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Vera_Pavlovna [14]
3 years ago
10

If f(x) =

Mathematics
1 answer:
givi [52]3 years ago
8 0

Answer:

i thin correct answer is b

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Consider the quadratic equations
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Consider the two functions as
<span>y1(x) =3x^2 - 5x, 
y2(x) = 2x^2 - x - c

The higher the value of c, father apart the two equations will be.
They will touch when the difference, i.e. y1(x)-y2(x)=x^2-4*x+c has a discriminant of 0.
This happens when D=((-4)^2-4c)=0, or when c=4.
(a)
So when c=4, the two equations will barely touch, giving a single solution, or coincident roots.
(b) 
when c is greater than 4, the two curves are farther apart, thus there will be no (real) solution.
(c) 
when c<4, then the two curves will cross at more than one location, giving two distinct solutions.

It will be more obvious if you plot the two curves in a graphics calculator using c=3,4, and 5.


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3 years ago
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Consider this system of linear equations:
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Answer:

m = -3 and b = 5

Step-by-step explanation:

y = -3x + 5

.......

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In rectangle MATH, AT = 8 and TH = 12. Find the length of
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The question is not clear enough
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Justin plays basketball and scored 1/8 of the 48 points. How many points did he score?
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Jorge is asked to build a box in the shape of a rectangular prism. The maximum girth of the box is 20 cm. What is the width of t
MariettaO [177]

Answer:

The width of the box is 6.7 cm

The maximum volume is 148.1 cm³

Step-by-step explanation:

The given parameters of the box Jorge is asked to build are;

The maximum girth of the box = 20 cm

The nature of the sides of the box = 2 square sides and 4 rectangular sides

The side length of square side of the box = w

The length of the rectangular side of the box = l

Therefore, we have;

The girth = 2·w + 2·l = 20 cm

∴ w + l = 20/2 = 10

w + l = 10

l = 10 - w

The volume of the box, V = Area of square side × Length of rectangular side

∴ V = w × w × l = w × w × (10 - w)

V = 10·w² - w³

At the maximum volume, we have;

dV/dw = d(10·w² - w³)/dw = 0

∴ d(10·w² - w³)/dw = 2×10·w - 3·w² = 0

2×10·w - 3·w² = 20·w - 3·w² = 0

20·w - 3·w² = 0 at the maximum volume

w·(20 - 3·w) = 0

∴ w = 0 or w = 20/3 = 6.\overline 6

Given that 6.\overline 6 > 0, we have;

At the maximum volume, the width of the block, w = 6.\overline 6 cm ≈ 6.7 cm

The maximum volume, V_{max}, is therefore given when w = 6.\overline 6 cm = 20/3 cm  as follows;

V = 10·w² - w³

V_{max} = 10·(20/3)² - (20/3)³ = 4000/27 = 148.\overline {148}

The maximum volume, V_{max} = 148.\overline {148} cm³ ≈ 148.1 cm³

Using a graphing calculator, also, we have by finding the extremum of the function V = 10·w² - w³, the coordinate of the maximum point is (20/3, 4000/27)

The width of the box is;

6.7 cm

The maximum volume is;

148.1 cm³

5 0
3 years ago
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