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1 year ago

Pls Help 20 Points

Mathematics

2 answers:

Furkat [3]1 year ago

6 0

i think the answer would be 33%

Finger [1]1 year ago

5 0

Answer may be 30%

Be free to Ask more questions:))

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George found that he blinked 85 times in 15 minutes. At this rate, how many seconds passed during 51 blinks?

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540 seconds which is equal to 9 minutes.

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3 years ago

The population of West Algebra can be modeled by the equation

AleksAgata [21]

Answer:

65.7

Step-by-step explanation:

Given the population of West Algebra can be modeled by the equation

P = 30. 1.04^T

If T is the number of years since 2000 and P is the population in millions, in 2020, T = 2020 - 2000 = 20

Substitute T = 20 into the expression and get T

P = 30. 1.04^20

P = 30(2.1911)

P = 65.73

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3 years ago

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2 years ago

A triangle with two 8

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Step-by-step explanation:

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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