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3 years ago

What is the vertex of the quadratic function?

Mathematics

1 answer:

Debora [2.8K]3 years ago

3 0

Answer:

f(x) = x² - 4x - 12

Step-by-step explanation:

f(x) = (x - 6)(x + 2)

f(x) = x² + 2x - 6x - 12

f(x) = x² - 4x - 12

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A box with a square base and open top must have a volume of 296352 c m 3 . We wish to find the dimensions of the box that minimi

mestny [16]

Answer:

Base Length of 84cm
Height of 42 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume, $V=x^2h=296352$

$h=\dfrac{296352}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{296352}{x^2}\\A(x)=x^2+4x\left(\dfrac{296352}{x^2}\right)\\A(x)=\dfrac{x^3+1185408}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+1185408}{x}\\A'(x)=\dfrac{2x^3-1185408}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84$

Step 4: Verify that x=84 is a minimum value

We use the second derivative test

$A''(x)=\dfrac{2x^3+2370816}{x^3}\\$When x=84$\\A''(x)=6$

Since the second derivative is positive at x=84, then it is a minimum point.

Recall:

$h=\dfrac{296352}{x^2}=\dfrac{296352}{84^2}=42$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 84cm
Height of 42 cm.

5 0

3 years ago

-(-2) <br> ...................................................

stepladder [879]

the 2 negatives get rid of eachother so its 2

4 0

3 years ago

The circle is inscribed in Triangle AEC. Which are congruent line segments

Harlamova29_29 [7]

Answer:

E-F and E-D

C-B and C-D

Step-by-step explanation:

The circle and triangle are as shown. The options are:

A-B and C-B
E-F and E-D
E-D and C-D
A-F and E-F
C-B and C-D

By drawing radius lines from the center of the circle to the tangent points B, D, and F, we can divide the triangle into 3 kites. Therefore, only segments that are legs of the same kite are congruent. So the answer must be E-F and E-D, and C-B and C-D.