A solid cylinder of iron whose diameter is 18 cm and height 12 cm is melted and turned into a solid sphere. Find the diameter of the sphere so formed​

The value of bh for b =12 and h = 2 is 24

The expression is given as:

bh

Where

b = 12 and h = 2

So, we have:

bh = 12 * 2

Evaluate

bh = 24

Hence, the value of bh for b =12 and h = 2 is 24

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Answer: -2 and 4

Step-by-step explanation:

Let [tex]t=x/2[/tex].

[tex]f(t)=0 \longrightarrow t=-1, 2\\\\\therefore x=-2, 4[/tex]

Answer:switch mikes Luke’s and Alex in between each other and let winter keep her own

Step-by-step explanation:

Answer:7 people

Step-by-step explanation:

Answer:

7

Step-by-step explanation:

The equation that can be used to find the height of the tree is as follows:

[tex]\frac{h}{sin 29} =\frac{40}{sin147}[/tex]

The height of the tree is 35.6 m

Th equation that can be used to find the height of the tree uses the principle of sine rule.

Therefore,

[tex]\frac{h}{sin 29} =\frac{40}{sin(180-4-29)}[/tex]

[tex]\frac{h}{sin 29} =\frac{40}{sin147}[/tex]

Therefore,

[tex]\frac{h}{sin 29} =\frac{40}{sin147}[/tex]

cross multiply

h sin 147 = 40 sin 29°

h = 40 sin 29° / sin 147

h = 19.3923848099 / 0.54463903501

h = 35.6015636045

h = 35.6 m

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The 1st table represents a linear as well as the quadratic relationship, the 2nd table represents neither and the 3rd table also represents neither relationship.

Hence, the 1st table represents a linear as well as a quadratic relationship, the 2nd table represents neither and the 3rd table also represents neither relationship.

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1. The solution to the first percent puzzle is as follows:

                              75%         66²/₃         11¹/₉

          33¹/₃%        25%          22¹/₅%      3⁷/₁₀%

         50%         37¹/₂%          33¹/₃%      5¹/₂%

        50%          37¹/₂%          33¹/₃%      5¹/₂%

2. The solution to the second percent puzzle is as follows:

                           10%             90%

         50%           5%              45%

         50%           5%              45%

A percent puzzle is a puzzle that engages students to practice working with percentages.

The percent puzzle involves a matrix format, in which students multiply the column percent by the row percent to find the percent they must put in the missing boxes to complete the puzzle.

The multiplication results for the missing boxes can be either stated in decimals, fractions, or a combination.

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The expressions are illustrated below.

3x = 12

x = 12/3

x = 4

x + 4 = 17

x = 17 - 4

x = 13

9y + 3 = 21

9y = 21 - 3

9y = 18

y = 18/9

y = 2

x ÷ 2 = 4

x = 4 × 2

x = 8

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Answer: B

Step-by-step explanation:

1: 5 × 1.25 = 6.25

 2 × 1.25 = 2.5

 1.7 × 1.25 = 2.125

2: L × W × H = 6.25 × 2.5 × 2.125 = 33.20 cubic

Answer:

Step-by-step explanation:

i didnt do anything bro why u delete my answer

Answer:

a.$4165

b.$4956.35

Step-by-step explanation:

a)19% × 3500 = 665.

3500+665 =4165

b)19% × 4165 = 791.35

4165+791.35= 4956.35

The value for x that is a double root of this function is x = -1

The function is given as:

f(x) = x^3 - 3x - 2

From the graph of the above function (see attachment), we have the following highlights:

The point that it touches the x-axis is the double root point

Hence, the value for x that is a double root of this function is x = -1

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See below for how the domain and the range are calculated

Function, y = -x^4 + 4

The function is given as:

y = -x^4 + 4

See attachment for the graph

From the attached graph, we have:

Function, y = 2x^3 - 1

The function is given as:

y = 2x^3 - 1

See attachment for the graph

From the attached graph, we have:

Function, y = ∛(2x + 8)

The function is given as:

y = ∛(2x + 8)

See attachment for the graph

From the attached graph, we have:

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The largest size of a team according to the task content is; 20 kids.

It follows from the task content that the population is; 20 students in Mrs. Tanya's class and 24 kids in Mrs. Olga's class.

