Given the two rectangles below. Find the area of the shaded region.

Mathematics

1 answer:

Julli [10]3 years ago

3 0

Answer:

10 *2= 20

2 * 9 =10

2*2=4

4+18+20 =42

42 units squared

Step-by-step explanation:

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Tell whether the triangle with the given sides lengths is a right triangle

Andre45 [30]

If this is a right triangle, then $a^{2}+ b^{2}= c^{2}$ where c is the hypotenuse which is ALWAYS THE LONGEST SIDE. Our longest side is 5, so we will set up Pythagorean's Theorem to see if the addition of the squares of a and b = the square of c: $(1.4)^{2} +(4.8) ^{2} =(5)^{2}$ . Squaring all of that and adding on the left gives us 25=25. Because this is true, we have proven the converse of the Pythagorean Theorem here and this is a right triangle.

3 0

4 years ago

Solve using the quadratic formula 2x^2+x+67=0

telo118 [61]

Answer:

$\displaystyle x=\frac{-1 \pm i\sqrt{535}}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Multiple Roots
Standard Form: ax² + bx + c = 0
Quadratic Formula: $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Algebra II

Imaginary Root i = √-1

Step-by-step explanation:

Step 1: Define

Identify

2x² + x + 67 = 0

a = 2

b = 1

c = 67

Step 2: Solve for x

Substitute in variables [Quadratic Formula]: $\displaystyle x=\frac{-1 \pm \sqrt{1^2-4(2)(67)}}{2(1)}$
Multiply: $\displaystyle x=\frac{-1 \pm \sqrt{1^2-4(2)(67)}}{2}$
[√Radical] Evaluate exponents: $\displaystyle x=\frac{-1 \pm \sqrt{1-4(2)(67)}}{2}$
[√Radical] Multiply: $\displaystyle x=\frac{-1 \pm \sqrt{1-536}}{2}$
[√Radical] Subtract: $\displaystyle x=\frac{-1 \pm \sqrt{-535}}{2}$
[√Radical] Simplify: $\displaystyle x=\frac{-1 \pm i\sqrt{535}}{2}$