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3 years ago

Fill in the blanks, solve x2+6x +y2- 4y = 23

Mathematics

1 answer:

Aleks [24]3 years ago

6 0

Answer:

Step-by-step explanation:

See completed figure attached.

Note that some blanks are missing colours from the question.

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A new battery is expected to have a storage capacity of 60 ampere-hour (A h). A sample is tested, resulting in a 2-side confiden

antoniya [11.8K]

Answer:

Option B) The battery storage capacity is significantly different than 60 Ah, at a confidence level of 95 %

Step-by-step explanation:

We are given the following in the question:

Mean storage capacity = 60 ampere-hour

95% confidence interval =

$63\leq \mu \leq 72$

Thus, the correct answer based on the confidence interval is

Option B) The battery storage capacity is significantly different than 60 Ah, at a confidence level of 95 %

Since it is a two sided test we test whether the mean storage capacity is 60 Ah or different than 60 Ah

Thus, the battery storage is different from 60 Ah because the mean storage capacity does not lie in the 95% confidence interval.

4 0

3 years ago

Verify the following identity and show steps (Cos2 θ)/(1+sin2 θ)= (cot θ-1)/(cot θ+1)

Paul [167]

Answer:

Verified below

Step-by-step explanation:

We want to show that (Cos2θ)/(1 + sin2θ) = (cot θ - 1)/(cot θ + 1)

In trigonometric identities;

Cot θ = cos θ/sin θ

Thus;

(cot θ - 1)/(cot θ + 1) gives;

((cos θ/sin θ) - 1)/((cos θ/sin θ) + 1)

Simplifying numerator and denominator gives;

((cos θ - sin θ)/sin θ)/((cos θ + sin θ)/sin θ)

This reduces to;

>> (cos θ - sin θ)/(cos θ + sin θ)

Multiply top and bottom by ((cos θ + sin θ) to get;

>> (cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ)

In trigonometric identities, we know that;

cos 2θ = (cos² θ - sin²θ)

cos²θ + sin²θ = 1

sin 2θ = 2sinθcosθ

Thus;

(cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ) gives us:

>> cos 2θ/(1 + sin 2θ)

This is equal to the left hand side.

Thus, it is verified.

5 0

3 years ago

Which of the following numbers has the commas in the correct positions

Genrish500 [490]

BBB because it has commas after the right units

6 0

3 years ago

In the accompanying diagram, lines a and b are parallel, and lines c and d are transversals. 5 a 6 3 4 7/8 be Which angle is con

OLga [1]

Answer:

Step-by-step explanation:

7 0

3 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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ we have evaluated the given limit.

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Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

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

Unit: Limits

3 0

2 years ago