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Flauer [41]
3 years ago
14

What point would you have when (-7, 3) is reflected over the line y=x?

Mathematics
2 answers:
chubhunter [2.5K]3 years ago
7 0
I believe it’s C.
The line y=x looks like a even, diagonal line going through (0, 0) on a graph. If you have (-7,3) , your in the first quadrant (in the upper left square of the graph). If you reflect it diagonally, the digits themselves would remain that same, but the value would flip (if it’s positive, it’ll become negative, if it’s negative, it’ll become positive). So if we do that with (-7, 3), it becomes (7, -3).
Solution: (7, -3)
Mekhanik [1.2K]3 years ago
4 0

Answer:

I believe its A.

Step-by-step explanation:

It usually the same numbers but backwards sometimes it just a positive or a negative different.

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7nadin3 [17]
I think its called a Two step equation
8 0
3 years ago
In the diagram provided, line I is parallel to line m. Select which of the following statements could be used to prove that
atroni [7]
<h3>Answer: There is only one answer and it is choice B</h3><h3>Angle 1 and angle 4 are alternate interior angles</h3>

========================================================

Explanation

  • A. This is false because it should be angle 4 + angle 5 = 180 without the angle 6. Adding on angle 6 results in some angle larger than 180. Note how angle 5 = (angle 3)+(angle 6).
  • B. This is true and useful to showing that the three angles of a triangle add to 180 degrees. This is because you'll use the fact that angles 4, 5 and 6 combine to 180 degrees.
  • C. While this is a true statement by the exterior angle theorem, it is not useful to the proof. It is better to state that angle 2 and angle 6 are congruent because they are alternate interior angles.
  • D. Like choice C, it is true but not useful. It's better to say that angle 1 is congruent to angle 4. See choice B above.

Note how it's not enough for a statement to be true. It also needs to be relevant or useful to the context at hand. A more simpler example of this could be stating that x+x = 2x.

8 0
3 years ago
What is 1\10 x 1.2 &gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;
My name is Ann [436]

Answer:

1/10 x 1.2 is 0.12

Step-by-step explanation:

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4 0
2 years ago
Find the derivative of following function.
Aleks04 [339]

Answer:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x}  + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}

General Formulas and Concepts:
<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
\displaystyle (cu)' = cu'

Derivative Property [Addition/Subtraction]:
\displaystyle (u + v)' = u' + v'

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
\displaystyle (uv)' = u'v + uv'

Derivative Rule [Quotient Rule]:
\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}

Derivative Rule [Chain Rule]:
\displaystyle [u(v)]' = u'(v)v'

Step-by-step explanation:

*Note:

Since the answering box is <em>way</em> too small for this problem, there will be limited explanation.

<u>Step 1: Define</u>

<em>Identify.</em>

\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}

<u>Step 2: Differentiate</u>

We can differentiate this function with the use of the given <em>derivative rules and properties</em>.

Applying Product Rule:

\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '

Differentiating the first portion using Quotient Rule:

\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}

Apply Derivative Rules and Properties, namely the Chain Rule:

\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}

Differentiating the second portion using Quotient Rule again:

\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Apply Derivative Rules and Properties, namely the Chain Rule again:
\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Substitute in derivatives:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Simplify:

\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}

We can rewrite the differential by factoring and common mathematical properties to obtain our final answer:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x}  + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}

∴ we have found our derivative.

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Learn more about derivatives: brainly.com/question/26836290

Learn more about calculus: brainly.com/question/23558817

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

Unit: Differentiation

8 0
2 years ago
Read 2 more answers
What is 21/49 sinple
NikAS [45]
21/49 in simplist from would be 3/7.
Your answer is <em>3/7</em>
6 0
3 years ago
Read 2 more answers
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