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3 years ago

Write an equation in point-slope form for the line that has a slope of 4/5 and contains the point (−5, −3).

Mathematics

1 answer:

ankoles [38]3 years ago

5 0

y = mx + b, where m = slope, and b = y-intercept.

Since you are not given the y-intercept, you have to solve for it through the equation, y = (4/5)x (which is the slope you are given) + b, and replace x and y with values given, which are (-5,-3)

y = (4/5)x + b

-3 = (4/5)(-5) + b

-3 = -4 + b

1 = b

Then replace b in the first equation to get the answer

y = (4/5)x + 1

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I'm struggling in this subject so if you'd give a step-by-step explanation that'd be great thx

defon

A. First you must start by plugging in your equation solve Y = 2(5) + 1 then you must simplify by solving your 2×5+ 1 and then you end up with Y equals 11

B. Again plug in your equation Y = 2(-5) + 1 then you must simplify by solving your 2×-5+ 1 and then you end up with Y equals -10+ one which simplifies to y = -9

I hope this helps

8 0

2 years ago

Read 2 more answers

Find two power series solutions of the given differential equation about the ordinary point x = 0. y'' + xy = 0

nalin [4]

Answer:

First we write y and its derivatives as power series:

y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2

Next, plug into differential equation:

(x+2)y′′+xy′−y=0

(x+2)∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

x∑n=2∞n(n−1)anxn−2+2∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

Move constants inside of summations:

∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0

∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0

Change limits so that the exponents for x are the same in each summation:

∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0

Pull out any terms from sums, so that each sum starts at same lower limit (n=1)

∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0

Combine all sums into a single sum:

4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0

Now we must set each coefficient, including constant term =0 :

4a2−a0=0⟹4a2=a0

2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an=0

We would usually let a0 and a1 be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since a0=4a2 , I’ll choose a1 and a2 as the two arbitrary constants. We can still express all other constants in terms of a1 and/or a2 .

an+2=−(n+1)nan+1+(n−1)an2(n+2)(n+1)

a3=−(2⋅1)a2+0a12(3⋅2)=−16a2=−13!a2

a4=−(3⋅2)a3+1a22(4⋅3)=0=04!a2

a5=−(4⋅3)a4+2a32(5⋅4)=15!a2

a6=−(5⋅4)a5+3a42(6⋅5)=−26!a2

We see a pattern emerging here:

an=(−1)(n+1)n−4n!a2

This can be proven by mathematical induction. In fact, this is true for all n≥0 , except for n=1 , since a1 is an arbitrary constant independent of a0 (and therefore independent of a2 ).

Plugging back into original power series for y , we get:

y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯

y=4a2+a1x+a2x2−13!a2x3+04!a2x4+15!a2x5−⋯

y=a1x+a2(4+x2−13!x3+04!x4+15!x5−⋯)

Notice that the expression following constant a2 is =4+ a power series (starting at n=2 ). However, if we had the appropriate x -term, we would have a power series starting at n=0 . Since the other independent solution is simply y1=x, then we can let a1=c1−3c2, a2=c2 , and we get:

y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)

y=c1x+c2∑n=0∞(−1)n+1n−4n!xn

Learn more about constants here:

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6 0

1 year ago

How to factor "25x ^2 +79+64

avanturin [10]

Answer: The question might be faulty because the question is to be solve using different of two squares.

Step-by-step explanation: something might be missing, please check and resend.

3 0

3 years ago

A teacher gives a test to a large group of students. The results are closely approximated by a normal curve. Teh mean is 73, wit

zavuch27 [327]

Answer:

The bottom for an A is 75. Round to the nearest whole number as needed.

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the scores on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=73,\sigma=7)$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

2) The bottom for an A is _____. Round to the nearest whole number as needed.

And we want the top 5.5% of the scores, so we need a value a such that:

$P(X>a)=0.055$ or $P(X$

We need on the right tail of the distribution a value a that gives to us 94.5% of the area below and 5.5% of the area above. Both conditions are equivalent.

Let's use the condition $P(X$ , the best way to solve this problem is using the z score with the following formula:

$z=\frac{x-\mu}{\sigma}$

So we need a value from the normal standard distribution that accumulates 0.945 of the area on the left and 0.055 on the right. This value on this case is 1.598 and we can founded with the following code in excel:

"=NORM.INV(0.945,0,1)"

If we apply the z score formula to our case we have this:

$P(X$