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2 years ago

I need help! I can't figure out how to do this! So, can you help me?

Mathematics

2 answers:

Tanzania [10]2 years ago

8 0

Answer:

5 - x=90°

6- angle c= 75°

7- angle a = 90°

Ksenya-84 [330]2 years ago

8 0

Answer:

c = 75

Step-by-step explanation:

All triangles add up to 180. Add the two angles which you have, and then subtract that from 180. that will be your third angle value.

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Over [174]

62 / 15.5 = 92 / x

to find x cross multiply:-

62x = 15.5 * 92
x = (15.5*92) / 62
= 23

8 0

2 years ago

Thabo invests R50000 at the rate of 5% per annum compound interest periodically every six months.How many years does it take to

lina2011 [118]

Answer:

50000-5%=49500 pls like

7 0

3 years ago

There were three quizzes for the week. The first quiz was worth twice as much as the second quiz. The third quiz was 10 points l

iren2701 [21]

X+y+z=70
x=2y
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6 0

2 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

I need help! I can't figure out how to do this! So, can you help me?​

I need help! I can't figure out how to do this! So, can you help me?