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masya89 [10]
3 years ago
6

If 5 gallons cost $ 6.60 how much do 29 gallons of gasoline cost​

Mathematics
1 answer:
ikadub [295]3 years ago
5 0

Answer:

Should be 38.28

Step-by-step explanation:

Dividing 6.60 by 5 is $1.32, aka the cost of one gallon. multiply by 29, answer is that it costs $38.28 to get 29 gallons of gasoline.

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Because of their small size, the sea lion pups are more vulnerable to changes in their environment and are less able to adapt to
Wittaler [7]
? Whats the question? This is a statement??
7 0
3 years ago
The following parts refer to the letters: LAMEFIREALARM. Recall that "word" means distinguishable letter arrangements. (a) How m
larisa86 [58]

Answer:

Step-by-step explanation:

The word "LAMEFIREALARM" word consists of

2L , 2M, 2E , 2R , 3A , F and I

no of words with the above letters such that M's are not together

=Total No of words- M's are together

Total no of words=\frac{13!}{4\times 2!\times 3!}=64,864,800

When M's are together . considering 2 M's as one

therefore there are \frac{12!}{3\times 2!\times 3!}=9,979,200

No of ways=\frac{13!}{4\times 2!\times 3!}-\frac{12!}{3\times 2!\times 3!}=54,885,600

(b)No of ways such that M's are separated by at least 2 letters

at least  2 letter means 2 letter, 3 letter ......11 letters

So we have to subtract no of ways where there are 1 letter in between M's from total no of ways where 2 M's are not next to each other

No of ways in which there is 1 letter between 2 M's

This can be done by considering 11 cases

In first case Place first M in Starting Position and 2 M on third place

Second case place First M in 2 nd Position and second M on 4 th place

Similarly For 11th case

Place first M in 11th place and second M on 13th place

Total ways

=\frac{11\times 11!}{3!\times 2!\times 2!\times 2!}

So , the total number of arrangements of the letters of the word LAMEFIREALARM where the two M's are separated by atleast two letters is

=\frac{13!}{4\times 2!\times 3!}-\frac{12!}{3\times 2!\times 3!}-\frac{11\times 11!}{3!\times 2!\times 2!\times 2!}

=45738000

7 0
3 years ago
GIVEN: LJ || WK || AP, PL || AG.
galina1969 [7]

Answer:

∠1 ≅ ∠2 ⇒ proved down

Step-by-step explanation:

#12

In the given figure

∵ LJ // WK

∵ LP is a transversal

∵ ∠1 and ∠KWP are corresponding angles

∵ The corresponding angles are equal in measures

∴ m∠1 = m∠KWP

∴ ∠1 ≅ ∠KWP ⇒ (1)

∵ WK // AP

∵ WP is a transversal

∵ ∠KWP and ∠WPA are interior alternate angles

∵ The interior alternate angles are equal in measures

∴ m∠KWP = m∠WPA

∴ ∠KWP ≅ ∠WPA ⇒ (2)

→ From (1) and (2)

∵ ∠1 and ∠WPA are congruent to ∠KWP

∴ ∠1 and ∠WPA are congruent

∴ ∠1 ≅ ∠WPA ⇒ (3)

∵ WP // AG

∵ AP is a transversal

∵ ∠WPA and ∠2 are interior alternate angles

∵ The interior alternate angles are equal in measures

∴ m∠WPA = m∠2

∴ ∠WPA ≅ ∠2 ⇒ (4)

→ From (3) and (4)

∵ ∠1 and ∠2 are congruent to ∠WPA

∴ ∠1 and ∠2 are congruent

∴ ∠1 ≅ ∠2 ⇒ proved

8 0
3 years ago
The curve
kherson [118]

Answer:

Point N(4, 1)

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  

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>

  1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                                   \displaystyle y = (x - 3)^{\frac{1}{2}}
  2. Chain Rule:                                                                                                        \displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]
  3. Basic Power Rule:                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)
  4. Simplify:                                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1
  5. Multiply:                                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}
  6. [Derivative] Rewrite [Exponential Rule - Rewrite]:                                          \displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}
  7. [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>

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

<em>y-coordinate</em>

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

4 0
3 years ago
A game stop membership cost $20 and includes one game A month for five dollars. Nonmembers can get one more game a month for sev
miss Akunina [59]

Answer:

C(x) = 5x + 20 (for members)

C(x) = 3.5x (for non-members)

Step-by-step explanation:

Cost of membership = $20

Price of a game per month = $5

So, the linear equation to compute the total cost for a member can be computed by:

C(x) = 5x + 20

where x is the number of games per month

On the other hand, non-members can get one more game per month for $7 which means they get 2 games for $7. The price for a single game is $7/2 = $3.5 a month.

The linear equation to compute the total cost for a non-member is:

C(x) = 3.5x

where x is the number of games per month.

The following system of equations can be used to decide whether to become a member or not, by substituting the number of games in place of x and finding out the total cost.

C(x) = 5x + 20 (for members)

C(x) = 3.5x (for non-members)

3 0
3 years ago
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