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2 years ago

Find the area of the shaded figure. To do so, subtract the area of the smaller

Mathematics

1 answer:

Vikentia [17]2 years ago

3 0

Answer:

x⁴ - 8x² + 9

Step-by-step explanation:

The Area of the smallest square is A1 = x . x = x²

The Area of the second square is A2 = ( x² - 3 )(x² - 3)

A2 = x⁴ - 3x² - 3x² + 9

A2 = x⁴ - 6x² + 9

The Area of the Shaded shape is A2 - A1

x⁴ - 6x² + 9 - x²

x⁴ - 8x² + 9

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Yanni threw his paper airplane 15 1/2 feet. Adrian threw his paper airplane 3/4 of yanni's distance. What is the distance Adrian

vagabundo [1.1K]

Answer:

19.125

Step-by-step explanation:

multiply 15.5 by 3 and then divide the answer by 4

4 0

3 years ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

3 0

3 years ago

Find the value of x and y in the following figure where ABCD is a parallelogram

asambeis [7]

Answer:

x = 3

y = 2

Step-by-step explanation:

Diagonals of a parallelogram bisect each other into two equal segments. Therefore:

3x - 1 = 2(x + 1)

Solve for x

3x - 1 = 2x + 2

Collect like terms

3x - 2x = 1 + 2

x = 3

Also:

5y + 1 = 6y - 1

Collect like terms

5y - 6y = -1 - 1

-y = -2

Divide both sides by -1

y = -2/-1

y = 2

5 0

3 years ago

In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and

Umnica [9.8K]

Answer:

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

A little more detail

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0

3 years ago

Each basketball comes in a cubic box with side lengths of 1 foot. The locker is 3 feet long, 2 feet wide, and 5 feet high. How m

LuckyWell [14K]

Answer:

30 basketballs

Step-by-step explanation:

one layer would be 3 x 2 = 6

to find how much you would need to cover the bottom of the locker, you would use the numbers 3 and 2, then multiply them and you get 6. You would need 6 basketballs to cover the bottom of the locker. That would be your first layer

5 layers would be 6 x 5 = 30

So the bottom would be filled with 6 basketballs. Now since the locker is 5 feet high multiply 6 x 5 and that would get you your total of basketballs

4 0

3 years ago