Does the point ( 2, 1 ) satisfy the equation y = 2x?

1 answer:

6 0

Answer:

No.

Step-by-step explanation:

Y=2x goes up twice and to the right once, it passes through the origin, therefore it won't touch or go pass by point (2,1)

You might be interested in

jonny [76]

${x}^{2} - 2 {x}^{2} - 4 \\ - {x}^{2} - 4$

The Quadratic Function has the domain as the set of all real numbers.

For the range, start from minimum value to maximum value.

But because the parabola is downward as a < 0. Thus, there are no minimum value but the maximum value instead.

Therefore the range is y <= -4

7 0

3 years ago

ch4aika [34]

Answer:

The number that belongs in the green box is equal to 909.

General Formulas and Concepts:
Algebra I

Equality Properties

Trigonometry

[Right Triangles Only] Pythagorean Theorem:
$\displaystyle a^2 + b^2 = c^2$

Step-by-step explanation:

Step 1: Define

Identify given variables.

a = 30

b = 3

c = x

Step 2: Find x

Let's solve for the general equation that allows us to find the hypotenuse:

[Pythagorean Theorem] Square root both sides [Equality Property]:
$\displaystyle \begin{aligned}a^2 + b^2 = c^2 \rightarrow c = \sqrt{a^2 + b^2}\end{aligned}$

Now that we have the formula to solve for the hypotenuse, let's figure out what x is equal to:

[Equation] Substitute in variables:
$\displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2}\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2} \\& = \boxed{ \sqrt{909} } \\\end{aligned}$