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2 years ago

Rectangle WXYZ was dilated to create W'X'Y'Z'.

Mathematics

2 answers:

Snezhnost [94]2 years ago

4 0

You did not attach any picture to solve this problem. We cannot calculate for the value W’X’ without the correct illustrations. However, I think I found the correct one (see attached), please attach it next time.

So the first thing we have to do is to calculate for the dilation factor. Taking point G as the reference point, we can see that the distance of point G from rectangle W’X’Y’Z’ is 1.5 while the distance from rectangle WXYZ is (1.5 + 7.5), therefore the dilation factor to use is:

dilation factor = 1.5 / (1.5 + 7.5) = 1.5 / 9 = 1/6

Since WX has an initial measure of 3 units, therefore the measure of W’X’ is:

W’X’ = 3 units * (1/6) = 0.5 units

Answer:

<span>0.5 units</span>

levacccp [35]2 years ago

4 0

Answer:

Answer is .5, just took the test and got it correct.

Step-by-step explanation:

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y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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2 years ago

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3 years ago

Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card an

poizon [28]

Answer:

We have that $B = 0.3$ .

$0.3 = b + (A \cap B)$

However, b is a probability, which means that it cannot be negative. So no, P(A ∩ B) cannot be 0.5. It can, at most, be 0.3.

Step-by-step explanation:

Event A:

Probability that a student has a Visa card.

Event B:

Probability that the student has a MasterCard.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student has a Visa card but not a MasterCard and $A \cap B$ is the probability that a student has both these cards.

By the same logic, we have that:

$B = b + (A \cap B)$

In this problem, we have that:

$A = 0.6, B = 0.3$

(a) Could it be the case that P(A ∩ B) = 0.5?

We have that $B = 0.3$ .

$0.3 = b + (A \cap B)$

However, b is a probability, which means that it cannot be negative. So no, P(A ∩ B) cannot be 0.5. It can, at most, be 0.3.

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3 years ago