However, since, the team forming condition states that students be placed on teams of equal size with students from only their own class.

It therefore follows that the largest size of a team is 20 as only 20 students are in Mrs. Tanya's class hence leaving 4 kids in Mrs. Olga's class without a team.

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Using the t-distribution, the critical value needed is: t = 2.2622.

The critical value is found using a calculator, with two parameters:

For this problem, we have a confidence level of 95% and 10 - 1 = 9 df, hence the critical value is t = 2.2622.

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Using the hypergeometric distribution, the probabilities are given as follows:

a) 0.2 = 20%.

b) 0.5539 = 55.39%.

The formula is:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

Researching this problem on the internet, it is found that the council has 15 members, of which 7 are Democrats and 8 are Republicans.

Item a:

The parameters are:

N = 15, n = 2, k = 7.

The probability is P(X = 2), hence:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,15,2,7) = \frac{C_{7,2}C_{8,0}}{C_{15,2}} = 0.2[/tex]

Item b:

At most one Democrat, with n = 3, hence:

[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]

In which:

[tex]P(X = 0) = h(0,15,3,7) = \frac{C_{3,0}C_{8,3}}{C_{15,3}} = 0.1231[/tex]

[tex]P(X = 1) = h(1,15,3,7) = \frac{C_{3,1}C_{8,2}}{C_{15,3}} = 0.4308[/tex]

Then:

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1231 + 0.4308 = 0.5539[/tex]

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The number of board game sorted is 14

The games have been sorted into video games and board games.

The number of video games is 28 less than 6 times the number of board games.

Therefore,

the number of board games = x

number of video game = 6x - 28

The games are 80% video games. Hence,

80 / 100 × x + 6x - 28 = 6x - 28

4 / 5 × 7x - 28 = 6x - 28

4 / 5 (7x - 28) = 6x -28

28 / 5 x - 112 / 5  = 6x - 28

28 / 5 x  - 6x = -28 + 112 / 5

28x - 30x / 5 =  - 28 + 112 / 5

-2x / 5  = -28 / 5

-10x = -140

x = 14

Therefore, the number of board game sorted is 14.

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Answer:

14

Step-by-step explanation:

80 / 100 × x + 6x - 28 = 6x - 28

4 / 5 × 7x - 28 = 6x - 28

4 / 5 (7x - 28) = 6x -28

28 / 5 x - 112 / 5  = 6x - 28

28 / 5 x  - 6x = -28 + 112 / 5

28x - 30x / 5 =  - 28 + 112 / 5

-2x / 5  = -28 / 5

-10x = -140

x = 14

Answer: False

Step-by-step explanation:

The only number that makes this impossible to solve is x = 0

Answer:

Answer:

B is correct

Using a system of equations, it is found that there were 100 students at the bus stop.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

Considering the first paragraph, the equation is:

[tex]\frac{x}{y - 25} = \frac{3}{2}[/tex]

Considering the second paragraph, the equation is:

y - 25 = x - 15.

Hence:

y = x + 10.

Replacing on the first equation:

[tex]\frac{x}{x - 15} = \frac{3}{2}[/tex]

3x - 45 = 2x

x = 45.

y = x + 10 = 55.

The total number was:

45 + 55 = 100.

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Using the z-distribution, the 95% confidence interval of the population mean is: (17.5, 24.5).

The confidence interval is:

[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]

In which:

In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.

For this problem, the other parameters are:

[tex]\overline{x} = 21, \sigma = 26.8, n = 225[/tex]

Hence the bounds of the interval are:

[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 21 - 1.96\frac{26.8}{\sqrt{225}} = 17.5[/tex]

[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 21 + 1.96\frac{26.8}{\sqrt{225}} = 24.5[/tex]

The interval is: (17.5, 24.5).

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The matrix [tex]\vec C[/tex] resulting from the multiplication between matrices [tex]\vec B[/tex] and [tex]\vec A[/tex] is equal to the matrix [tex]\left[\begin{array}{ccc}37&55&- 20\\23&11&- 20\\25&28&-9\end{array}\right][/tex]. (Correct choice: C)

In this question we must apply the operation of multiplication of two matrices, whose definition is shown below:

[tex]\vec B\, \cdot \, \vec A[/tex], where [tex]\vec B \in \mathbb {R}_{n \times p}[/tex] and [tex]\vec A \in \mathbb{R}_{p \times n}[/tex]     (1)

Where each element of the resulting matrix is equal to the following expression:

[tex]c_{(i, j)} = b_{(i, p)} \,\bullet\, a_{(p, j)}[/tex]     (2)

Where the operator "[tex]\bullet[/tex]" represents a dot product.

