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Romashka [77]
4 years ago
10

Solve for x. z = 6 π x y z/6πy =x z/6π =x x = 6 π y z z/6 =x

Mathematics
1 answer:
IceJOKER [234]4 years ago
6 0
The answer should be X = Z / 6piy
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Please help its a quiz
Aleks04 [339]

Answer:

C.

Step-by-step explanation:

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3 years ago
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WHAT THREE EQUAL NUMBERS COULD YOU ADD TO GET 180
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60 60 and 60. Hope I helped!
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write The equation of the line that passes through the point parentheses (6 , 4) and is parallel to the equation 2X plus 3Y equa
aliina [53]

Answer:

2x + 3y - 24 = 0

Step-by-step explanation:

2x + 3y = 18

3y = -2x + 18

y = \frac{-2}{3}x + 6

m = -2/3

y - 4 = -2/3(x - 6)

3y - 12 = -2x + 12

2x + 3y - 24 = 0

6 0
3 years ago
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; (x^2-4)^2y'' + (x + 2)y' + 7y = 0

The above equation can be better expressed as:

y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

p(x) = \dfrac{(x+2)}{(x^2-4)^2} \

p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \

p(x) = \dfrac{1}{(x+2)(x-2)^2}

Also;

q(x) = \dfrac{7}{(x^2-4)^2}

q(x) = \dfrac{7}{(x+2)^2(x-2)^2}

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

\lim \limits_{x \to-2} (x+ 2) p(x) =  \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{1}{(x-2)^2}

\implies \dfrac{1}{16}

\lim \limits_{x \to-2} (x+ 2)^2 q(x) =  \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{7}{(x-2)^2}

\implies \dfrac{7}{16}

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0
3 years ago
How to find the volume of L blocks?
choli [55]
 <span>Volume = L * W * H cubic units 

Surface Area = 2 * (LW + LH + WH) square units</span>
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4 years ago
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