What is the length of PQ¯¯¯¯¯?

Mathematics

2 answers:

timofeeve [1]4 years ago

5 0

Since you don't have any angle and its opposite side, you cannot establish the ratio for the law of sines, so you must use the law of cosines.

$c^2 = a^2 + b^2 - 2ac \cos C$

Let's change it for this problem.

$r^2 = p^2 + q^2 - 2pq \cos R$

$r^2 = 9^2 + 6^2 - 2(9)(6) \cos 34^\circ$

$r^2 = 9^2 + 6^2 - 2(9)(6) \cos 34^\circ$

$r^2 = 27.46$

$r = 5.24$

$PQ = 5.2~km$

serg [7]4 years ago

3 0

Answer:

Hello! The correct answer is 5.2 km

Step-by-step explanation:

I am just confirming!

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Using the point (3,1) and a slope of 3,

Mariulka [41]

Answer:

y= -1= 3(x -3)

Step-by-step explanation:

y -y1= m(x -x1)

y -1= 3(x -3)

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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:

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

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