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ss7ja [257]
3 years ago
6

Write the equation of the line, in SLOPE-INTERCEPT FORM, that has a slope of -2 and passes through the point (20, 40)

Mathematics
1 answer:
solmaris [256]3 years ago
5 0

Answer:

y=-2x+80

Step-by-step explanation:

40=(-2*20)+b

40=-40+b

80=b

y=(-2*20)+80

y=-40+80

y=40

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Find the value of x..helppp
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For the diagram to be a parallelogram, then the measure of x is 4

Step-by-step explanation:

If the diagram is a parallelogram, then the diagonals bisect each other

What this mean is that they divide each other into 2

Mathematically, this means that the values AC is divided into two equal parts AE and CE

That means;

5x + 28 = 3x + 36

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8 0
3 years ago
Nina and her friends have $40 to
DochEvi [55]

Answer:

2s+5p=40

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6 0
3 years ago
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
Zina [86]

Answer:

Step-by-step explanation:

The given differential equation is:

x^3y'' + 2x^2y' + 4y

the main task here is to determine the singular points of the given differential equation and Classify each singular point as regular or irregular.

So, for a regular singular point ;  x=x_o is  located at the first power in the denominator of P(x) likewise at the Q(x) in the second power of the denominator. If that is not the case, then it is termed as an irregular singular point.

Let first convert it to standard form by dividing through with x³

y'' + \dfrac{2x^2y'}{x^3} + \dfrac{4y}{x^3} =0

y'' + \dfrac{2y'}{x} + \dfrac{4y}{x^3} =0

The standard form of the differential equation is :

\dfrac{d^2y}{dy} + P(x) \dfrac{dy}{dx}+Q(x)y =0

Thus;

P(x) = \dfrac{2}{x}

Q(x) = \dfrac{4}{x^3}

The zeros of x,x^3  is 0

Therefore , the singular points of above given differential equation is 0

Classify each singular point as regular or irregular.

Let p(x) = xP(x)    and q(x) = x²Q(x)

p(x) = xP(x)

p(x) = x*\dfrac{2}{x}

p(x) = 2

q(x) = x²Q(x)

q(x) = x^2 * \dfrac{4}{x^3}

q(x) =\dfrac{4}{x}

The function (f) is analytic if at a given point a it is represented by power series in x-a either with a positive or infinite radius of convergence.

Thus ; from above; we can say that q(x) is not analytic  at x = 0

Q(x) = \dfrac{4}{x^3}  do not satisfy the condition,at most to the second power in the denominator of Q(x).

Thus, the point x =0 is an irregular singular point

6 0
3 years ago
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