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vodomira [7]
3 years ago
11

What is the distance between the points (5,1) and (-3,-5)?​

Mathematics
1 answer:
Charra [1.4K]3 years ago
6 0

Answer

\boxed{10 \: \:  units}

Step by step explanation

Let the points be A and B

A ( 5 , 1 ) ⇒ ( x₁ , y₁ )

B ( -3 , -5 )⇒ ( x₂ , y₂ )

Now, let's find the distance between theses two points:

Distance = \mathsf{ \sqrt{ {(x2 - x1)}^{2} +  {(y2 - y1)}^{2}  } }

Plug the values

\mathsf{ \sqrt{ {( - 3 - 5)}^{2} +  {( - 5 - 1)}^{2}  } }

Calculate

\mathsf{ \sqrt{ { ( - 8)}^{2}  +  {( - 6)}^{2} } }

Evaluate the power

\mathsf{ \sqrt{64 + 36} }

Add the numbers

\mathsf{ \sqrt{100} }

Write the number in exponential form with a base of 10

\mathsf  {\sqrt{ {10}^{2} } }

Reduce the index of the radical and exponent with 2

\mathsf{10 \: units}

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

The distance formula is used to determine the distance ( d ) between two points. If the co-ordinates of the two points are ( x₁ , y₁) and ( x₂ , y₂ ) , the distance equals the square root of x₂ - x₁ squared + y₂ - y₁ squared.

Hope I helped!

Best regards!

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amid [387]
No, the correct identity is \tan^2(x)+1=\sec^2(x). It is obtained by dividing the both sides of the Pythagorean Trigonometric Identity by cos²(x):

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6 0
4 years ago
Kendra is considering two different long-distance phone plans. Phone plan A charges a $100 sign-up fee and 3 cents per minute. P
Fed [463]

Answer:

Part 1) The number of minutes in a month must be greater than 50 in order for the plan A to be preferable

Part 2) The number of minutes in a month must be equal to 50 minutes

Step-by-step explanation:

<u><em>The question is</em></u>

Part 1) How many minutes would Kendra have to use in a month in order for the plan A to be preferable? Round your answer to the nearest minute

Part 2) Enter the number of minutes where Kendra will pay the same amount for each long distance phone plan

Part 1)

Let

x ---> the number of minutes

we have

<em>Cost Plan A</em>

3x+100

<em>Cost Plan B</em>

5x

we know that

In order for plan A to be cheaper than plan B, the following inequality must hold true.

cost of plan A < cost of plan B

substitute

3x+100 < 5x

solve for x

subtract 3x both sides

100< 5x-3x\\100

divide by 2 both sides

50 < x

Rewrite

x> 50\ min

therefore

The number of minutes in a month must be greater than 50 in order for the plan A to be preferable

Part 2)

Let

x ---> the number of minutes

we have

<em>Cost Plan A</em>

3x+100

<em>Cost Plan B</em>

5x

we know that

In order for plan A cost the same than plan B, the following equation must hold true.

cost of plan A = cost of plan B

substitute

3x+100= 5x

solve for x

5x-3x=100\\2x=100\\x=50\ min

therefore

The number of minutes in a month must be equal to 50 minutes

5 0
3 years ago
Write and solve an equation to find the number n. Six times the sum of a number and 15 is -42.
nata0808 [166]

Answer:

8

Step-by-step explanation:

6*(n + 15) = -42

Divide both sides by 6

6*(n - 15)/6 = -42/6

n - 15 = -7

n = -7 + 15

n = 8

3 0
3 years ago
PLEASE HELP ME THANK YOU IF YOU DO ❤️
docker41 [41]

Step-by-step explanation:

Here,

we have

the speed = 105 miles in 3 hours

Therefore, 105÷3

= 35

Hence, It's average speed is 35 miles.

if it helps don't forget to like and mark me

4 0
3 years ago
Read 2 more answers
How do u solve this?
malfutka [58]

you can work it out or put in random numbers until the two sides are equal

3 0
4 years ago
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