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vazorg [7]
3 years ago
15

Multiply and simplify

%20" id="TexFormula1" title=" \frac{2}{10} \times \frac{40}{150} " alt=" \frac{2}{10} \times \frac{40}{150} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Vitek1552 [10]3 years ago
3 0

\frac{2}{10}  \times \frac{40}{150}  \\  =  \frac{2 \div 2}{10 \div 2}  \times  \frac{40 \div 10}{150 \div 10}  \\  =  \frac{1}{5}  \times  \frac{4}{15}  \\  =  \frac{4}{75}
The answer is 4/75
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(x² + x - 17) ÷ (x - 4)
ludmilkaskok [199]
Answer:
(X^2 + X-17) /x-4
5 0
3 years ago
Please help ill give brainlist
nordsb [41]

Answer:

c. m∠2 = m∠5

Step-by-step explanation:

We can check 1 by 1 and find the false one.

For answer A, measures 2 and 4 are vertical angles, which means they are equal, so that's not false.

For answer b, measures 4 and 8 are corresponding angles, since X and Y are parallel, so that's also not false.

For answer c, measures 2 and 5 are supplementary to each other, and since the transversal is not perpendicular to X nor Y, they are not equal, therefore, this is false, making that the answer.

For answer d, measure 2 and 6 are corresponding angles, so that's also not false.

Therefore, the answer must be c.

7 0
3 years ago
Find an equation of the line through (3,7) and parallel to y=3x-1
iren2701 [21]

Answer:

y = 3x -2

Step-by-step explanation:

If the line is parallel, that means it has the same slope. The given equation had a slope of 3, so the new line must also have a slope of 3.

You can plug the given coordinates into the slope-intercept form with a slope of 3 to find your answer.

y = m*x + b

7 = 3*3 + b

7 = 9 + b

-2 = b

y = 3x -2

7 0
3 years ago
A.) Find the product (-1.15) x 3.2 b.) Explain in words HOW you determined the sign of your answer.
Deffense [45]

(A)

we are given

(-1.15) x 3.2

Since, sign of 1.15 is negative

and sign of 3.2 is positive

so, we put negative sign in front

(-1.15)*(3.2)=-(1.15*3.2)

now, we can multiply them

and we get

(-1.15)*(3.2)=-3.68.............Answer

(B)

Suppose, we are given product of two numbers 'a' and 'b'

so, we use this property

case-1: When both positive , then we put positive in front

a*b=+(a*b)

case-2: When both negative  , then we put positive in front

-a*-b=+(a*b)

case-3: When first is  negative and second is positive   , then we put negative  in front

-a*b=-(a*b)


case-4: When first is  positive and second is negative    , then we put negative  in front

a*-b=-(a*b)


7 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
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