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Semmy [17]
3 years ago
8

Build an equation with at least three different operations that has solution x=2

Mathematics
1 answer:
spayn [35]3 years ago
8 0

Answer: (10x+7)/2 = -13/2

This is not the only answer possible. There are infinitely many possible answers.

=================================

Explanation:

We start with the given equation x = -2

Multiply both sides by some number. I'll pick 10

So if we multiply both sides by 10, we get

10*x = 10*(-2)

10x = -20

Now let's add something to both sides. Let's say we add 7 to both sides

10x+7 = -20+7

10x+7 = -13

Finally, let's divide both sides by 2

(10x+7)/2 = -13/2

You could keep going to add, subtract, multiply, or divide both sides by some number. This process can continue indefinitely. I'll stop here since we have three operations done.

-------------------

If we were to solve (10x+7)/2 = -13/2, we would get to x = -2

To solve (10x+7)/2 = -13/2, we follow the steps done in the previous section in reverse.

So we'll undo the "divide both sides by 2" operation by multiplying both sides by 2.

Then we'll undo the "add 7 to both sides" operation to subtract 7 from both sides

Lastly, we'll undo the "multiply both sides by 10" by dividing both sides by 10.

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Answer:

X+ 0.5X or 1.5X

Step-by-step explanation:

the sum means that it want an addition problem. and so it is saying X + 1/2 (0.5)X which gives you 1.5X or you can leave it has X+0.5X

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4 years ago
Let c be the path c(t) = (2t, t2, log(t)), defined for t > 0. find the arc length of c between the points (2, 1, 0) and (12,
V125BC [204]
The path starts at t=1 and ends at t=6, so we have an arc length of


\displaystyle\int_{\mathcal C}\mathrm dS=\int_{t=1}^{t=6}\|\mathbf c'(t)\|\,\mathrm dt
=\displaystyle\int_1^6\sqrt{2^2+(2t)^2+\left(\frac1t\right)^2}\,\mathrm dt
=\displaystyle\int_1^6\sqrt{4t^2+4+\frac1{t^2}}\,\mathrm dt
=\displaystyle\int_1^6\frac1t\sqrt{4t^4+4t^2+1}\,\mathrm dt
=\displaystyle\int_1^6\frac1t\sqrt{(2t^2+1)^2}\,\mathrm dt
=\displaystyle\int_1^6\frac{2t^2+1}t\,\mathrm dt
=\displaystyle\int_1^6\left(2t+\frac1t\right)\,\mathrm dt
=t^2+\log t\bigg|_{t=1}^{t=6}
=(6^2+\log6)-(1^2+\log1)=35+\log6
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4 years ago
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