Help I need answers!! please

Find missing angles and variables

shutvik [7]

Answer:

e= 2x-4

Step-by-step explanation:

the opposite of e is 2x-4.

7 0

3 years ago

Anyone know the answer?

OLga [1]

3. Both are supplementary to angle BCE
4. Angle 2 = Angle 1

8 0

3 years ago

Hey loves, plz help if you have time:)

aksik [14]

Answer:

ΔABX and ΔEDX

Step-by-step explanation:

Given: AX ≅ EX

BX ≅ DX

SAS implies a congruent relation with respect to Side-Angle-Side between the two required triangle.

AX ≅ EX (given)

BX ≅ DX (given)

<BXA ≅ <DXE (vertical opposite angles)

Therefore,

ΔABX ≅ ΔEDX (Side-Angle-Side, SAS, congruence property)

3 0

3 years ago

What is the graph of the solution to the following compound inequality?

ohaa [14]

Answer:

Step-by-step explanation:

<, > - open circle

≤, ≥ - solid circle

<, ≤ - draw to the left

>, ≥ - draw to the right

$3-x\geq2\qquad\text{subtract 3 from both sides}\\\\-x\geq-1\qquad\text{change the signs}\\\\\boxed{x\leq1}\\\\4x+2\geq10\qquad\text{subtract 2 from both sides}\\\\4x\geq8\qquad\text{divide both sides by 4}\\\\\boxed{x\geq2}$

3 0

3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2