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3 years ago

Step by step explanation please!!

Mathematics

1 answer:

elixir [45]3 years ago

3 0

Answer:

what do you mean on the question?

Step-by-step explanation:

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Prove that using: X ∩ (Y ∪ Z ) =(X∩Y) ∪ (Y ∩ Z )

sertanlavr [38]

There should be a typo in the question: the correct formula would be

$X\cap(Y\cup Z)=(X\cap Y)\cup (X\cap Z)$

Let's compute $X\cap (Y\cup Z)$ first. We have

$(Y\cup Z)=\{1, 4, 5, 6, 7, 8, 9\}$

If we intersect this set with X, we have

$X\cap (Y\cup Z) = \{5, 8, 9\}$

So, that would be the left hand side. For the right hand side, we have

$X\cap Y = \{8\},\quad X\cap Z = \{5,9\}$

And their union is again $\{5,8,9\}$

5 0

3 years ago

Solve the equation -4 + 4a = 12 for a.

WARRIOR [948]

Hello there! The answer to your question is a = 4.

To solve for a in the equation -4 + 4a = 12 you want to isolate, or separate, a from all of the equation's other values.

Our first step is to add 4 to both sides so the 4 cancels out on the left side.

-4+4 + 4a = 12+4

4a = 16

Next, we divide both sides by 4 to finish isolating a.

4/4a = 16/4

a = 4

Want to verify that this is correct? Place the value we found for a (4) into the equation in place of a and if it comes out as a true statement, the answer is correct.

-4 + 4(4) = 12

-4 + 16 = 12

12 = 12

Since 12 is in fact equal to 12, i can guarantee you that i have provided you with the correct answer. I hope this helps & have a great rest of your day! :)

6 0

3 years ago

Read 2 more answers

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

<u>Answer-</u>

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$