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krek1111 [17]
3 years ago
13

Select the correct answer. Joe wants to enlarge the rectangular pumpkin patch located on his farm. The pumpkin patch is currentl

y 40 meters wide and 60 meters long. The new pumpkin patch will be 3x meters wider and 5x meters longer than that of the original pumpkin patch. Which of the following functions will give the area of the new pumpkin patch in square meters? A. f(x) = 15x2 + 420x + 2,400 B. f(x) = 15x2 C. f(x) = 15x2 + 2,400 D. f(x) = 15x2 + 380x + 2,400
Mathematics
1 answer:
Butoxors [25]3 years ago
3 0

Answer:

Option D

f(x)=15x^{2}+380x+2400

Step-by-step explanation:

Original dimensions are as follows

Length=60 m

Width=40 m

Area=60*40=2400 m^{2}

New dimensions

Length=(60+5x) m

Width=(40+3x) m

Area=(60+5x)\times (40+3x)

Area=60(40+3x)+5x(40+3x)=2400+180x+200x+15x^{2} and collecting like terms we obtain

Area=15x^{2}+380x+2400

Therefore,

f(x)=15x^{2}+380x+2400

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Answer:

ion know

Step-by-step explanation:

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3 years ago
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Type the correct answer in each box. Use numerals instead of words.
lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17   eq(1)

y = -1/2x            eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a = 5/4, b =0 and c = -17

x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}

Finding value of y:

y = -1/2x

y=-1/2(\frac{\pm2\sqrt{85}}{5})

y=\frac{\pm\sqrt{85}}{5}

System A has 4 real solutions.

System B

y = x^2 -7x + 10    eq(1)

y = -6x + 5            eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a= 1, b =-1 and c =5

x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9    eq(1)

8x - y = -17        eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17    

8(-2) - y = -17    

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So, x= -2 and y = 1

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Answer:

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Step-by-step explanation:

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f(x)=-2x-18

f(-8)=-2(-8)-18

f(-8)=16-18

f(-8)=-2

The first value of the range of the function is -2

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f(x)=-2x-18

f(-2)=-2(-2)-18

f(-2)=4-18

f(-2)=-14

The second value of the range is -14

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f(x)=-2x-18

f(4)=-2(4)-18

f(4)=-8-18

f(4)=-26

Th third values of the range is -26

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f(x)=-2x-18

f(16)=-2(16)-18

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Answer:

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Step-by-step explanation:

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