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svet-max [94.6K]
1 year ago
9

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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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

Hence the amount of people that will be there in 2020 is 65.7million people

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3 years ago
Piper has a balance of $1,500 on
saul85 [17]

Answer: $408.96

Step-by-step explanation:

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2 years ago
A triangle with two 8
tekilochka [14]

Answer:

One unique triangle

Step-by-step explanation:

3 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

Read more about the plane

brainly.com/question/1962726

#SPJ4

5 0
1 year ago
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