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2 years ago

I NEED HELP I DONT UNDERSTAND

Mathematics

1 answer:

wel2 years ago

3 0

Answer:

i have no idea, Im very much sorry

Step-by-step explanation:

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The grocery store sells kumquats for $5.00 a pound and Asian pears for $3.25 a pound. Write an equation in standard form for the

aniked [119]

Okay so your equation is going to be : 18= 5k + 3.25p

3 0

3 years ago

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I WILL MARK BRAINIEST <br> PLEASE HELP ASAP

Leni [432]

Answer:

1: 4:15

2: 4/15

Step-by-step explanation:

I think Im right. I hope this helps you out!

7 0

3 years ago

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Suppose Marcy's rectangular laptop measures 12 inches by 9 inches. Find the diagonal measurement from corner A to corner B.

astraxan [27]

Answer:

the diagonal measurement from corner A to corner B=15 inches

Step-by-step explanation:

as we know that the loptop has the shape of a rectangle that means all it's angles are right angles. so we can use the pythogoras theorem to find out the diagonal of the rectangle.

let us denote the diagonal of rectangle by D and the sides of rectangle be denoted by X=12 and Y=9

so by using pythogoras theorem we have,

$D^{2} =X^{2} +Y^{2}$

$D^{2}$ =225

D=15

Hence the diagonal measurement from corner A to corner B is 15 inches.

7 0

3 years ago

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How to solve this problem

insens350 [35]

What’s da problem that u have?

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3 years ago

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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