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2 years ago

11

40,000 ft.2 to hectares (1 hectare = 10,000 m2)

2 answers:

melisa1 [442]2 years ago

7 0

Answer:

0.37 hectares

Step-by-step explanation:

40,000 sq ft X 9.29 / 1,000,000 = 0.37 hectares

frozen [14]2 years ago

4 0

Answer:

1.2192 hectares

Step-by-step explanation:

1 foot² = .3048 meter²

40000 feet² * .3048 = 12192 meters²

12192/10000 = 1.2192

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If 3x+54+(65 × 32)+ 902- (43/12) =y, then what is y?

IceJOKER [234]

Answer:

We cannot solve without a value for x. The simplified form is $3x+\frac{36,389}{12} =y$

Step-by-step explanation:

If the equation is $3x+54+(65)(32)+902-\frac{43}{12}=y$ then we cannot solve it until we know the value of x. We can simplify it using order of operations such as PEMDAS.

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2 years ago

Which expression is equivalent to x - 5/3

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$\frac{1}{3}(3x-5)$

Step-by-step explanation:

If you expand an equation you get its equivalent expression

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3 years ago

Consider the matrices -2 -1 1 4 1 A = B = 4 -2 4 -2 -2 4 -2 1 -1 (a) Verify that they are diagonalizable and that they commute.

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Explanation to answers are shown in the attachment

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3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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Mrac [35]

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3 years ago