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horrorfan [7]
3 years ago
11

What is 37÷6572 because you know to know for the forth grade

Mathematics
2 answers:
eduard3 years ago
7 0
37÷6572 = 0.00566876
Montano1993 [528]3 years ago
3 0
The answer is 0.0056299452
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  B.  Step 2 uses the associative property, and step 3 uses the commutative property.

Step-by-step explanation:

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The commutative property lets you change the order of any pair of terms involved in addition. It appears that in Step 3, the order of the terms within the group has been swapped.

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<em>Comment on associative and commutative properties</em>

What applies to terms in addition applies to factors in multiplication.

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When finding the area of a quadrilateral, what operation would I use and why?
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Use the formula width x height = area. If the width is 4 inches and the height is 2 inches, the area equals 8 inches squared.

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Carl earned grades of 62 78 59 and 89 on four math tests what is the mean if his grades
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A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman
Yanka [14]

Answer:

W\approx 8.72 and L\approx 15.57.

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

2L+W+\frac{1}{2}(2\pi r)=38

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

2L+W+\frac{1}{2}(2r\pi)=38

2L+W+\frac{\pi}{2}W=38

Let us find area of window equation as:

\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)

\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})

\text{Area}=W\cdot L+\frac{\pi}{8}W^2

Now, we will solve for L is terms W from perimeter equation as:

L=38-(W+\frac{\pi }{2}W)

Substitute this value in area equation:

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2

Since we need the area of window to maximize, so we need to optimize area equation.

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2  

A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2  

Let us find derivative of area equation as:

A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W  

A'=38-2W-\pi W+\frac{\pi}{4}W    

A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W

A'=38-2W-\frac{3\pi W}{4}

To find maxima, we will equate first derivative equal to 0 as:

38-2W-\frac{3\pi W}{4}=0

-2W-\frac{3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}*4=-38*4

-8W-3\pi W=-152

8W+3\pi W=152

W(8+3\pi)=152

W=\frac{152}{8+3\pi}

W=8.723210

W\approx 8.72

Upon substituting W=8.723210 in equation L=38-(W+\frac{\pi }{2}W), we will get:

L=38-(8.723210+\frac{\pi }{2}8.723210)

L=38-(8.723210+\frac{8.723210\pi }{2})

L=38-(8.723210+\frac{27.40477245}{2})

L=38-(8.723210+13.70238622)

L=38-(22.42559622)

L=15.57440378

L\approx 15.57

Therefore, the dimensions of the window that will maximize the area would be W\approx 8.72 and L\approx 15.57.

8 0
3 years ago
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