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ella [17]
2 years ago
6

Is 9.3 X 100^9 in scientific notation

Mathematics
1 answer:
den301095 [7]2 years ago
8 0

Answer: Yes

Step-by-step explanation: A scientific notation is when a huge number is written in a decimal form, for example 600,000 can be written as

6 X 10^5

Thanks and have a great day!

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Plz help quick. This question is on a quiz in Edg 2020 whoever answered first will get brainliest and I will be giving 50 points
lys-0071 [83]

Answer:

A B and D

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
You are purchasing office supplies. Because of the volume of your orders your company receives 15% off all purchases. You buy a
vova2212 [387]
 First add 24.95 and 35.99. You get 60.94 (Remember This!)

Now you take 49.95 divided by 2. you should get: 24.975.

add 24.975 with 60.94. You get 85.915

take 85.915 and multiply it by 0.15. You get 12.88725. 

Subtract 85.915 by 12.88725. You should get 73.02775.

Now multiply that by 0.045. you should get 3.28624875. 

Add that onto 73.02775, and you get 76.31399875.

Round that to the nearest hundredth, and your done! (The answer is: 76.31!) 

The answer will be the B.
6 0
3 years ago
Find the equation of the line parallel to y=2x+6 passing through (4,5)
ozzi

Answer:

y=2x-3

when two lines are parallel, they have the same slope.

in the formula y=mx+b, m=slope of the line

we know 2 will be the slope for the line passing through (4,5)

next plug in (4,5) into the formula with 2 as the slope

4=x 5=y

next 5=2(4)+b

and you end up with

y=2x-3 when you simplify

3 0
3 years ago
The population of Hillsborough County in Florida in 2010 was 1,229,226 and the land
Kruka [31]

Answer:

About 1205.12352941 people per square mile in 2010

About 1329.36306853 people per square mile in 2015

Explanation:

take population and divide by land area

for 2015 pop do total pop divided by 9.7 plus total pop

6 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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