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emmainna [20.7K]
3 years ago
15

Khan academy question. Please help!

Mathematics
1 answer:
Zepler [3.9K]3 years ago
7 0
It’s B
because i said so
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Use (<, >, =) to compare 2⁄3 and 3⁄4. A. None of the above B. 2⁄3 > 3⁄4 C. 2⁄3 < 3⁄4 D. 2⁄3 = 3⁄4
Vinvika [58]
C 3/4 >2/3 bc 3/4=.75 and 2/3 = .66
4 0
3 years ago
Determine the equation of the line that passes through the given points. (If you have a graphing calculator, you can use the tab
expeople1 [14]

<span>
You can write the equation in point-slope form, which has the format <em>y-y</em>subscript1=<em>m</em>(<em>x-x</em>subscript1), with <em>y</em>subscript1 and <em>x</em>subscript1 being the y and x coordinates for a point on the line, and <em>m</em> being the slope. </span>

<span /><span>Substitute a y and x coordinate into the equation so you have <em>y</em>-6=<em>m</em>(<em>x</em>-2)</span>

<span /><span><span>Then find the slope so you can replace <em>m</em>. The slope formula is <em />(<em>y</em>subscript2-<em>y</em>subscript1)/(<em>x</em>subscript2-<em>x</em>subscript1). </span><span>Substitute the coordinates in so you have <em>m</em>=(16-6)/(4-2), which simplifies to 10/2 and then 5.</span></span>

<span><span /></span><span>Now the equation is <em>y</em>-6=5(<em>x</em>-2)</span>

<span />If you want a different form, for example slope-intercept form, you can change it to that:

<span><em>y</em>-6=5(<em>x</em>-2)</span>

<span><em>y</em>=5x-4</span>

6 0
3 years ago
This is to help my wonderful mother. What is 1/3rd of 4 1/2s and 8 1/4s?
crimeas [40]
4 1/2 = 9/2
8 1/4 = 33/4

1/3 x 9/2 = 3/2
1/3 x 33/4 = 33/12
5 0
3 years ago
What is the area of the polygon below
kvasek [131]
The area is 84.

What I do it cut the polygon into something easier like a square and rectangle.

Rectangle: 6•12=72
Square: 4•3=12

Add them together and you have 84
5 0
3 years ago
Solve the system of equations by transforming a matrix representing the system of equation into reduced row echelon form.
Gekata [30.6K]

Take the augmented matrix,

\left[\begin{array}{ccc|c}2&1&-3&-20\\1&2&1&-3\\1&-1&5&19\end{array}\right]

Swap the row 1 and row 2:

\left[\begin{array}{ccc|c}1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]

Add -2(row 1) to row 2, and -1(row 1) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]

Multiply through row 3 by 1/9:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]

Add 5(row 3) to row 2:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]

Multiply through row 2 by -1/3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]

Add -2(row 2) and -1(row 3) to row 1:

\left[\begin{array}{ccc|c}1&0&0&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]

So we have \boxed{x=-3,y=-2,z=4}.

3 0
3 years ago
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