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Nookie1986 [14]
3 years ago
9

Compare. Select <, >, or=. 3.568 ? 3.577

Mathematics
1 answer:
alexira [117]3 years ago
5 0

Answer:

3.568 < 3.577

Step-by-step explanation:

3.568 < 3.577

So 3.577 is greater than 3.568

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You are designing a rectangular poster to contain 7575 in2 of printing with a 33​-in margin at the top and bottom and a 11​-in m
erma4kov [3.2K]

Answer:

The dimensions of the rectangular poster is 15 in by 5 in.

Step-by-step explanation:

Given that, the area of the rectangular poster is 75 in².

Let the length of the rectangular poster be x and the width of the rectangular poster be y.

The area of the poster = xy in².

\therefore xy=75

\Rightarrow y=\frac{75}{x}....(1)

1 in margin at each sides and 3 in margin at top and bottom.

Then the length of printing space is= (x-2.3) in

                                                           =(x-6) in

The width of printing space is = (y-2.1) in

                                                  =(y-2) in

The area of the printing space is A =(x-6)(y-2) in²

∴ A =(x-6)(y-2)

Putting the value of y

\Rightarrow A =(x-6)(\frac{75}{x}-2)

\Rightarrow A = 87-\frac{450}{x}-2x

Differentiating with respect to x

A '= \frac{450}{x^2}-2

Again differentiating with respect to x

A''=-\frac{900}{x^3}

To find the minimum area of printing space, we set A' = 0

\therefore \frac{450}{x^2}-2=0

\Rightarrow 450 =2x^2

\Rightarrow x^2=225

\Rightarrow x=\pm 15

Now putting x=±15 in A''

A''|_{x=15}=-\frac{900}{15^3}

A''|_{x=-15}=-\frac{900}{(-15)^3}=\frac{900}{(15)^3}>0

Since at x=15 , A"<0 Therefore at x=15 , the area will be minimize.

From (1) we get

y=\frac{75}{x}

Putting the value of x

y=\frac{75}{15}

   =5 in

The dimensions of the rectangular poster is 15 in by 5 in.

4 0
3 years ago
Statistics would not be useful if they were not presented with 100% certainty. T/F
ExtremeBDS [4]

Answer: False

Nothing is 100% certain no matter how sure it seems. This is especially true with statistics. There's always going to be some amount of error. The goal is to minimize the error as much as possible in the best cost-effective manner.

7 0
3 years ago
Transversal t cuts parallel lines a and bas shown in the diagram. Which equation is necessarily true?
irina1246 [14]

Answer:

Option D is correct

Step-by-step explanation:

Using the given diagram, we want to know the equation that is true

Option A is wrong as both are on a straight line and in fact should add up to equal 180 and not be equal to each other

Option B is not correct as both are supplementary and does not equal each other

Option C is not correct, both are corresponding to each other and should not add up to 90

Option D is correct

Both angles are supplementary as they are exterior angles that add up to 180

4 0
3 years ago
Log, 7 1/2<br> Solve in exponential form
shutvik [7]
What is the log for?
6 0
3 years ago
Read 2 more answers
Behind my math class help pls
vladimir2022 [97]

Answer:

Please check the explanation.

Step-by-step explanation:

Part a)

Given that the two parallel lines are crossed by a transversal line.

Given that

m∠2 = 2x + 54 and m∠6 = 6x - 11

Angle ∠2 and ∠6 are corresponding angles.

Corresponding angles are congruent.

Thus,

m∠2 = m∠6

2x + 54 = 6x - 11

flipe the equation

6x - 11 = 2x + 54

subtract 2x from both sides

6x - 2x - 11 = 2x - 2x + 54

4x - 11 = 54

adding 11 to both sides

4x - 11 + 11 = 54 + 11

4x = 65

dvide both sides by 4

4x/4 = 65/4

x = 16.2500    (round to 4 decimal places)

Part b)

We have already determined

x = 16.2500

Given

m∠2 = 2x + 54

substitute x = 16.2500 in the euation

        = 2(16.2500) + 54

        = 86.5°

As angle ∠2 and angle ∠1 lie on a straight line. Hence, the sum of their angles must be 180°.

i.e.

m∠1 + m∠2 = 180°

substituting m∠2 =  86.5° in the equation

m∠1 + 86.5°  = 180°

subtracting 86.5° from both sides

m∠1 + 86.5° - 86.5° = 180° - 86.5°

m∠1 = 93.5°

Therefore, the measure of angle m∠1 is:

  • m∠1 = 93.5°
4 0
3 years ago
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