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2 years ago

If the vertices of a right triangle are located at (2,6), (2,3), and (6,3), what is the length of the hypotenuse?

Mathematics

1 answer:

Sergio [31]2 years ago

7 0

Answer:

5 units

Step-by-step explanation:

ez pz lemon squeezy. hippity hoppity your math is now my property. splish splash your opionion is trash

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PLEASE HELP ASAP I am kinda stuck on this because my teacher didn’t teach me this

wariber [46]

Answer:

the first option

a = -2

b = 3

Step-by-step explanation:

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2 years ago

Is 1/4 and 3/8 proportional

Sunny_sXe [5.5K]

Answer:

no it's not because is not

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2 years ago

Read 2 more answers

Which of the following are solutions to the equation. 6x^2-2x+36=5x^2+10x

Oduvanchick [21]

Answer:

x = 6

Step-by-step explanation:

Given

6x² - 2x + 36 = 5x² + 10x ( subtract 5x² + 10x from both sides )

x² - 12x + 36 = 0 ← in standard form

This is a perfect square of the form

(x - a)² = x² - 2ax + a²

36 = 6² ⇒ a = 6 and 2ax = (2 × 6)x = 12x, hence

(x - 6)² = 0

x - 6 = 0 ⇒ x = 6

6 0

2 years ago

Find the distance from P to l.

ki77a [65]

$Equation\ of\ a\ line\ from\ 2\ points (x_1;\ y_1);\ (x_2;\ y_2):\\\\y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\(-4;\ 2)\ and\ (3;-5)\\subtitute:\\\\y-2=\dfrac{-5-2}{3-(-4)}\cdot(x-(-4))\\\\y-2=\dfrac{-7}{7}\cdot(x+4)\\\\y-2=-(x+4)\\\\y-2=-x-4\ \ \ |add\ x\ and\ 4\ to\ both\ sides\\\\x+y+2=0\\-------------------\\$

$Point-line\ distance\\l:Ax+By+C=0;\ (x_0;\ y_0)\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\\\\x+y+2=0\to A=1;\ B=1;\ C=2\\(1;\ 2)\to x_0=1;\ y_0=2\\subtitute\\\\d=\dfrac{|1\cdot1+1\cdot2+2|}{\sqrt{1^2+1^2}}=\dfrac{|1+2+2|}{\sqrt2}=\dfrac{|5|}{\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\\\\=\boxed{\dfrac{5\sqrt2}{2}}$

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2 years ago

What is a-2÷5a+a+2÷5a

Andrej [43]

Answer:

=2a

Step-by-step explanation:

a-2/5a and a+2/5a cancel each other out and your left with a+a aka 2a

4 0

3 years ago