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dsp73
3 years ago
11

Re-write the expression in rational form.

Mathematics
1 answer:
Nuetrik [128]3 years ago
5 0

Given:

x^{\frac{5}{4}}=100

x=\sqrt[a]{100}^b

To find:

The values of a and b.

Solution:

We have,

x^{\frac{5}{4}}=100

It can be written as

x^{\frac{5}{4}\times \frac{4}{5}}=100^{\frac{4}{5}}

x^{1}=\left[100^{\frac{1}{5}}\right]^4       [\because a^{mn}=(a^m)^n]

x=\left[\sqrt[5]{100}\right]^4       [\because a^{\frac{1}{n}}=\sqrt[n]{a}]

x=\sqrt[5]{100}^4

On comparing this with x=\sqrt[a]{100}^b, we get

a=5

b=4

Therefore, the value of a is 5 and value of b is 4.

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The equation 2x2 − 12x + 1 = 0 is being rewritten in vertex form. Fill in the missing step. Given 2x2 − 12x + 1 = 0 Step 1 2(x2
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Answer:

Part 1) 2(x-3)^{2}-17=0  (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

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Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

Part 8) (x + 3)^2 − 4

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Step-by-step explanation:

Part 1) we have

2x^{2} -12x+1=0

Convert to vertex form

step 1  

Factor the leading coefficient and complete the square

2(x^{2} -6x)+1=0

2(x^{2} -6x+9)+1-18=0

step 2

2(x^{2} -6x+9)+1-18=0

2(x^{2} -6x+9)-17=0

step 3

Rewrite as perfect squares

2(x-3)^{2}-17=0

Part 3) we have

f(x)=-x^{2}+16x-60

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

f(x)+60=-x^{2}+16x

Factor the leading coefficient

f(x)+60=-(x^{2}-16x)

Complete the squares

f(x)+60-64=-(x^{2}-16x+64)

f(x)-4=-(x^{2}-16x+64)

Rewrite as perfect squares

f(x)-4=-(x-8)^{2}

f(x)=-(x-8)^{2}+4

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

-2(x-2)^{2}+5=0

-2(x-2)^{2}=-5

(x-2)^{2}=2.5

square root both sides

(x-2)=(+/-)1.58

x=2(+/-)1.58

x=2(+)1.58=3.58

x=2(-)1.58=0.42

Part 5) we have

f(x)=-x^{2}+50x-264

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

so

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

-x^{2}+50x-264=0

-x^{2}+50x=264

Factor -1 of the leading coefficient

-(x^{2}-50x)=264

Complete the squares

-(x^{2}-50x+625)=264-625

-(x^{2}-50x+625)=-361

(x^{2}-50x+625)=361

Rewrite as perfect squares

(x-25)^{2}=361

square root both sides

(x-25)=(+/-)19

x=25(+/-)19

x=25(+)19=44

x=25(-)19=6

Part 6) we have

-2x^{2}+28x+20

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

-2(x^{2}-14x)+20

Complete the square

-2(x^{2}-14x+49)+20+98

-2(x^{2}-14x+49)+118

Rewrite as perfect square

-2(x-7)^{2}+118

The vertex is the point (7,118)

therefore

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Part 7) we have

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Convert to vertex form

f(x)+10=-x^{2}+8x

Factor -1 the leading coefficient

f(x)+10=-(x^{2}-8x)

Complete the square

f(x)+10-16=-(x^{2}-8x+16)

f(x)-6=-(x^{2}-8x+16)

Rewrite as perfect square

f(x)-6=-(x-4)^{2}

f(x)=-(x-4)^{2}+6

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therefore

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Part 8) we have

x^{2}+6x+5

Convert to vertex form

Group terms

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Complete the square

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(x^{2}+6x+9)-4

Rewrite as perfect squares

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Part 9) we have

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This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

2(x^{2}-2x)-2=0

Complete the square

2(x^{2}-2x+1)-2-2=0

2(x^{2}-2x+1)-4=0

Rewrite as perfect squares

2(x-1)^{2}-4=0

2(x-1)^{2}=4

The vertex is the point (1,-4)

Part 10) we have

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This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

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Complete the square

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Rewrite as perfect squares    

8(x-4)^{2}+592

the vertex is the point (4,592)

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