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Ganezh [65]
3 years ago
6

Which lines is a parallel to the line y=1/2x+5 and passes through the point (-2,1)?

Mathematics
1 answer:
Archy [21]3 years ago
5 0

Answer:

The answer is: y = 1/2x + 2

Step-by-step explanation:

Given equation: y = 1/2x + 5

Given point: (-2, 1)

The slope of the given line is 1/2. Parallel lines have the same slope.

Use the point slope form and solve for y:

y - y1 = m(x - x1)

y - 1 = 1/2(x - (-2))

y - 1 = 1/2(x + 2)

y - 1 = 1/2x + 1/2 * 2

y - 1 = 1/2x + 1

y = 1/2x + 2

Proof:

f(-2) = 1/2(-2) + 2

= -1 + 2

= 1, giving point (-2, 1)

Hope this helps!! Have an Awesome Day!! :-)

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Serggg [28]
I think it should be option d
3 0
3 years ago
If g(x)=3x-11, what is x is g(x)=289
Liula [17]

Answer:x=100

Step-by-step explanation:

Since you know that g(x)=3x-11 all you have to do is substitute that into g(x)=289. So you get 3x-11=289. To get x you have to add 11 on both sides.

3x-11=289

+11 +11

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

3x = 300

Divide 3 on both sides to get x by itself

3x =300

---- -------

3 3

x=100

8 0
3 years ago
Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions
monitta
I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take z=y', so that z'=y'' and we're left with the ODE linear in z:

y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2

Now suppose y has a power series expansion

y=\displaystyle\sum_{n\ge0}a_nx^n
\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Then the ODE can be written as

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0

\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0

All the coefficients of the series vanish, and setting x=0 in the power series forms for y and y' tell us that y(0)=a_0 and y'(0)=a_1, so we get the recurrence

\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}

We can solve explicitly for a_n quite easily:

a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}

and so on. Continuing in this way we end up with

a_n=\dfrac{a_1}{n!}

so that the solution to the ODE is

y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x

We also require the solution to satisfy y(0)=a_0, which we can do easily by adding and subtracting a constant as needed:

y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}
4 0
3 years ago
Help me please ^^ much appreciated
Reil [10]

Answer:

C

Step-by-step explanation:

Using the sine ratio in the right triangle.

sinC = \frac{opposite}{hypotenuse} = \frac{AB}{BC} = \frac{16}{28} , thus

∠ C = sin^{-1} (\frac{16}{28} ) ≈ 34.85° ( to 2 dec. places )

8 0
2 years ago
ed ReviewA company makes bars of soap.soapThe is mixed in large vats, then poured intomolds to make 560 bars of soap. Themixture
inysia [295]
Given:
weight of mixture : 1792 ounces
output of mixture : 560 bars of soap

to get the weight per unit, we simply divide by total weight by the total number of output.

1792 oz / 560 bars = 3.2 ounces per bar.
6 0
3 years ago
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