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N76 [4]
2 years ago
7

Sarah's math grade average is 93.3333(repeating)%. What is the repeating decimal part of her grade as a fraction?

Mathematics
2 answers:
UNO [17]2 years ago
8 0

Answer:

280/3

Step-by-step explanation:

multiply the number by 10x

subtract it by the original number on both sides

it should give u 9x to whatever answer u got multiply by 10 then divide by 90

Reil [10]2 years ago
7 0
It is 1/3....................
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PLS HELP ASAP PLSSSS!!!!!!!!!!!!!!!!!!!!!!!!!
Lunna [17]

Answer:

2800 in^3

Step-by-step explanation:

Calculate the volume as normal, but take 1/10 of the result. Since volume is length*width*height, we have 50*28*20 which is 28000. 1/10 of that is 2800. So the volume when scaled down by a factor of 1/10 is 2800 in^3

5 0
3 years ago
Whats 12 in exponent form<br>​
nexus9112 [7]

Answer:

2 2 3

Step-by-step explanation:

7 0
3 years ago
The numbers on the line plot
densk [106]
The picture isn't very clear
4 0
3 years ago
NEED THE ANSWER ASAP, PLEASE HELP!! What are the coordinates of the focus of the parabola?
qwelly [4]

Answer:

Vertex is (-8, 4)  not one of the answers offered!

Step-by-step explanation:

Factor -1/8 out of the first two terms. Leave a space inside parentheses in which to add a number.

y=-\frac{1}{8}(x^2 + 16x + \text{-----})-4

Complete the square by squaring half the coefficient of <em>x </em>and adding it in the space.

(\frac{16}{2})^2=64

Add 64 in the space you made, then compensate for that at the end of the expression by <u>subtracting</u>  -\frac{1}{8}(64)=-8

y=-\frac{1}{8}(x^2+16x+64)-4+8\\y=-\frac{1}{8}(x+8)^2+4

The x-coordinate of the vertex is the <u>opposite</u> of +8, the number inside the parentheses.  The y-coordinate of the vertex is the number at the end of the expression.

V(-8, 4)

8 0
3 years ago
Hello again! This is another Calculus question to be explained.
podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

Functions

  • Function Notation
  • Exponential Property [Rewrite]:                                                                   \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Property [Root Rewrite]:                                                           \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

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:

  1. f(x) = cxⁿ
  2. 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:

We are given the following and are trying to find the second derivative at <em>x</em> = 2:

\displaystyle f(2) = 2

\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}

When we differentiate this, we must follow the Chain Rule:                             \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]

Use the Basic Power Rule:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]

Simplifying it, we have:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]

We can rewrite the 2nd derivative using exponential rules:

\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}

To evaluate the 2nd derivative at <em>x</em> = 2, simply substitute in <em>x</em> = 2 and the value f(2) = 2 into it:

\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}

When we evaluate this using order of operations, we should obtain our answer:

\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219

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

Unit: Differentiation

5 0
2 years ago
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