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3 years ago

What is 100 over 90 times 81 - 9

Mathematics

1 answer:

Blizzard [7]3 years ago

3 0

Answer:

Step-by-step explanation:

set it up like 100/90*81-9

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CAN SOMEONE PLSSSSS ANSWER BOTH OF THESE QUESTIONS I WILL MARK BRAINLIEST

Margaret [11]

Answer: B and C

Step-by-step explanation:

404 + 50 + 50 = 504

you are buying 144 magazines total

504/144 = cost of one magazine

cost of one = 3.50

you are paying $60 a year for 12 years

60 x 12 = 720

720/144 = 5

5 - 3.50 = 1.50

120x + 180x = 1080

300x = 1080

x = 1080/300

x = 3.6

3.6 x 60 = 216

3 0

2 years ago

A student earned a grade of 80% on a math test that had 20 problems. How many problems on this test did the student answer corre

boyakko [2]

Answer:

16 problems.

Step-by-step explanation:

100% divided by 20 questions.

100/20 is equal to 5, therefore each question is worth 5 points.

Grade of 80% divided by points per question.

80/5 is 16 so the student answered 16 questions correctly.

6 0

2 years ago

Read 2 more answers

Please someone help me to prove this.

morpeh [17]

Answer: see proof below

Step-by-step explanation:

Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ

Use the Sum/Difference Identities:

sin(α + β) = sinα · cosβ + cosα · sinβ

cos(α - β) = cosα · cosβ + sinα · sinβ

Use the Unit circle to evaluate: sin45 = cos45 = √2/2

Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ

Use the Pythagorean Identity: cos²Ф + sin²Ф = 1

Proof LHS → RHS

LHS: 2sin(45 + 2A) · cos(45 - 2A)

Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)

Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]

Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]

Distribute: cos²2A + 2cos2A·sin2A + sin²2A

Pythagorean Identity: 1 + 2cos2A·sin2A

Double Angle: 1 + sin4A

LHS = RHS: 1 + sin4A = 1 + sin4A $\checkmark$

6 0

2 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$