Find the distance between the points (-4, -5 and (3, -1.

2 answers:

7 0

≡ We know that:
⇔ $(x_{1},y_{1})$ → $(-4,-5)$
⇔ $(x_{2},y_{2})$ → $(3,-1)$

≡ Solution:
⇒ $\sqrt{[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]}$
⇒ $\sqrt{[(3-(-4))^{2}+(-1-(-5))^{2}]}$
⇒ $\sqrt{[7^{2}+4^{2}]}$
⇒ $\sqrt{[49+16]}$
⇒ $\boxed{\sqrt{65}}$

tatyana61 [14]2 years ago

7 0

Answer: $\sqrt{65} \\$

i got it right in my quiz

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The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func

katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle ( $\frac{1}{3},\frac{1}4$ ).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A( $\frac{1}{3},\frac{1}4$ )

Let O be the origin i.e. (0,0)

Point B will be ( $\frac{1}{3},0$ )

Now, let us consider the right angled triangle $\triangle OBA$ :

Sides:

$Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}$

Using Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}$