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Elden [556K]
3 years ago
7

The ratio of 7th grade students to 8th grade students in a soccer league is 17:23. If there are 200 students in all, how many ar

e in 7th grade?
Mathematics
2 answers:
Setler [38]3 years ago
6 0
17:23....added = 40

17/40(200) = 3400/40 = 85 (7th graders)
23/40(200) = 4600/40 = 115 (8th graders)
pashok25 [27]3 years ago
3 0
The answer to your question is 85 7th graders.How is it 85?

A ratio of 17 : 23   means there are 40 equal parts

 

So   (17/40) * 200  =   85   7th graders

 

And  (23/40) * 200  =  115    8th graders

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PLEASE HELP ME GUYS OR I WONT PASS <br>this calculus!!!!​
KonstantinChe [14]

Answer:

b.  \displaystyle \frac{1}{2}

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> </u>

<u>Algebra I</u>

  • Functions
  • Function Notation
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Rule [Root Rewrite]:                                                                     \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}<u> </u>

<u>Calculus</u>

Derivatives

Derivative Notation

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:

<u>Step 1: Define</u>

<em>Identify</em>

<em />\displaystyle H(x) = \sqrt[3]{F(x)}<em />

<em />

<u>Step 2: Differentiate</u>

  1. Rewrite function [Exponential Rule - Root Rewrite]:                                      \displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}
  2. Chain Rule:                                                                                                        \displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]
  3. Basic Power Rule:                                                                                             \displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)
  4. Simplify:                                                                                                             \displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}
  5. Rewrite [Exponential Rule - Rewrite]:                                                              \displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}

<u>Step 3: Evaluate</u>

  1. Substitute in <em>x</em> [Derivative]:                                                                              \displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}
  2. Substitute in function values:                                                                          \displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}
  3. Exponents:                                                                                                        \displaystyle H'(5) = \frac{6}{3(4)}
  4. Multiply:                                                                                                             \displaystyle H'(5) = \frac{6}{12}
  5. Simplify:                                                                                                             \displaystyle H'(5) = \frac{1}{2}

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

Unit: Derivatives

Book: College Calculus 10e

5 0
3 years ago
X+2(x+3(x+4(x+5)))=153
Allushta [10]

Answer:

x=1

Step-by-step explanation:

Here first open the inner parentheses and use the distributive property,

4(x+5)=4x+20\\\\(x+4(x+5))=(x+4x+20)=(5x+20)\\\\3(x+4(x+5))=3(5x+20)=15x+60\\\\(x+3(x+4(x+5)))=(x+15x+60)=16x+60\\\\2(x+3(x+4(x+5)))=2(16x+60)=32x+120\\\\x+2(x+3(x+4(x+5)))=x+32x+120=33x+120\\\\\\Hence\ x+2(x+3(x+4(x+5)))=x+32x+120=33x+120\\\\\Rightarrow 33x+120=153\\\\33x=153-120\\\\33x=33\\\\x=1

3 0
3 years ago
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NeX [460]

Answer:

The 2 represents that each toppings costs $2.

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3 years ago
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