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3 years ago

An equation of the line that passes through (3, 1) and (0, 10).

Mathematics

1 answer:

MrMuchimi3 years ago

3 0

Answer:

So the equation of the line that passes through (3, 1) and (0, 10) is y=-3x+10

Step-by-step explanation:

(3, 1) and (0, 10)

m=y²-y¹/x²-x¹

m=10-1/0-3

m=9/-3

m=-3

y=mx+b

y-y=m(x-x1)

y-1=-3(x-3)

y-1=-3x+9

+1 +1

y=-3x+10

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Prove that a triangle with the sides a - 1 cm to root under a

vivado [14]

Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm

Hence,

${AB}^2 = (a -1)^2$

Now We know that

$(a- b)^2 = a^2+b^2 - 2ab$

So;

${AB}^2= a^2 + 1^2 -2\times a \times1$

${AB}^2 = a^2 + 1 -2a$

Now;

${BC}^2 = (2\sqrt{a})^2= 4a$

Also;

${CA}^2 = (a + 1)^2$

Now We know that

$(a+ b)^2 = a^2+b^2 + 2ab$

${CA}^2= a^2 + 1^2 +2\times a \times1$

${CA}^2 = a^2 + 1 +2a$

${CA}^2 = AB^2 + BC^2$

[By Pythagoras theorem]

$a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a$

Hence, ${CA}^2 = AB^2 + BC^2$

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

4 0

3 years ago

SIMPLIFY THE EXPRESSION -3(-5 + s)

Arturiano [62]

Answer

15 + (-3s) or 15 - 3s

Step-by-step explanation:

-3 times -5 is 15

-3 times s is -3s

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