There goes the right problem

qaws [65]

9) It helps to draw a picture. Since one angle is 60°, the other angle is 120°. When you draw the ddiagonal lines, you will see that you end up with a 30°-60°-90° triangle where √3 is the hypotenuse. The longest diagonal will be across from the 60° angle, which will be length a√3. The length of the hypotenuse is 2a. (Remember: a, a√3, 2a are lengths for this special triangle)
2a = √3 → a = √3/2.
side across from 60° = a√3 = (√3/2)·√3 → side from across 60° = 3/2
Remember that this is only one-half of the diagonal of the rhombus so multiply that by 2.
Length of longest diagonal is 3.

10) Volume of a square pyramid = (1/3) B · h, where B is the area of the base
200 = (1/3)B(12)
50 = B
Since the base is a square, then each side is √50 (which equals 5√2).

Now, to find the lateral edge, use Pythagorean Theorem (a² + b² = c²).
Height (a) = 12, base (b) = 5√2, and lateral edge (c) is unknown.

(12)² + (5√2)² = c²
144 +   50   = c²
   194    = c²
   √194    = c

Lateral edge is √194.

3 0

3 years ago

HELPPPPPPP PLEASEEE FASTTTT SOMEONE

Katen [24]

Answer:

A. is your answer.

Step-by-step explanation:

Why A is your answer, is because there is just one 2 and that makes it 2^6.38.

A. is your answer.

6 0

3 years ago

Read 2 more answers

Percentage rate of increase or decrease y = 3500(1.079)

JulsSmile [24]

Answer:

Increase

Step-by-step explanation:

If you multiply the numbers it increases!

7 0

3 years ago

Hi can you help me in mate plis<br>simplify the following algebraic expressions

mixas84 [53]

Answer:

3b

Step-by-step explanation:

$\dfrac{24a^2b^4c^3}{8a^2b^3\sqrt[3]{c^9}} =$

$= 3a^{2 - 2}b^{4 - 3} \times \dfrac{c^3}{c^\frac{9}{3}}$

$= 3a^{0}b^{1} \times \dfrac{c^3}{c^3}$

$= 3bc^{3-3}$

$= 3bc^{0}$

$= 3b$

8 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