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2 years ago

Which theorem is it? Answer choice below in the picture:

Mathematics

1 answer:

Ymorist [56]2 years ago

5 0

Answer:

None of the above

Step-by-step explanation:

The figure3 shows 1 side and 1 angle of one triangle congruent to 1 side and 1 angle of the other triangle. We are not shown anything else, so the answer is:

None of the above

and you can't prove the triangles congruent.

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What is the value of 5 in 756 write and draw to explain how you know?

kykrilka [37]

In the attached diagram you can see Place Value Chart that is used to help understand the value of each digit based on the place or position.

For the number 756:

7 has a value of 7 hundreds, or 700;
5 has a value of 5 tens, or 50;
6 has a value of 6 ones, or 6.

Answer: 5 has a value of 5 tens, or 50

8 0

3 years ago

Read 2 more answers

The graph of y = –3x + 4 is:

MrRa [10]

Answer:

Step-by-step explanation:

It's a line with many solutions like

When x=1 y=1

X=2 y=-2

And so on all the pts on the line

8 0

2 years ago

B(h+w)=4h(11-j), solve for w

frozen [14]

Answer:

W=4h(11-j)/b-h

Step-by-step explanation:

Eliminate variables on the left side first.

b(h+w) = 4h (11-j) - so, divide b by both sides

(h+w)/b=4h(11-j)/b then, eliminate h by subtracting on both sides.

(h+w)-h=4h(11-j)/b - h (h is eliminated on the left side and now you are left with

W=4h(11-j)/b - h

4 0

3 years ago

53 pts! which of the following is equal to g(x)?

ziro4ka [17]

Answer:

B is the correct answer, as x - 2 in the exponent would move the graph to the right 2.

7 0

3 years ago

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

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

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

4 0

3 years ago