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3 years ago

Find three different ways to write the equation that represents the line in the plane that passes through points

Mathematics

1 answer:

Zinaida [17]3 years ago

5 0

Answer:

y - 2 = -3(x - 1).

y = -3x + 5.

3x + y = 5.

Step-by-step explanation:

The slope of the line is (-1-2) / (2 - 1)

= -3.

Using the point slope form of the equation of a straight line:

y - 2 = -3(x - 1) This is one way to write it.

Simplifying the above:

y = -3x + 3 + 2

y = -3x + 5. This is another way called the slope-intercept form.

Now we convert this to the standard form:

y = -3x + 5

y + 3x = 5

3x + y = 5.

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baherus [9]

Answer: see proof below

Step-by-step explanation:

Given: cos 330 = $\frac{\sqrt3}{2}$

Use the Double-Angle Identity: cos 2A = 2 cos² A - 1

$\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}$

Proof LHS → RHS:

LHS cos 165

Double-Angle: cos (2 · 165) = 2 cos² 165 - 1

⇒ cos 330 = 2 cos² 165 - 1

⇒ 2 cos² 165 = cos 330 + 1

Given: $2 \cos^2 165 = \dfrac{\sqrt3}{2} + 1$

$\rightarrow 2 \cos^2 165 = \dfrac{\sqrt3}{2} + \dfrac{2}{2}$

Divide by 2: $\cos^2 165 = \dfrac{\sqrt3+2}{4}$

$\rightarrow \cos^2 165 = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}$

$\rightarrow \cos^2 165 = \dfrac{2\sqrt3+4}{8}$

Square root: $\sqrt{\cos^2 165} = \sqrt{\dfrac{4+2\sqrt3}{8}}$

Scratchwork: $\cos^2 165 = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2$

$\rightarrow \cos 165 = \pm \dfrac{\sqrt3+1}{2\sqrt2}$

Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE

$\rightarrow \cos 165 = - \dfrac{\sqrt3+1}{2\sqrt2}$

LHS = RHS $\checkmark$

4 0

4 years ago

Round to the nearest tenth if necessary.

Andrei [34K]

Answer:

B) 25

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)

Algebra II

Distance Formula: $\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

Step 1: Define

Identify

Point (-1, -8)

Point (-4, -4)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Substitute in points [Distance Formula]: $\displaystyle d = \sqrt{(-4--1)^2+(-4--8)^2}$
[√Radical] (Parenthesis) Subtract: $\displaystyle d = \sqrt{(-3)^2+(4)^2}$
[√Radical] Evaluate exponents: $\displaystyle d = \sqrt{9+16}$
[√Radical] Add: $\displaystyle d = \sqrt{25}$
[√Radical] Evaluate: $\displaystyle d = 5$

4 0

3 years ago