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2 years ago

The point (6, - 5) is reflected across the y axis what is the new point

Mathematics

1 answer:

jonny [76]2 years ago

8 0

Answer:

(-6, -5)

Step-by-step explanation:

Reflection across the y-axis leaves the point on the same horizontal line, but with the sign of its x-coordinate changed.

(x, y) ⇒ (-x, y) . . . . . reflection across the y-axis

(6, -5) ⇒ (-6, -5)

The image point is (-6, -5).

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Select the correct answer.

Alex_Xolod [135]

The answer is D

Explanation:

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3 years ago

Click on the measure of angle x. х 38° » 152 1)) 142° () 38° )) 52°

GaryK [48]

Answer:

142°

Step-by-step explanation:

$180^{\circ} - 38^{\circ} = 142^{\circ}$

7 0

2 years ago

13/15 - 1/4 = Please help me

podryga [215]

13/15-1/4

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2 years ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago