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3 years ago

Please someone help me with this! It’s due tomorrow! First person to tell me the answer and why gets brainlest

Mathematics

1 answer:

vichka [17]3 years ago

8 0

1. 0.95
2. -0.125
3. 3.4
4. 0.7 (repeating 7)
5. 0.73 (repeating 3)
6. 2.6 (repeating 6)

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PLz help :( i will mark barinliest

RUDIKE [14]

Answer:

Step-by-step explanation:

It's D.

A theorem is true by some sort of mathematical logic.

It is not a definition. Not C

It is not in the end true because of inductive reasoning (a theorem may lend itself to inductive reasoning, but it must be proved.) Not A

It certainly not assumed to be true without proof. Not B

3 0

2 years ago

What is the same as 2/3

Paul [167]

Answer:

4/6 and 6/9 us simlar as 2/3

7 0

1 year ago

Read 2 more answers

What is the value of b in the equation (y^-9)^b=y^45?

nevsk [136]

Answer:

b = (log(y^45))/log(1/y^9) + (2 i π n)/log(1/y^9) for n element Z

Step-by-step explanation:

Solve for b:

(1/y^9)^b = y^45

Take the logarithm base 1/y^9 of both sides:

Answer: b = (log(y^45))/log(1/y^9) + (2 i π n)/log(1/y^9) for n element Z

5 0

3 years ago

A loan is paid off in 30 years with a total of $99,000. It had a 3% interest rate that compounded monthly. What was the principa

gayaneshka [121]

Answer:

The principle will be "40295.63". A further explanation is given below.

Step-by-step explanation:

The given values are:

Total amount,

A = 99,000

Rate of interest,

R = 3%

Time period,

T = 30 years

= 360 months

As we know,

⇒ $A=P(1+\frac{{\frac{R}{12} }}{100} )^T$

On substituting the values, we get

⇒ $99000=P(1+\frac{\frac{3}{12}}{100})^{360}$

⇒ $99000=P(1+\frac{\frac{1}{4} }{100}) ^{360}$

⇒ $P=\frac{99000}{(1.0025)^{360}}$

⇒ $P=40295.63$

4 0

2 years ago

Can you find the limits of this

Pavel [41]

Answer:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Constant]: $\displaystyle \lim_{x \to c} b = b$

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

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

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

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}$

Let's substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}$

Evaluating this, we arrive at an indeterminate form:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}$

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}$

Substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}$

Evaluating this, we get:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}$

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

2 years ago