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3 years ago

Find the distance between the two points in simplest radical form. (7,-1) and (9,-9)

Mathematics

2 answers:

storchak [24]3 years ago

7 0

Answer:

$\sqrt{68}$

Step-by-step explanation:

(7 , -1) = (x1 , y1)

(9 , -9) = (x2 , y2)

distance formula = $\sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

= $\sqrt{(9 - 7)^2 + (-9 - (-1))^2}$

= $\sqrt{2^2 + (-9 + 1)^2}$

= $\sqrt{4 + (-8)^2}$

= $\sqrt{4 + 64}$

= $\sqrt{68}$

sweet [91]3 years ago

4 0

Answer:

$Distance = \sqrt{ 68 } \\\\or \\\\Distance =2 \sqrt{17}$

Step-by-step explanation:

$Let \ (x_1 , y_1) = ( 7 , - 1 ) \ and \ (x_2, y _ 2 ) = ( 9 , - 9 )\\\\\\Distance = \sqrt{(x_2 - x_ 1)^2 + (y_ 2 - y_ 1)^2}\\\\$

$= \sqrt{(9 - 7)^2 + ( -9 --1)^2} \\\\=\sqrt{2^2 + ( -9 + 1)^2 }\\\\=\sqrt {4 + (-8)^2 }\\\\=\sqrt{ 4 + 64 }\\\\=\sqrt{68}\\\\= \sqrt { 4 \times 17}\\\\=\sqrt{2^2 \times 17}\\\\=2 \sqrt{17}$

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P and Q are sunsets of the universal set U={n:1<n<10} and n is an integer. P consists of even numbers less than 7 and Q co

Sauron [17]

Part A

<h3>Answer: P' u Q' = {3,4,5,6,7,8,9}</h3>

---------------------------------

Work Shown:

The universal set here is U = {2,3,4,5,6,7,8,9} so we're listing all the integers between 2 and 9. We don't include 1 or 10 since n is larger than 1, and smaller than 10.

P = even numbers less than 7, subset of U

P = {2,4,6,8}

P ' = {3,5,7,9}

Note how the set P' is the set of everything in U that is not in set P. It's the opposite of set P. We call this the complement of set P.

Q = prime numbers less than 10, subset of U

Q = {2,3,5,7}

Q ' = {4,6,8,9}

Applying the union set operation on the sets P' and Q' leads to

P' u Q' = {3,5,7,9, 4,6,8,9}

All I did was combine the two sets of numbers under one umbrella. From here we toss out the duplicate entry of 9. This next step is optional, but sorting the values is standard convention.

Doing all this leads to

P' u Q' = {3,4,5,6,7,8,9}

This set represents all the values that are in set P' or in set Q' or both.

=======================================================

Part B

<h3>Answer: P' n Q' = {9}</h3>

---------------------------------

Work Shown:

Now we're looking at the intersection of sets P' and Q'

List out those sets

P ' = {3,5,7,9}

Q ' = {4,6,8,9}

We see that only the value 9 is in common

Therefore,

P' n Q' = {9}

=======================================================

Part C

Answer:

(P' n Q')' = {2,3,4,5,6,7,8}

---------------------------------

Work Shown:

You start with the result from part B. Then you erase that item (9) from the universal set. Everything in this answer set is not found in the set {9}

Put another way: we're finding the complement or opposite of P' n Q'

=======================================================

Part D

<h3>Answer: (P u Q)' = {9}</h3>

---------------------------------

Work Shown:

P = {2,4,6,8}

Q = {2,3,5,7}

P u Q = {2,3,4,5,6,7,8}

Everything in set P u Q is either found in P, Q, or both. Any duplicates are tossed out.

Take the opposite of this to get

(P u Q)' = {9}

It is not a coincidence we get the same result as part B. It turns out that the two equations are true for any two sets P and Q

(P u Q)' = P' n Q'
(P n Q)' = P' u Q'

For more information, check out De Morgan's Laws. Use of a Venn Diagram may help visualize what is going on, so you can organize the values.

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3 years ago

The base of a triangle is shrinking at a rate of 1cm/min and the height if the triangle is increasing at a rate of 5cm/min. Find

lapo4ka [179]

Answer: Rate of change of Area of the triangle = 2.75cm~/min

Step-by-step explanation:

If the present height of the triangle is 22cm and the height increases by 5cm/min. It means the height was 2cm four minutes earlier: 5cm for 1st min + 5cm for 2nd min + 5cm for 3rd min + 5cm for 4th min = 20cm after 4min.

Therefore height equals 22 - 20 = 2cm four minutes earlier.

Using the same four minutes for the base. Since the base shrinks at 1cm/min, for the base to shrink to 10cm it means it was 14cm four minutes earlier.

The Area of a triangle is given by

A = 1/2 × base × height

Therefore Area= 1/2× 10×22 = 110cm~

Since it took four minutes to get to this area, the rate of change of the area will be 110÷4 = 2.75cm~/min

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The answer is the option d, which is: d) $\frac{-77+21i}{130}$

The explanation for this problem is shown below:

1. Smplify the denominator and rewrite the numerator in this form:

$\frac{7i}{7-2i-4-9i} \\\frac{7i}{3-11i}$

2. Multiply the denominator and the numerator by the conjugated $(3+11i)$ and simplify the expression, as following:

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ZanzabumX [31]

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