White the number 876,543,210.25 in expanded notation

1 answer:

4 0

<span>We need to write the expanded notation of the given number.
The given number is => 876 543 210 .25
Expanded notation is an expression where you give the expanded form of a certain number.
=> 876 543 210 .25 = 800 000 000 + 70 000 000 + 6 000 000 + 500 000 + 40 000 + 3 000 + 200 + 10 + .20 + .05
Notice that we show the expanded form by adding each value of given number.</span>

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Can someone help me with these I don’t understand at all

cupoosta [38]

Answer:

#1: ????

#2: $\frac{3}{4}$ x $15$

#3: $\frac{7}{8}$ ÷ $\frac{3}{4}$

Step-by-step explanation:

For #2:

$\frac{3}{4}$ x $15$ .

It's that because there are 3/4 cups of honey for one cake. To use for 15 cakes, you would have to multiply it by 15.

For #3:

$\frac{7}{8}$ ÷ $\frac{3}{4}$ .

It's that because the baker has 7/8 cups of honey. It takes 3/4 cups of honey for one cake. So you have to divide it by 3/4 cups in order to find out how many cakes.

Hope this helped! <3

7 0

3 years ago

Melissa puts 2/5 of her babysitting money in savings and uses 1/2 of the remaining money to go to the movies. If she has $12 lef

ser-zykov [4K]

If you work backwards, she had 24 before going to the movies. This is 3/5 of what she had originally. Use this to find 1/5 the but dividing 24 by 4 = 8 then multiply by 5 to get back to the original value 8 x 5 = 40

8 0

3 years ago

Read 2 more answers

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.

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