Simplify 9+2i ------- 1-4i

Compare and Contrast a square and an equilateral triangle? Be sure to communicate and discuss using mathematical language, ideas

ella [17]

The comparison and contrast between a square and an equilateral triangle are stated and explained below:

What is a square? A square is a plane shape with an equal length of sides. This implies that the measure of its length is the same as that of its breadth.

What is an equilateral triangle? A triangle is a plane shape with three sides, and a sum of its internal angles to equal $180^{o}$ . Thus an equilateral triangle is a type of triangle with equal lengths of sides, and therefore equal internal angles.

Comparison: i. The two shapes have equal lengths of sides.

ii. An equilateral triangle and a square are examples of plane shapes.

Contrast: i. An equilateral triangle has three sides, while a square has four sides.

ii. An equilateral triangle has three internal angles, while a square has four internal angles which are right angles.

iii. The sum of internal angles of an equilateral triangle is $180^{o}$ , while that of a square is $360^{o}$ .

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6 0

2 years ago

The difference between the roots of the quadratic equation x^2−14x+q=0 is 6. Find q.

Advocard [28]

The value of q is 40

Step-by-step explanation:

In the quadratic equation ax² - bx + c = 0, where

a = 1
b is the sum of its two roots
c is the product of its two roots

Assume that the roots of the quadratic equations are x and y

∵ The quadratic equation is x² - 14x + q = 0

∴ a = 1 , b = 14 and c = q

∵ b is the sum of its two roots

∵ Its roots are x and y

∴ x + y = 14 ⇒ (1)

∵ The difference between the roots is 6

∴ x - y = 6 ⇒ (2)

Now we have a system of equation to solve it

Add equations (1) and (2)

∴ 2x = 20

- Divide both sides by 2

∴ x = 10

Substitute the value of x in equation (1) or (2) to find y

∵ 10 + y = 14

- Subtract 10 from both sides

∴ y = 4

∵ c is the product of the two roots

∴ c = 10 × 4 = 40

∵ c = q

∴ q = 40

The value of q is 40

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6 0

4 years ago

Which expression is equivalent to (jk)l?

ivanzaharov [21]

I Think the answer your looking for here is

-2.3 6

------ = . ( -----

-2+3 1

-6

5 0

3 years ago

Find the maxima and minima of the function <img src="https://tex.z-dn.net/?f=f%28x%2Cy%29%3D2x%5E%7B2%7D%20%2By%5E%7B4%7D" id="T

NARA [144]

Using the second partial derivative test to find extrema in D :

Compute the partial derivatives of f(x, y) = 2x² + y⁴.

∂f/∂x = 4x

∂f/∂y = 4y³

Find the critical points of f, where both partial derivatives vanish.

4x = 0 ⇒ x = 0

4y³ = 0 ⇒ y = 0

So f has only one critical point at (0, 0), which does belong to the set D.

Compute the determinant of the Hessian matrix of f at (0, 0) :

$H = \begin{bmatrix}\dfrac{\partial^2f}{\partial x^2} & \dfrac{\partial^2f}{\partial y\partial x} \\ \\ \dfrac{\partial^2f}{\partial x\partial y} & \dfrac{\partial^2f}{\partial y^2}\end{bmatrix} = \begin{bmatrix}4 & 0 \\ 0 & 12y^2 \end{bmatrix}$

We have det(H) = 48y² = 0 at the origin, which means the second partial derivative test fails. However, we observe that 2x² + y⁴ ≥ 0 for all x, y because the square of any real number cannot be negative, so (0, 0) must be a minimum and we have f(0, 0) = 0.

Using the second derivative test to find extrema on the boundary of D :

Let x = cos(t) and y = sin(t) with 0 ≤ t < 2π, so that (x, y) is a point on the circle x² + y² = 1. Then

f(cos(t), sin(t)) = g(t) = 2 cos²(t) + sin⁴(t)

is a function of a single variable t. Find its critical points, where the first derivative vanishes.

g'(t) = -4 cos(t) sin(t) + 4 sin³(t) cos(t) = 0

⇒ cos(t) sin(t) (1 - sin²(t)) = 0

⇒ cos³(t) sin(t) = 0

⇒ cos³(t) = 0 or sin(t) = 0

⇒ cos(t) = 0 or sin(t) = 0

⇒ [t = π/2 or t = 3π/2] or [t = 0 or t = π]

Check the values of g'' at each of these critical points. We can rewrite

g'(t) = -4 cos³(t) sin(t)

Then differentiating yields

g''(t) = 12 cos²(t) sin²(t) - 4 cos⁴(t)

g''(0) = 12 cos²(0) sin²(0) - 4 cos⁴(0) = -4

g''(π/2) = 12 cos²(π/2) sin²(π/2) - 4 cos⁴(π/2) = 0

g''(π) = 12 cos²(π) sin²(π) - 4 cos⁴(π) = -4

g''(3π/2) = 12 cos²(3π/2) sin²(3π/2) - 4 cos⁴(3π/2) = 0

Since g''(0) and g''(π) are both negative, the points (x, y) corresponding to t = 0 and t = π are maxima.

t = 0 ⇒ x = cos(0) = 1 and y = sin(0) = 0 ⇒ f(1, 0) = 2

t = π ⇒ x = cos(π) = -1 and y = sin(π) = 0 ⇒ f(-1, 0) = 2

Both g''(π/2) and g''(3π/2) are zero, so the test fails. These values of t correspond to

t = π/2 ⇒ x = cos(π/2) = 0 and y = sin(π/2) = 1 ⇒ f(0, 1) = 1

t = 3π/2 ⇒ x = cos(3π/2) = 0 and y = sin(3π/2) = -1 ⇒ f(0, -1) = 1

but both of the values of f at these points are between the minimum we found at 0 and the maximum at 2.

So over the region D, max(f) = 2 at (±1, 0) and min(f) = 0 at (0, 0).

3 0

3 years ago

Use the image to answer the question.

coldgirl [10]

The area of the rectangle is 240 square inches

<h3>How to determine the area?</h3>

The dimensions are given as:

Length = 15 inches

Width = 4/3 feet

Start by converting feet to inches

Width = 4/3 * 12 inches

Width = 16 inches

The area is then calculated as:

Area = Length * Width

This gives

Area = 15 * 16

Evaluate

Area = 240

Hence, the area of the rectangle is 240 square inches