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CaHeK987 [17]
3 years ago
11

Hannah buys ice cream and onions at the store

Mathematics
1 answer:
Snezhnost [94]3 years ago
3 0

Answer:

x=5.12

Step-by-step explanation:

35.16= 4.44+6x

30.72=6x

5.12=x

x=5.12

Hope this helps :)

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Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4
Alik [6]

Answer:

\alpha=3

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

(x-h)^2+(y-k)^2=r^2

We are given the equation of a circle as

3x^2+3y^2-6\alpha x+12y-3\alpha=0

Note we have corrected it by adding the square to the y. Simplify by 3

x^2+y^2-2\alpha x+4y-\alpha=0

Complete squares and rearrange:

x^2-2\alpha x+y^2+4y=\alpha

x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4

(x-\alpha)^2+(y+2)^2=r^2

We can see that, if r=4, then

\alpha+\alpha^2+4=16

Or, equivalently

\alpha^2+\alpha-12=0

There are two solutions for \alpha:

\alpha=-4,\ \alpha=3

Keeping the positive solution, as required:

\boxed{\alpha=3}

8 0
3 years ago
What is a polynomial?
Alex777 [14]

Answer:

Now, I ain't tryin' to sound mean or standoff-ish but I'm pretty sure Wikipedia could answer this question for you

3 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.

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3 years ago
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viva [34]

Answer:

This is a acute angle in the digits 1,4,6 should help the answer is acute

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Joey wants to get a 90 test average in Science. He has already had two tests and he earned
kiruha [24]

I NEED A FREE BRAINLIEST ASAP

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