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velikii [3]
2 years ago
13

7B2%7D%20" id="TexFormula1" title="3 \frac{1}{2} + 2 \frac{1}{4} - 1 \frac{1}{2} " alt="3 \frac{1}{2} + 2 \frac{1}{4} - 1 \frac{1}{2} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Svetach [21]2 years ago
5 0

Answer:

Step-by-step explanation:

Step 1: Convert into improper fractions

Step 2: Find  least common denominator and find equivalent fractions

Step 3 : Do the required operation

3\frac{1}{2}+2\frac{1}{4}-1\frac{1}{2}=\frac{7}{2}+\frac{9}{4}-\frac{3}{2}\\

Least common denominator of 2 , 4 = 4

   =\frac{7*2}{2*2}+\frac{9}{4}-\frac{3*2}{2*2}\\\\=\frac{14}{4}+\frac{9}{4}-\frac{6}{4}\\\\=\frac{14+9-6}{4}\\\\=\frac{17}{4}\\\\=4\frac{1}{4}

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Answer:

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Step-by-step explanation:

The point of interest is ...

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  P = (4, 7)

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The point that divides the segment into the ratio a:b is the weighted average of the endpoints, with the weights being "b" and "a". The weight of the first end point corresponds to the length of the far end of the segment.

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(\sqrt{3x}+\sqrt5)\cdot(\sqrt{15x}+2\sqrt{30})\\\\=(\sqrt{3x})(\sqrt{15x})+(\sqrt{3x})(2\sqrt{30})+(\sqrt5)(\sqrt{15x})+(\sqrt5)(2\sqrt{30})\\\\=\sqrt{(3x)(15x)}+2\sqrt{(3x)(30)}+\sqrt{(5)(15x)}+2\sqrt{(5)(30)}\\\\=\sqrt{45x^2}+2\sqrt{90x}+\sqrt{75x}+2\sqrt{150}\\\\=\sqrt{9x^2\cdot5}+2\sqrt{9\cdot10x}+\sqrt{25\cdot3x}+2\sqrt{25\cdot6}\\\\=\sqrt{9x^2}\cdot\sqrt5+2\sqrt9\cdot\sqrt{10x}+\sqrt{25}\cdot\sqrt{3x}+2\sqrt{25}\cdot\sqrt6
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Alex787 [66]

Answer:

y=-\frac{2}{5}x-\frac{1}{5}

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

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  • b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

m = -\frac{2}{5}

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-\frac{2}{5}x + b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

  • (-3,1). When x of the line is -3, y of the line must be 1.
  • (2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: y=-\frac{2}{5}x + b. b is what we want, the --\frac{2}{5} is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

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See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points  (-3,1) and (2,-1) is y=-\frac{2}{5}x-\frac{1}{5}

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