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musickatia [10]

3 years ago

5

A soccer team wins 65% of its matches, and 15% of its matches end in a draw. If the team is scheduled to play 20 matches, about

how many matches is it expected to lose?

1 answer:

Finger [1]3 years ago

3 0

The team plays 20 matches.

65% of the matches, the teams wins
= 20 * 0.65
= 13 matches

15% of the matches, the games ends in draw.
= 20 * 0.15
= 3 matches

The team is expected to lose in
= 20 - 13 - 3
= 4 matches

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Whts X+y x 9 +3 +4 ?

babunello [35]

Expression simplified: $x+9y+7$

Step-by-step explanation:

The expression that we have in this problem is:

$x+y\cdot 9 +3 +4$

The first type of operation that we have to solve in such expression are multiplication/division.

In this expression, there is only one multiplication, that is:

$y\cdot 9 = 9y$

So we can rewrite the expression as

$x+9y+3+4$

Now we can evaluate the sum of the similar terms: in this case, similar terms are only 3 and 4, and their sum is

$3+4=7$

Therefore, the expression can be rewritten as

$x+9y+7$

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3 years ago

kherson [118]

Answer:

Point N(4, 1)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

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>

<u /> $\displaystyle y = \sqrt{x - 3}$ <u />

<u /> $\displaystyle y' = \frac{1}{2}$ <u />

<u />

<u>Step 2: Differentiate</u>

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = (x - 3)^{\frac{1}{2}}$
Chain Rule: $\displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)$
Simplify: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1$
Multiply: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x - 3}}$

<u>Step 3: Solve</u>

<em>Find coordinates</em>

<em />

<em>x-coordinate</em>

Substitute in <em>y'</em> [Derivative]: $\displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle 1 = \frac{1}{\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply √(x - 3) on both sides: $\displaystyle \sqrt{x - 3} = 1$
[Equality Property] Square both sides: $\displaystyle x - 3 = 1$
[Addition Property of Equality] Add 3 on both sides: $\displaystyle x = 4$

<em>y-coordinate</em>

Substitute in <em>x</em> [Function]: $\displaystyle y = \sqrt{4 - 3}$
[√Radical] Subtract: $\displaystyle y = \sqrt{1}$
[√Radical] Evaluate: $\displaystyle y = 1$

∴ Coordinates of Point N is (4, 1).

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

Unit: Derivatives

Book: College Calculus 10e

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2 years ago

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Papessa [141]

Answer:

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Step-by-step explanation:

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garik1379 [7]

1/2 / 4

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BlackZzzverrR [31]

Answer:

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Step-by-step explanation:

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