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3 years ago

PLEASE HELP PLEAsee before It’s due

Mathematics

1 answer:

xz_007 [3.2K]3 years ago

6 0

The measure of BOC is 47

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15 - (-3) - 4 -16 22 -8 14

uranmaximum [27]

In this equation, you have to treat the number in the bracket first on the basis of BODMAS
15 - [-3]- 4
Note that when two minuses come together the product is a plus sign.
15 +3 - 4
You have to add before you subract
18 - 4 =14
Therefore, 15- [-3] - 4 = 14.

8 0

3 years ago

PUT THESE NUMBERS IN ORDER GREATEST TO LEAST ✔✔ I WILL ADD BRAINLIST

Fiesta28 [93]

Answer:

2 12/16, 2 18/25, and lastly 2.

Step-by-step explanation:

Since 2 is least, we can put that at the end. That leaves us with 2 18/25 and 2 12/16. Since 12/16 is greater, we will put that as the greatest. 2 18/25 is in the middle. Hope this helps!

6 0

3 years ago

You work at a bike shop. A customer purchases 6 new bike tires at $6.99 each. If the customer gives you $40, how much is still o

Mariulka [41]

Answer:

194 cents is the correct answer

Step-by-step explanation:

6.99 dollars is 41.94

7 0

3 years ago

Read 2 more answers

Please, Help. the answer isn't 30°

vagabundo [1.1K]

The missing angle of the right triangle ABC has a measure of 30°. (Correct answer: A)

<h3>How to find a missing angle by triangle properties</h3>

Triangles are geometrical figures formed by three sides and whose sum of internal angles equals 180°. There are two kind of triangles existing in this question: (i) Right triangles, (ii) Isosceles triangles.

Right triangles are triangles which one of its angles equals 90° and isosceles triangles are triangles which two of its sides have equal measures.

According to the statement, we know that triangle BQR is an isosceles triangle, whereas triangles ABC, ANB and NBC are right triangles. Based on the figure attached below, we have the following system of linear equations based on right triangles ABC and NBC:

2 · x + 90 + θ = 180 (1)

(90 - x) + 90 + θ = 180 (2)

By equalizing (1) and (2) we solve the system for x:

2 · x = 90 - x

3 · x = 90

x = 30

And by (1) we solve the system for θ:

θ = 180 - 2 · x - 90

θ = 30

The missing angle of the right triangle ABC has a measure of 30°. (Correct answer: A) $\blacksquare$

To learn more on right triangles, we kindly invite to check this verified question: brainly.com/question/6322314

8 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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ 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

1 year ago