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2 years ago

Brainiest to whoever right

Mathematics

1 answer:

Anna35 [415]2 years ago

6 0

Answer:

since M lies between point A and B , we came to know that,

M(4,6) =(x,y)

A(-2,-1)=(x1 ,y1)

B( _, _ )=(x2,y2)

Now using mid point formula,

x=x1×x2÷2 y=y1+y2÷2

so, point B is (10,13)

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WILL MAKE BRAINLIEST!!! what is the slope of each line?

Scorpion4ik [409]

The correct answer is d, 2

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2 years ago

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Help me with answering this question

Ksju [112]

Answer:

x-intercept is (-4,0) and the y-intercept is (0,-2

Step-by-step explanation:

There are 2 ways to find the y-intercept. One way is to put the equation in slope-intercept form. However, there is a more efficient way as well.

Another way to do it is to plug 0 in for x and y. The x-intercept will have the coordinate of (x,0). So to find the x-intercept you can plug in 0 for y and solve for x. To find the y-intercept do the opposite and plug 0 in for x.

So, first, plug 0 in for y

x+2(0)=-4
x+0=-4
x=-4

Then plug in 0 for x

0+2y=-4
2y=-4
y=-2

Therefore, the x-intercept is (-4,0) and the y-intercept is (0,-2).

6 0

3 years ago

One angle of a triangle is 50 degrees greater than the smallest angle, and the third angle is 10 degrees less than twice the sm

tia_tia [17]

All 3 angles of a triangle must equal to 180. You can simply set up an equation to solve this and set it equal to 180.

Angle 1 - 50 + x
Angle 2 - x
Angle 3 - 2x -10

50+x+x+2x-10=180
4x+40=180
4x=140
x=35

Angle 1 = 50+35= 85
Angle 2 = 35
Angle 3 = 2(35) - 10 = 60

60+35+85=180

6 0

3 years ago

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Guys i need your help!! what is 600x300

PtichkaEL [24]

Answer: 180000

6 multiplied by 3 is 18 and just add 4 zeros because they’re the total zeros present in the question

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3 years ago

Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.)

olya-2409 [2.1K]

You're looking for a solution of the form

$\displaystyle y = \sum_{n=0}^\infty a_n x^n$

Differentiating twice yields

$\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n$

$\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n$

Substitute these series into the DE:

$\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0$

$\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0$

$\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0$

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

$\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0$

which indicates that the coefficients in the series solution are governed by the recurrence,

$\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}$

Use the recurrence to get the first few coefficients:

$\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}$

You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the n = 1 position being the odd one out. So we have

$\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots$

which looks a lot like the power series expansion for -7eˣ.

Fortunately, we can rewrite the linear term as

3x = 10x - 7x = 10x - 7/1! x

and in doing so, we can condense this solution to

$\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}$

Just to confirm this solution is valid: we have

y = 10x - 7eˣ ==> y (0) = 0 - 7 = -7

y' = 10 - 7eˣ ==> y' (0) = 10 - 7 = 3

y'' = -7eˣ

and substituting into the DE gives

-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0

as required.

8 0

3 years ago