Now we proceed to calculate each element of the matrix:

[tex]c_{(1, 1)} = \left[\begin{array}{ccc}5&7&3\end{array}\right] \,\bullet\,\left[\begin{array}{c}1\\5\\- 1\end{array}\right][/tex]

5 · 1 + 7 · 5 + 3 · (- 1)

37

[tex]c_{(1, 2)} = \left[\begin{array}{ccc}5&7&3\end{array}\right] \,\bullet\,\left[\begin{array}{c}7\\2\\2\end{array}\right][/tex]

5 · 7 + 7 · 2 + 3 · 2

55

[tex]c_{(1, 3)} = \left[\begin{array}{ccc}5&7&3\end{array}\right] \,\bullet\,\left[\begin{array}{c}- 3\\1\\- 4\end{array}\right][/tex]

5 · (- 3) + 7 · 1 + 3 · (- 4)

- 20

[tex]c_{(2, 1)} = \left[\begin{array}{ccc}- 3&7&9\end{array}\right] \,\bullet\,\left[\begin{array}{c}1\\5\\- 1\end{array}\right][/tex]

(- 3) · 1 + 7 · 5 + 9 · (- 1)

23

[tex]c_{(2, 2)} = \left[\begin{array}{ccc}- 3&7&9\end{array}\right] \,\bullet\,\left[\begin{array}{c}7\\2\\2\end{array}\right][/tex]

(- 3) · 7 + 7 · 2 + 9 · 2

11

[tex]c_{(2, 3)} = \left[\begin{array}{ccc}- 3&7&9\end{array}\right] \,\bullet\,\left[\begin{array}{c}- 3\\1\\- 4\end{array}\right][/tex]

(- 3) · (- 3) + 7 · 1 + 9 · (- 4)

- 20

[tex]c_{(3, 1)} = \left[\begin{array}{ccc}2&5&2\end{array}\right] \,\bullet\,\left[\begin{array}{c}1\\5\\- 1\end{array}\right][/tex]

2 · 1 + 5 · 5 + 2 · (- 1)

25

[tex]c_{(3, 2)} = \left[\begin{array}{ccc}2&5&2\end{array}\right] \,\bullet\,\left[\begin{array}{c}7\\2\\2\end{array}\right][/tex]

2 · 7 + 5 · 2 + 2 · 2

28

[tex]c_{(3, 3)} = \left[\begin{array}{ccc}2&5&2\end{array}\right] \,\bullet\,\left[\begin{array}{c}- 3\\1\\- 4\end{array}\right][/tex]

2 · (- 3) + 5 · 1 + 2 · (- 4)

- 9

The matrix [tex]\vec C[/tex] resulting from the multiplication between matrices [tex]\vec B[/tex] and [tex]\vec A[/tex] is equal to the matrix [tex]\left[\begin{array}{ccc}37&55&- 20\\23&11&- 20\\25&28&-9\end{array}\right][/tex]. (Correct choice: C)

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Using the monthly payment formula, he needs to deposit A = $1,342.12 a month to reach his goal.

It is given by:

[tex]A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}[/tex]

In which:

For this problem, the parameters are:

P = 250000, r = 0.05, n = 12 x 30 = 360.

Then:

r/12 = 0.05/12 = 0.004167.

Then:

[tex]A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}[/tex]

[tex]A = 250000\frac{0.004167(1+0.004167)^{360}}{(1+0.004167)^{360} - 1}[/tex]

A = $1,342.12

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Answer:

log(a^x^2-27(radicalx))

Step-by-step explanation:

Assuming you have written your question correctly, this is it.

The function is illustrated below based on the information.

When x <= 8000

The cost remains constant at 0.35 when x increases from 0 to 8000. The slope of the cost function over this part is 0

When 8000 < x <= 20000

The cost remains constant at 0.75 when x increases from 8000 to 20000 and the slope of the cost function over this part is 0.

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