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3 years ago

12

Using the distance formula, find the distance between two points. (6, 0) (-5, 4)

1 answer:

Ipatiy [6.2K]3 years ago

3 0

Answer:

11.7046999

Step-by-step explanation:

using the distance formula (d=√(x₂ -x₁)² + (y₂-y₁)²), we can plug in the two points to find the distance between them:

d=√(6-(-5)² + (0-4)²

d=√(11)²+ (-4)²

d=√121+16

d=√137

d approximately equal to (≈) 11.7046999 or 11.7 or 11

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How do you solve 5/6 ×20

Alex73 [517]

Ok so remember that if you had
1/2 times 3/4 that equals (1 ties 3)/(2 times 4)
just mutply top number with top number and bottom number with bottom number so

remember that 20=20/1 so
5/6 times 20/1=
5 times 20=100
6 times 1=6

5/6 times 20/1=100/6 (can be simplified to 50/3 or 16 and 2/3)

5 0

3 years ago

One student says -5 is bigger than -4 and uses money as the analogy: “if i owe $5, i have a bigger debt than if i owe $4.” what

makkiz [27]

Answer:

-5 is smaller than -4 bc -4 is closer to zero meaning it has more value than 5

Step-by-step explanation:

8 0

3 years ago

Melanie has 10 packets of 20 biscuits. She puts 3/10 of them on a plate to serve. How many biscuits are being served?

jeka57 [31]

From the problem statement Melanie has 10 packets of 20 biscuits

60 biscuits

Solution

Let us find 3/10 of 10 packets

= 3 packets

In one packet there are 20 biscuits,

Hence in 3 packets there will be

20*3 = 60 biscuits

From the solution above Melanie served 60 biscuits in the plate

Learn more about fraction here:

brainly.com/question/78672

4 0

2 years ago

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

Is 23 in bigger than 2ft?

denpristay [2]

No

23 in equal 1 ft & 11 in

4 0

3 years ago