A community theater group produced 8 plays over the last two years. The
group's goal for the next two years is to produce at least 1 times as many
plays as they did in the two previous years. How many plays does the group
want to produce in the next two years?

Answer:

Step-by-step explanation:

diameter = 6 in

r = 6/2 = 3 in

h = 8 in

[tex]\textbf{Area of cylinder = $2\pi r(r+h)$ }[/tex]

                              [tex]= 2*3.14*3*(3+8)\\\\= 2*3.14*3*11\\\\= 207.24 \ in^{2}[/tex]

Step-by-step explanation:

please mark me as brainlest

Answer:

0.7

Step-by-step explanation:

5(9n-6)              

5×9n-5×6

45n-30

45n/45  -30/45

or 45n÷45-30÷45

n= 0.6666

0.7

hope dis helps!!!!!    

Answer:

[tex]\displaystyle Dependency\:[Infinitely\:Many\:Solutions][/tex]

Explanation:

In the bottom equation, moving [tex]\displaystyle -2x[/tex]to the oppocite side will give you the linear standard equation above it, therefore numerous solutions will occur.

I am joyous to assist you at any time.

The two numbers are illustrations of system of equations

The two numbers are 2.7 and 37.3

Let the numbers be x and y.

So, we have the following equations

x + y = 40

xy = 100

To determine the value of the variables, we simply plot the graph of both equations

From the graph, we have the following result

(x,y) = (2.7, 37.3)

Hence, the two numbers are 2.7 and 37.3

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keeping in mind that the vertex will be half-way between the directrix and the focus point, in this case the directrix is below the focus point, meaning is a vertical parabola, Check the picture below.

The volume of the container is the amount of can in it

The contents weigh 1047.1 ounces

The given parameters are:

Volume = 28.3 cubic feet

Average weight of can = 37 ounces per cubic foot

So, the weight of the container would be:

Weight = Volume * Average weight of can

Substitute known values

Weight = 28.3 cubic feet * 37 ounces per cubic foot

Evaluate the product

Weight = 1047.1 ounces

Hence, the contents weigh 1047.1 ounces

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Answer:

The Scale Factor is 5.3

Step-by-step explanation:

Here's the reason why:

You are trying to make the rectangle much larger.

The 14yd by 6yd is trying to get to the 74.2yd by 31.8yd.

We need to remember the importance of length and width.

If 14 yards is our length then 74.2 yards is our length on the enlarged rectangle.

You must do 74.2/14 because we need to work backwords.

**You can't find something to multiply 14 by to get to 74.2 unless you do division.**

74.2/14 is 5.3

Now do the same with the width.

31.8/6

That is 5.3 as well.

Therefore the scale factor is 5.3

Answer:

Step-by-step explanation:

Answer: All three sides are congruent

The problem is with A. It's true as well. At least 2 of the sides are congruent. The way to eliminate it is to say that does not mean specifically that all three sides are congruent.

I would argue that this is a faulty question, and watch someone squirm as they try to explain to you why a is not the answer. I don't know how to comply with your request. How are the choices designated

A

B

C

or

1  

2

3

Answer:

C. All three sides are congruent.

Step-by-step explanation:

Answer:

1. Degree= 6, trinomial

2. degree= 4, binomial

Step-by-step explanation:

1. The largest exponent is 6 making it the degree and there are three terms because (m^5n^2), (mn^2), (n^6).

2. Since (9a^2bc^2) has the more than 1 exponent, you add them and the  degree is 4. There are two terms because (2a^3b), (9a^2bc^2).

From the table, it is found that there is a [tex]\frac{5}{12}[/tex] probability that parents will have 1 boy and 1 girl if they have a total of two children.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, out of 12 experiments, 5 resulted in one of each, hence:

[tex]p = \frac{5}{12}[/tex]

There is a [tex]\frac{5}{12}[/tex] probability that parents will have 1 boy and 1 girl if they have a total of two children.

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Answer:

105/330 = 21/66 = 7/22

7/22 is the answer I think

Step-by-step explanation:

n(U) = 330

n(A) = 85

n(B) = 200

n(A n B) = 60

Now,

n(A U B) = n(A) + n(B) - n(A n B)

or, n(A U B) = 85 + 200 - 60

or, n(A U B) = 285 - 60 = 225

Now,

n(A U B) compliment = n(U) - n(A U B)

n(A U B) compliment = 330 - 225

so, n(A U B) compliment = 105

Now,

probability = 105/330

Answer:

(400 × 3 ) + (70 × 3) + (5 × 3)

Answer:

3 feet

Step-by-step explanation:

Based on the cost function, the company that produces cell phones can minimize the cost function at 13.4 cell phones.

