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3 years ago

Choose the expression that represents “divide 0.04 by n.”

Mathematics

1 answer:

zheka24 [161]3 years ago

6 0

Answer:

.04/n=4/100n or 4÷100n

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What will you get when you add two integers

dybincka [34]

Well, it depends if its positive or negative. 2 positive integers will equal a positive, 2 negatives will equal a negative. A negative and a positive will depend.

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3 years ago

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Twice the sum of the number and 10 is 60 what is the number?

n200080 [17]

Let the number be n

$\\ \sf\longmapsto 2(n+10)=60$

$\\ \sf\longmapsto n+10=60/2$

$\\ \sf\longmapsto n+10=30$

$\\ \sf\longmapsto n=30-10$

$\\ \sf\longmapsto n=20$

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3 years ago

Factor this polynomial completely x2 + 3x – 18 A. (x-2)(x + 9) B. (x+3)(x-6) оо C. (x - 3)(x+6) D. (x - 3)(x-6)

Umnica [9.8K]

Answer:

C. $(x-3)(x+6)$

Step-by-step explanation:

I think the equation meant $x^2+3x-18$ ?

Anyways, to factor these kinds of quadratic, keep into consideration:

$ax^2+bx+c=(x+w)(x+v)$

ONLY if:

$w+v=b\\wv=c$

Start off by finding factors of c, which in this case, -18:

±(1, 2, 3, 6, 9, 18)

If one of the numbers is negative then the other number must be positive.

Find which two factors will sum up to b, which in this case, is 3.

$1+(-18)\neq 3\\2+(-9)\neq 3\\3+(-6)\neq 3\\6+(-3)=3\\9+(-2)\neq 3\\18+(-1)\neq 3\\$

The only two factors that work are 6 and -3.

Replace them into the factored form:

$(x+w)(x+v)\\(x+6)(x-3)\\(x-3)(x+6)$

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2 years ago

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Answer:

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a line segment is M(-3,2)one end point of the segment isP(1,-3)find the coordinates of the end point

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2 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.

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

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

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3 years ago