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Nataly [62]
3 years ago
14

Alisa's family planted 7 trees in their yard. The

Mathematics
1 answer:
stealth61 [152]3 years ago
4 0

Answer:

Neither of Alisa's guesses were correct.

Step-by-step explanation:

7 × 11 = 77

7 × 31 = 217

To find how many times more trees the park has, you divide 147 by 7 and you get 21.

Hope this helps!

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Find all values of k so that each trinomial <br> 2 x^2+ kx+12
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K can be 10,11,14,25

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(2x+4)(X+3)
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The measure of a side of a square is x units. A new square is formed with each side 6 units longer than the original squares sid
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We know that
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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
Help with these questions I´ll CASHAPP YOU 55$$
miskamm [114]

Answer:

umm ok srry this is confusing

Step-by-step explanation:

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