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lbvjy [14]
2 years ago
10

Solve for x: √x + 3-4=7

Mathematics
1 answer:
nalin [4]2 years ago
7 0

Answer:

x = 64

Step-by-step explanation:

<em>Solve for x: √x + 3-4=7</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>√x + 3-4=</em><em>7</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>√</em><em>x</em><em> </em><em>+</em><em> </em><em>1</em><em> </em><em>=</em><em> </em><em>7</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>√</em><em>x</em><em> </em><em>=</em><em> </em><em>7</em><em> </em><em>+</em><em> </em><em>1</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>√</em><em>x</em><em> </em><em>=</em><em> </em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>(</em><em>√</em><em>x)</em><em>²</em><em> </em><em>=</em><em> </em><em>8</em><em>²</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x </em><em>=</em><em> </em><em>6</em><em>4</em>

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Baxter is thinking about buying a car. The table below shows the projected value of two different cars for three years.
MAVERICK [17]

Answer:

A) car 1: exponential

   car 2: linear

B) car 1: f(x)=18500(0.94)^{x-1}

    car 2: f(x)=-1000x+19500

C) Yes, there will be a significant difference of $1,100.40 which is approx 10% of the value of the cars after 10 years.

Step-by-step explanation:

<u>Part A</u>

Car 1: exponential as the projected value is not decreasing by the same amount each year

Car 2: linear as the projected value decreases by the same amount each year ($1000)

<u>Part B</u>

<u>Car 1</u>

General form of exponential function: y=ab^x

a = initial value = 18500

b = growth factor = \dfrac{17390}{18500}=0.94

\implies f(x)=18500(0.94)^{x-1}

NB We have to change x to "x-1" since we are told in the table that <u>after</u> 1 year the car's value is 18500.

<u>Car 2</u>

General form of linear function:  y=mx+b

m=\dfrac{\textsf{change in} \ y}{\textsf{change in} \ x}=\dfrac{17500-18500}{2-1}=-1000

y-y_1=m(x-x_1)

\implies y-17500=-1000(x-2)

\implies f(x)=-1000x+19500

<u>Part C</u>

Car 1 after 10 years: \implies f(10)=18500(0.94)^{10-1}=\$10,600.40

Car 2 after 10 years: \implies f(10)=-1000(10)+19500=\$9,500

Difference = 10600.40 - 9500 = 1100.40

Yes, there will be a significant difference of $1,100.40 which is approx 10% of the value of the cars after 10 years.

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