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iVinArrow [24]
3 years ago
15

A fish and game worker catches, tags, and releases 120 trout in a lake. Later, she catches 84 trout and finds tags on 30 of them

. Estimate the number of trout in the lake.
A. 43
B. 234
C. 336
D. 3600
Mathematics
1 answer:
goldenfox [79]3 years ago
5 0
The final answer is B because you just add all three together. Hope you get it!
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You make $10.50 an hour at your part-time job and Andy makes $9.75 an hour at his full-time job. If you work a full week of 40 h
Marianna [84]
The awnser is $420.00 (10.50 X 40 = 4200)
7 0
2 years ago
Use the theorem in Sec. 28 to show that if f (z) is analytic and not constant throughout a domain D, then it cannot be constant
alexandr1967 [171]

Answer:

The value of f(z) is not constant in any neighbourhood of D. The proof is as explained in the explaination.

Step-by-step explanation:

Given

For any given function f(z), it is analytic and not constant throughout a domain D

To Prove

The function f(z) is non-constant constant in the neighbourhood lying in D.

Proof

1-Assume that the value of f(z)  is analytic and has a constant throughout some neighbourhood in D which is ω₀

2-Now consider another function F₁(z) where

F₁(z)=f(z)-ω₀

3-As f(z) is analytic throughout D and F₁(z) is a difference of an analytic function and a constant so it is also an analytic function.

4-Assume that the value of F₁(z) is 0 throughout the domain D thus F₁(z)≡0 in domain D.

5-Replacing value of F₁(z) in the above gives:

F₁(z)≡0 in domain D

f(z)-ω₀≡0 in domain D

f(z)≡0+ω₀ in domain D

f(z)≡ω₀ in domain D

So this indicates that the value of f(z) for all values in domain D is a constant  ω₀.

This contradicts with the initial given statement, where the value of f(z) is not constant thus the assumption is wrong and the value of f(z) is not constant in any neighbourhood of D.

6 0
2 years ago
Jamal's deck is in the shape of a polygon and is shown on the grid below.(-8,6)(6,6)o[(-8, -4),(6,-4)What is the area of Jamal's
PSYCHO15rus [73]

Let;

A(-8,6) B(6,6) C(6, -4) D(-8, -4)

Let's find the length AB

x₁= -8 y₁=6 x₂=6 y₂=6

We will use the distance formula;

d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}

=\sqrt[]{(6+8)^2+(6-6)^2}=\sqrt[]{14^2+0}=14

Next, we will find the width BC

B(6,6) C(6, -4)

x₁= 6 y₁=6 x₂=6 y₂=-4

substitute into the distance formula;

d=\sqrt[]{(6-6)^2+(-4-6)^2}

=\sqrt[]{(-10)^2}=\sqrt[]{100}=10

Area = l x w

= 14 x 10

= 140 square units

7 0
1 year ago
A sanitary landfill has available space of 16.2 ha at an average depth of 10 m. Seven hundred sixty-five (765) cubic meters of s
AfilCa [17]

Answer:

16.3 years

Step-by-step explanation:

1 ha = 10,000 m^2

The total capacity of the landfill is:

C = 16.2*10,000*10 = 1,620,000\ m^3

If waste is compacted to twice its delivered density, its volume is half of the delivered volume. Assuming that a year has 52 weeks, the volume of compacted solid waste dumped per year is:

V=52*(5*0.5*765)\\V=99,450\ m^3

The expected life of the landfill is given by its capacity divided by the yearly volume:

L=\frac{C}{V}=\frac{1,620,000}{99,450} \\L=16,3\ years

The landfill has an expected life of 16.3 years.

6 0
3 years ago
A rectangle has a length at most two more than seven times the width. If the perimeter is less than or equal to 52 units, find t
ella [17]

7x + 2 \leqslant 52
7 0
3 years ago
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