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djyliett [7]
3 years ago
11

Helpppppppp please, I need it fast​

Mathematics
2 answers:
nirvana33 [79]3 years ago
8 0
The answer would be 15 minutes
professor190 [17]3 years ago
3 0
The answer would be A. <3 hope this helps!
You might be interested in
F(x)=3x+2 what is f(5)
mario62 [17]

Answer:

17

Step-by-step explanation:

f(5)=3*5+2 where X is equal to 5

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=8%20%2B%203%20%5Ctimes%204" id="TexFormula1" title="8 + 3 \times 4" alt="8 + 3 \times 4" align
Ira Lisetskai [31]

Answer:

20 is correct answer

Step-by-step explanation:

8+3×4

=8+12

=20

<em>Reme</em><em>mber</em><em> </em><em>always</em><em> </em><em>foll</em><em>ow</em><em> </em><em>bodmas</em><em> </em><em>rule</em><em> </em>

hope it helped you:)

thanks!

8 0
3 years ago
Read 2 more answers
Is -40 the solution to -12=<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bu%7D%7B8%7D" id="TexFormula1" title="\frac{u}{8}" alt="\f
valentinak56 [21]
Answer: Yes -40 is the answer

Explanation:

-12 = u/8 - 7
8(-12)/8 = u/8 - 8(7)/8
-96/8 = u/8 - 56/8
-96 = u - 56
-u = -56 + 96
-u = 40
u = -40
4 0
3 years ago
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
What is the reciprocal of 5/9
Nat2105 [25]

Answer:

 \frac{9}{5}

Step-by-step explanation:

To find the reciprocal, divide 1 by the number given'

 

                                                   \frac{1}{\frac{5}{9} }

<em>                            (if you can't see it, it is 1 over 5/9)</em>

<em />

Simplify;

Multiply the numerator by the reciprocal of the denominator.

                                                    1(\frac{5}{9} )

<em>                         (if you can't see it, it is 1 × 5/9)</em>

<em />

Multiply \frac{9}{5} by 1.

                                                   \frac{9}{5}

<em />

So our answer is   \frac{9}{5}  !

We did it!

5 0
3 years ago
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