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Kobotan [32]
2 years ago
8

Find the surface area of the cylinder

Mathematics
2 answers:
pishuonlain [190]2 years ago
5 0

Answer:

V=pr²t

=3,14 ×8×8×3

=602.88

Step-by-step explanation:

IVA

faltersainse [42]2 years ago
3 0

Answer: surface area

A = 56 π cm^2

Step-by-step explanation:

A = 56 π cm^2

you can stop here and write our answer as 56π Or we can plug in 3.14 and multiply. 56*3.14 = 175.84cm^2

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A RSS Clothing manufacturer makes two types of jogging pants, design A and design B. The design A jogging pants sells to the ret
nikdorinn [45]

Answer:

  120 of design B

Step-by-step explanation:

The profit per dollar cost for design A is (2500/1750 -1) ≈ 0.43. For design B, it is (2100/1200 -1) ≈ 0.75. Design B is more profitable and less expensive to make, so its production should be maximized.

For maximum profit, 120 jogging pants of design B should be made each week.

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The budget of 150,000 would allow 150,000/1,200 = 125 pants of design B to be made. The production limit of 120 pants does not use the entire budget.

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3 years ago
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Montano1993 [528]

11 times 12 is 132

Becuas she is spending her money it would be negative

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3 years ago
Witch has a greater absolute value 33 dollars and -52 dollars
lakkis [162]

Answer:

-52

Step-by-step explanation:

the absolute value is always positive and so 33 would remain 33, but -52 would change to 52 which is greater than 33.

6 0
3 years ago
Read 2 more answers
16x4<br> what is the answer?
Viefleur [7K]

Answer:

64

Step-by-step explanation:

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3 years ago
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Determine if the columns of the matrix form a linearly independent set. Justify your answer. [Start 3 By 4 Matrix 1st Row 1st Co
Volgvan

Answer:

Linearly Dependent for not all scalars are null.

Step-by-step explanation:

Hi there!

1)When we have vectors like v_{1},v_{2},v_{3}, ... we call them linearly dependent if we have scalars a_{1},a_{2},a_{3},... as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+a_{3}\vec{v_{3}}+...a_{m}\vec{v_{m}}=0  

When all scalar coefficients are equal to zero, we can call them linearly independent

2)  Now let's examine the Matrix given:

\begin{bmatrix}1 &-2  &2  &3 \\ -2 & 4 & -4 &3 \\ 0&1  &-1  & 4\end{bmatrix}

So each column of this Matrix is a vector. So we can write them as:

\vec{v_{1}}=\left \langle 1,-2,1 \right \rangle,\vec{v_{2}}=\left \langle -2,4,-1 \right \rangle,\vec{v_{3}}=\left \langle 2,-4,4 \right \rangle\vec{v_{4}}=\left \langle 3,3,4 \right \rangle Or

Now let's rewrite it as a system of equations:

a_{1}\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_{2}\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_{3}\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_{4}\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.

 \left ( \left.\begin{matrix}1 &-2  &2  &3 \\ -2 &4  &-4  &3 \\ 0 & 1 &-1  &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )R_{1}\times2 +R_{2}\rightarrow R_{2}\left ( \left.\begin{matrix}1 &-2  &2  &3 \\ 0 &0 &9  &0\\ 0 & 1 &-1  &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\ R_{2}\Leftrightarrow  R_{3}\left ( \left.\begin{matrix}1 &-2  &2  &3 \\ 0 &1  &-1  &4 \\ 0 &0 &9  &0 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\left\{\begin{matrix}1a_{1} &-2a_{2}  &+2a_{3}  &+3a_{4}  &=0 \\  &1a_{2}  &-1a_{3} &+4a_{4}  &=0 \\  &  &  &9a_{4}  &=0 \end{matrix}\right.\Rightarrow a_{1}=0, a_{2}=a_{3},a_{4}=0

S=\begin{bmatrix}0\\ a_{3}\\ a_{3}\\ 0\end{bmatrix}

3 0
3 years ago
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