All categories

3 years ago

a line segment has endpoints A (-8,6) and B (4,6). What is the distance in units between point A and B?

Mathematics

1 answer:

lyudmila [28]3 years ago

5 0

Answer:

The distance between point A and point B is 12 units.

Step-by-step explanation:

d = √(x2 - x1)² + (y2 - y1)²

d = √(-8 - 4)² + (6 - 6)²

d = √144 + 0

d = 12

You might be interested in

Helppppppppp pleaseeeeee

pishuonlain [190]

Answer:

Yo solo pongo esta respuesta que no dice nada porque busco puntiooooooos

8 0

2 years ago

7.25 x 2 1/2= WHAT PS HELPPP!!!!

Levart [38]

Answer:

Its either 18.125 or 145 / 8

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Given the following functions f(x) and g(x), solve f[g(6)]. f(x) = 6x + 12 g(x) = x − 8

iogann1982 [59]

Answer:

Answer:

Option 2nd is correct.

=0.

Step-by-step explanation:

Given the function:

Solve:

First calculate:

f[g(x)]

Substitute the function g(x)

Replace x with x-8 in the function f(x) we get;

The distributive property says that:

Using distributive property:

⇒

Put x = 6 we get;

Therefore, the value of is 0.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Can someone please please help me?? :(

salantis [7]

Answer:

-60 3/4m

Step-by-step explanation:

The diver first dives down 50 1/4m, which is represented by the -50 1/4m. He then descends another 10 1/2m down. So -50 1/4m - 10 1/2m.

-50-10=-60

-(1/4)-(1/2)= (1/4)+(1/2)

(1/4+1/2)=(2/8+4/8) you can only add with the same denominators.

=6/8 (simplifies down to 3/4)

8 0

3 years ago

A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters

In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

$\Rightarrow y=\sqrt{x^2+100}$

Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

so, $\frac{dx}{dt}=-13$

$\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)$

Now

$\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)$

$=\frac{24\times (-13)}{\sqrt{676}}$

$=\frac{24\times (-13)}{26}$

= -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0

3 years ago