data:

The first point given iis (a, b)=(-6,2).

The second point given is (c,d)=(0, -6).

The expression for the slope is,

m=(d-b)/(c-a)

Substitute the given points in the above expression.

m=(-6-2)/(0-(-6))

=(-8)/(6)

=-4/3

Thus, the slope of the line is -4/3, so (C) option is correct.

8 0

1 year ago

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

2 years ago