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son4ous [18]
3 years ago
14

◆ Quadratic Equations ◆Please help !

Mathematics
1 answer:
Mila [183]3 years ago
3 0
I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
b = 2
c = 3
d = 4

\sf ax^2+(1-a(b+c))x+abc-d)

\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))

Simplify into standard form:

\sf x^2+(1-1(5))x+6-4

\sf x^2+(1-5)x+2

\sf x^2-4x+2

Use the quadratic formula to solve:

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

For functions in the form of \sf ax^2+bx+c. So in this case:

a = 1
b = -4
c = 2

Plug them in:

\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}

Solve for 'x':

\sf x=\dfrac{4\pm\sqrt{16-8}}{2}

\sf x=\dfrac{4\pm\sqrt{8}}{2}

\sf x\approx\dfrac{4\pm 2.83}{2}

\sf x\approx 0.59,3.41

So the answer would be A.
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The number of kilograms of water in a human body varies directly as the mass of the body. An 87​-kg person contains 58 kg of wat
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Answer:

50 kg water.

Step-by-step explanation:

We have been given that the number of kilograms of water in a human body varies directly as the mass of the body.

We know that two directly proportional quantities are in form y=kx, where y varies directly with x and k is constant of variation.

We are told that an 87​-kg person contains 58 kg of water. We can represent this information in an equation as:

58=k\cdot 87

Let us find the constant of variation as:

\frac{58}{87}=\frac{k\cdot 87}{87}

\frac{29*2}{29*3}=k

\frac{2}{3}=k

The equation y=\frac{2}{3}x represents the relation between water (y) in a human body with respect to mass of the body (x).

To find the amount of water in a 75-kg person, we will substitute x=75 in our given equation and solve for y.

y=\frac{2}{3}(75)

y=2(25)

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Therefore, there are 50 kg of water in a 75​-kg person.

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3 years ago
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Hey, I’m pretty sure the answer is y=1/2x + 0.

To explain, we can plug the x value into the equation for all of the x values, and we will get our y value.

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The second one,

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Answer:

None of the choices are correct.

Step-by-step explanation:

Let x be the width of the frame,

The framed art cannot me more than 320 square inches

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By using the quadratic formula

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

x = \frac{-52\pm \sqrt{(52^2-4(4)(-155))}}{2(4)}

x = \frac{-52\pm \sqrt{(2704+2480)}}{2(4)}

x = \frac{-52\pm \sqrt{(5184)}}{2(4)}

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x = 2.5

Frame must not be more than 2.5 inches wide.

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3 years ago
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