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3 years ago

The given line passes through the points (-4,-3) and (4,

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

Step-by-step explanation:

first we gotta find the slope of the first line

(-4,-3),(4,1)

slope = (y2 - y1) / (x2 - x1) = (1- (-3) / (4 - (-4) = (1 + 3) / (4 + 4) = 4/8 = 1/2

so the slope is 1/2.....so we are looking for a line that is perpendicular....perpendicular lines have negative reciprocal slopes...all that means is flip the slope and change the sign....so the slope we need is :

1/2....flip it....2/1.....change the sign....-2.....we need a -2 slope

y - y1 = m(x - x1)

slope = -2

(-4,3)...x1 = -4 and y1 = 3

now sub

y - 3 = -2(x - (-4) =

y - 3 = -2(x + 4) <=====

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Selling Price = $70 Rate of Sales Tax = 6% What is the sales tax? What is the total price?

Hatshy [7]

Answer:

Tax = $4.2, Total price = $74.2

Step-by-step explanation:

Multiply 70 by 0.06 to get the price of the tax

4.2

Add 70 and 4.2 to get the total price

74.2

3 0

2 years ago

Power Series Differential equation

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$

4 0

3 years ago

Adam is comparing the graphs of y=4x and y=8x. Which of the following statements is TRUE?

Inga [223]

The first one is true your welcome hope thi helps

5 0

2 years ago

Taryn conducted a science experiment on saturation. She added sugar to sugar water solution at different intervals. The graph sh

dusya [7]

Equation of a line:
$y=mx+c$

m = gradient: The difference between two y points and two x points.
$m= \frac{\Delta y}{\Delta x}$

c = y-intercept: Where the line crosses the y-axis (x=0)

You have:
$y=\textunderscore\textunderscore \ x + \textunderscore$
so you are missing the m and the c.

To calculate m find two y coordinates -you have (12, 7) and (0, 1)- and subtract them. Then divide this by the subtracted values of the x coordinates -you have (12, 7) and (0, 1)- This gives:

$m= \frac{7-1}{12-0}$
$m= \frac{6}{12}$
$m=0.5$

To calculate the c, you just see where the line crosses the y-axis. Because you have the point (0, 1), you know that when x=0, y=1. Because x=0 is on the y-axis, you can tell that the line passes through y=1. This makes your c = 1:

$c=1$

When you plug these values into the equation you get your answer:

$y=0.5x+1$

6 0

3 years ago

PLEASE HELP ME ?!!!!!

cluponka [151]

Answer:

b & c for sure

Step-by-step explanation:

............

8 0

3 years ago