To minimize the cost function, you should differentiate it.

= 4x² - 107x + 4322

dC/dx = 0

8x - 107 = 0

Solving gives:
8x = 107

x = 107 / 8

= 13.4 cell phones

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The number of distinct subset of the set is an illustration of combination

There are 34 distinct subset of the set that have three digit sums

The set is given as:

S = {1,8,9,39,52,91}

Group the elements

S = {(1,8),9,39,52,91}

So, we have:

S ={s1,s2,s3,s4,s5}

To have a three-digit sum, the sum of the selected elements must be 100 or above.

The number of ways the above set can be selected is:

[tex]n_1 = 4 * [^3C_1 + ^3C_2 + ^3C_3] + 1[/tex]

Evaluate

[tex]n_1 = 4 * [3 + 3 + 1] + 1[/tex]

[tex]n_1 = 29[/tex]

The set can be further rearranged as:

S = {(52,39),9,(8,1)}

i.e.

S = {b1,b2,b3}.

The number of ways the above set can be selected such that the sum is 100 or above is:

[tex]n_2 = 2 * [^2C_1] + 1[/tex]

Evaluate

[tex]n_2 = 2 * 2 + 1[/tex]

[tex]n_2 = 5[/tex]

So, the total number of ways is:

[tex]n = n_1 +n_2[/tex]

This gives

[tex]n = 29 +5[/tex]

[tex]n = 34[/tex]

Hence, there are 34 distinct subset of the see that have three digit sums

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There are 34 distinct subsets of the see that have three-digit sums.

When we know that S is a subset of T, we place the circle representing S inside the circle representing T.

The set is given as:

S = {1,8,9,39,52,91}

Group the elements

S = {(1,8),9,39,52,91}

S ={s1,s2,s3,s4,s5}

To have a three-digit sum, the sum of the selected elements must be 100 or above.

The number of ways the above set can be selected is:

[tex]\rm n_1=4\times (^3C_1+^3C_2+^3C_3)+1\\\\n_1=4(3+3+1)+1\\\\n_1=4 \times 7+1\\\\n_1=28+1\\\\n_1=29[/tex]

The set can be further rearranged as:

S = {(52,39),9,(8,1)}

S = {b1,b2,b3}.

The number of ways the above set can be selected such that the sum is 100 or above is:

[tex]\rm n_2=2\times ^2C_1+1\\\\n_2= 2\times 2+1\\\\n_2=4+1\\\\n_2=5[/tex]

The distinct subsets of the set are;

[tex]\rm n_1+n_2=29+5 =34[/tex]

Hence, there are 34 distinct subsets of the see that have three-digit sums.

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Answer:

18 is 30% of 60

Step-by-step explanation:

Try to line up percents % and the actual numbers they represent.

number : precent

       60  is 100%

        18  is x %

cross multiply

x·60 = 100·18

divide both sides by 60 to isolate x

x = ((100·18)/60

x = 30

The Laplace transform of f(t) = cos² t is equal to the following expression: [tex]\mathcal {L} \{\cos^{2} t\} = \frac{1}{2\cdot s} - \frac{s}{2\cdot (s^{2}+2)}[/tex].

In this we need to rewrite the given function in terms of sines and cosines, whose Laplace transforms are well known. There is the following trigonometric formula:

[tex]\cos ^{2} t = \frac{1-\cos 2t}{2}[/tex]   (1)

Now we proceed to apply the Laplace transforms:

[tex]\mathcal {L} \{f(t)\} = \frac{1}{2}\cdot \mathcal {L} \left\{1 \right\}-\frac{1}{2}\cdot \mathcal {L}\left\{\cos 2t\right\}[/tex]

[tex]\mathcal {L} \{\cos^{2} t\} = \frac{1}{2\cdot s} - \frac{s}{2\cdot (s^{2}+2)}[/tex]

The Laplace transform of f(t) = cos² t is equal to the following expression: [tex]\mathcal {L} \{\cos^{2} t\} = \frac{1}{2\cdot s} - \frac{s}{2\cdot (s^{2}+2)}[/tex]. [tex]\blacksquare[/tex]

The statement is poorly formatted, the correct form is shown below:

Find [tex]\mathcal {L}\{f(t)\}[/tex] by first using a trigonometric expression. (Write your answer as a function of s). f(t) = cos² t

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Answer:

Step-by-step explanation:

A "shaded area" problem can be worked by directly computing the area of the shaded shape, or by decomposing the figure in a convenient way. Here the overall square can be decomposed into unshaded triangles and a shaded square. The triangles have dimensions given, but the smaller square does not. Work must necessarily proceed from given dimensions to arrive at the values associated with dimensions not given.

