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Step2247 [10]
3 years ago
5

The function c = 100 + 0.30m represents the cost c (in dollars) of renting a car after driving m miles. What would the cost be t

o rent the car and drive 100 miles?
Mathematics
1 answer:
aleksley [76]3 years ago
7 0

Answer:

c = $130

Step-by-step explanation:

Given:

c = 100 + 0.30m

Where,

c = cost of renting a car (in dollars)

m = number of miles driven (in miles)

If m = 100 miles

c = 100 + 0.30m

= 100 + 0.30(100)

= 100 + 30

= 130

c = $130

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scZoUnD [109]

Answer:

1) Inequality: 110x>150+85x

2) Solution: x>6, then Nadia would need to sell more than 6 ads per week in order for Choice A to be the better choice,


<u>Solution:</u>

Be x the number of ads Nadia sells

A) Choice A: $110 for each ad she sells. She would earn:

Ea=110x

B) Choice B: A weekly salary of $150, plus $85 for each ad she sells. She would earn:

Eb=150+85x

1) Write an inequality to determine the number of ads Nadia would need to seel per week in order for Choice A to be the better choice:

Ea>Eb

110x>150+85x

2) Solution

Solving for x: Subtracting 85x both sides of the equation:

110x-85x>150+85x-85x

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Dividing both sides of the equation by 25:

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3 years ago
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Answer:

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Step-by-step explanation:

4 0
3 years ago
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In the figure CD is the perpendicular bisector of AB . If the length of AC is 2x and the length of BC is 3x - 5 . The value of x
Dennis_Churaev [7]

Answer

Find out the value of x .

To proof

SAS congurence property

In this property two sides and one angle of the two triangles are equal.

in the Δ ADC and ΔBDC

(1) CD = CD (common side of both the triangle)

(2) ∠CDA = ∠ CDB = 90 °

( ∠CDA +∠ CDB = 180 ° (Linear pair)

as given in the diagram

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∠ CDB = 180 ° - 90°

∠ CDB = 90°)

(3) AD = DB (as shown in the diagram)

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by using the SAS congurence property .

AC = BC

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As given

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2x = 3x - 5

3x -2x =5

x = 5

The value of x is 5 .

Hence proved


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3 years ago
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velikii [3]
The graphs have precisely the same shape, but that of g(x) is that of f(x) translated 4 units DOWN.
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Sammy’s dad drove their car 150 miles in three hours. At this rate, how far would he drive in nine hours?
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Answer:

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