All categories

2 years ago

Find the distance between points (3,6) and (9,2)

Mathematics

1 answer:

andrew-mc [135]2 years ago

5 0

Answer:

A=(3,6)

B =(9,2)

The distance between AB=√(9-3)^2+(2-6)^2

=3√13

You might be interested in

Circle P is centered at the origin. Which of the following statements can be used to

a_sh-v [17]

the correct option is:

"The distance from the origin to point A is equal to the radius of circle P. Therefore, point A lies on circle P."

<h3>Which of the following statements can be used to construct an algebraic proof the point A lies on circle P?</h3>

A circle is defined as the set of equidistant points to a given point which is the center of the circle.

That distance to the center is called the radius.

Here, the center is at the point (0, 0), also called the origin, so if the distance between the point A and the origin, then point A lies on top of the circle P.

From that, we conclude that the correct option is:

"The distance from the origin to point A is equal to the radius of circle P. Therefore, point A lies on circle P."

That is the only proof that we can (and should) use to prove that a point lies on a circle.

If you want to learn more about circles:

brainly.com/question/1559324

#SPJ1

3 0

1 year ago

Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

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

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

6 0

2 years ago

Tamara paid $3.00 for cartons of milk. Teo paid $3.75 for a different number of cartons at the same store. How much is one carto

Rashid [163]

1- Price of each carton:
Since they both bought different number of cartons of milk from the same store, we can simply find the price of one carton from the difference between what they both paid.

Since Tamara paid 3$ and Toe paid 3.75$, we can simply say that one carton of milk costs 0.75$ and that Toe bought one carton more than Tamra.

2- How many cartons each bought:
Tamra bought : 3 / 0.75 = 4 cartons
Toe bought : 3.75 / 0.75 = 5 cartons

3 0

3 years ago

How do you write 1.5 × 104 in standard form?

rewona [7]

I'm assuming you meant 1.5 × 10⁴. Remember, when we write in standard form, you need to use ^ to denote it's 'to the power of'. So it would be 1.5 × 10^4.

I'm also assuming you meant in decimal form, because that's already in standard form.

However, the answer is 15,000. This is because you lose one zero when multiplying a decimal with one number after it. For each number after the decimal, you take off a zero. Therefore, the end number is 15,000.

5 0

2 years ago

Examine the system of equations.

Pachacha [2.7K]

Step-by-step explanation:

The lines have different slopes

.......,............

6 0

2 years ago