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Andrej [43]
3 years ago
8

What partial products are shown by the model below

Mathematics
2 answers:
alina1380 [7]3 years ago
6 0
16x34=544 rats the answer hope this would help
Lina20 [59]3 years ago
6 0

Answer:

300, 40, 180, 24

Step-by-step explanation:

This method of multiplication is called the Grid method multiplication where the larger numbers of a multiplication operation are split into a form which is easier to multiply. The split numbers are arranged in a grid and multiplied as shown below:

Here the two numbers are 16 and 34 which are split by (10+6) and (30+4)

We multiply them

(10+6)×(30+4)

=10×30+10×4+6×30+6×4

=300+40+180+24

=544

The partial products are 300, 40, 180, 24

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Since this is in point-slope form, you know one of the points, which is (1,7), and you know the slope, which is -3. So you can plot (1,7) on the graph and then use the slope to plot more points by going down 3 and right 1, or up 3 and left 1.
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3 years ago
The cost for 15 cases of oranges is $75. What is the cost for 1 case? *
postnew [5]

Answer:Ok I This question can be solved using a system of equation, which requires translating the above word problem into something that can be algebraically

Let O=oranges and  Let G=Grapefruit

5O +8G=235

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To solve, you must isolate one of the variables and then work from there. To do that multiply both sides of the bottom or second equation by four and then subtract the bottom equation from the top.

Which would look like this

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Then subtract the bottom from the top to get this

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<h3>Hope this helps have a awesome night/day❤️✨</h3>

Step-by-step explanation:

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Step-by-step explanation:

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Find a set of vectors {u⃗ ,v⃗ }{u→,v→} in r4r4 that spans the solution set of the equations {w−x+2y+4z5w+2x−y+3z==0,0. {w−x+2y+4
rusak2 [61]
Given

w-x+2y+4z=0 \\ 5w+2x-y+3z=0

We can rewrite it in matrix form as:

\left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right]   \left[\begin{array}{c}w\\x\\y\\z\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ -5R_1+R_2\rightarrow R_2
\left[\begin{array}{cccc}1&-1&2&4\\0&7&-11&-17\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \  \frac{1}{7} R_2\rightarrow R_2 \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ R_1+R_2\rightarrow R_1
\\  \\ \left[\begin{array}{cccc}1&0&\frac{3}{7}&\frac{11}{7}\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \Rightarrow w= -\frac{3}{7} y- \frac{11}{7} z \\ x=\frac{11}{7} y+ \frac{17}{7} z \\ y=free \\ z=free \\  \\ =y\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ +z\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \

Thus, the solution set is a span of \{\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ ,\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \ \}
7 0
3 years ago
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