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3 years ago

A(-3,2) B(5,-4) midpoint is (1,-1)

1 answer:

7 0

$CHECK:\\\\the\ midpoint\ AB:\left(\frac{-3+5}{2};\ \frac{2-4}{2}\right)=\left(\frac{2}{2};\ \frac{-2}{2}\right)=(1;-1)\\-------------------------------\\C(x;\ y)\\\\the\ midpoint\ AC:\left(\frac{-3+x}{2};\ \frac{2+y}{2}\right)\\\\B\ is\ the\ midpoint\ AC\ then:\\\\\frac{-3+x}{2}=5\ and\ \frac{2+y}{2}=-4\\\\-3+x=10\ and\ 2+y=-8\\\\x=10+3\ and\ y=-8-2\\\\x=13\ and\ y=-10\\\\Solution:C(13;-10).$

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What will be the linear equation for the shown table?<br><br>

vladimir1956 [14]

Answer:

y=2x+5

Step-by-step explanation:

To find a linear equation you must find the slope and y intercept to create an equation in y=mx+b form (slope-intercept form)

So slope is = y2-y1 / x2-x1 [2 and 1 are subscripts] so 11-7/3-1 so m=4/2 = 2

Slope = 2, now find b

So using y = mx + b, substitute what you know so 7 = 2(1) + b so b=7-2 = 5

Put it all together y = 2x + 5

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2 years ago

2.2 Your friend Sam, mistakenly wrote down the multiples of 10 as: 1; 2; 5; and 10.

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2 years ago

Simplify: 2/5y−4+7−9/10y

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3 years ago

Read 2 more answers

Find x if 2^x=8^3x-8

Anon25 [30]

Step-by-step explanation:

First we have to same the bases.

As we know 8=2^3.

Replace 8 with 2^3.

Like this .

2^3x=2^3(2x+1).

Now 3x=3(2x+1) distribute 3 by the Barker.

3x=6x+3.

6x+3=3x.

6x-3x=-3.

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2 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

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3 years ago