For a problem like this, working the last question first can make it easier.

The total area is the area of a square 6+3 = 9 units on a side. Its area is ...

  Total area = s² = (9 units)² = 81 units²

__

The area of each of the four non-shaded triangles is ...

  A = 1/2bh = 1/2(6 units)(3 units) = 9 units²

There are 4 triangles with that area that are not shaded, so ...

  Non-shaded region = 4 × (9 units²) = 36 units²

__

The shaded region is that portion of the total area that is not Non-shaded. Its area will be the difference ...

  Shaded area = Total area - Non-shaded area

  Shaded area = 81 units² -36 units² = 45 units²

Area :-

[tex]\\ \rm\rightarrowtail side^2[/tex]

[tex]\\ \rm\rightarrowtail 13^2[/tex]

[tex]\\ \rm\rightarrowtail 169[/tex]

Square root:-

Answer:

30.37

Step-by-step explanation:

Answer:

[tex]164 \div 5.40 ={\ 30.3704{}}[/tex]

The answer is 30.3704

Answer:

not

Step-by-step explanation:

Answer:

4/5 and -4/5 are not same because

4/5=

The Area of the shaded portion = area of 5 shaded squares = 80 in.².

Area of a square = s², where the edge length is s.

Edge length of each small square = 12/3 = 4 in.

Area of the shaded portion = area of 5 shaded squares = 5(s²)

s = 4 in.

Area of the shaded portion = 5(4²)

Area of the shaded portion = 80 in.²

Therefore, the Area of the shaded portion = area of 5 shaded squares = 80 in.².

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Answer:

ab and ac are congruent with each other

Answer:

BC > AC

Step-by-step explanation:

∠A > ∠B (Given)

Side opposite to larger angle is larger and side opposite to smaller angle is smaller.

->AC < BC

or

BC > AC

Answer:

-5 ; -3 ; -1.5 ; 0 ; 11 (**Quotient order**) -30/6 ; 21/-7 ; - (-3/-2) ; 0/-20 ; -44/-4  (** Division Problem Order which is the Original Order**)

Step-by-step explanation:

Here are your division problems:

-30/6 ; 0/-20 ; -44/-4 ; 21/-7 ; - (-3/-2)

First, do the -30 divided by 6

The rule when dividing with a negative and positive is you divide normally but have a negative for your answer.

30 divided by 6 is 5 but carry the negative due to the rule.

First one is -5

Next,

This one is simplistic,

It doesn't matter if it is negative anything divided by 0 is 0

Second one is 0

The next one is -44 divided by -4.

The rule when dividing with negative and negative is they cancel out and become a positive answer.

44 divided by 4 is 11

Third is 11

21 divided by -7 is the same as the first one, divide normally and carry negative.

21 divided by 7 is 3 but carry the negative.

Fourth is -3

Lastly,  we need to use the rules once again. the negatives cancel but you still have a negative. This shall be a negative answer.

Here's the twist. 3 can't divide into 2 easily. You need to get a decimal which is fine.

Fifth is -1.5

Now with all of our numbers:

-5, 0, 11, -3, and -1.5 you need to order them from least to greatest.

Rule: The farther you go out with negatives the smaller they are for example like -12.

-5 is first

-3 is second

-1.5 is third

0 is fourth

11 is fifth

-5 ; -3 ; -1.5 ; 0 ; 11 is the quotient order

-30/6 ; 21/-7 ; - (-3/-2) ; 0/-20 ; -44/-4 is the division problem order

Answer:

26

Step-by-step explanation:

Cellphone: 32 - 19 (students without tablet)

Tablet: 70 -57 (students without cellphone)

Both equations remain with only 13 students with both a cellphone and a tablet.

Then 13 + 13 = 26

Answer:

1

Step-by-step explanation:

[tex] \tan \bigg( \frac{5\pi}{4} \bigg) \\ \\ = \tan \bigg( \pi + \frac{\pi}{4} \bigg) \\ \\ = \tan \bigg(\frac{\pi}{4} \bigg) \\ \\ = 1[/tex]

[tex]\tan \left(\dfrac{5 \pi}4 \right)\\\\\\=\tan \left(2 \cdot \dfrac{\pi}2 + \dfrac \pi4\right)\\\\\\=\tan \left(\dfrac{\pi}4 \right)\\\\\\=1[/tex]