Ask question

Ask question

All categories

3 years ago

5

Question 8 of 10

2 answers:

noname [10]3 years ago

6 0

The federal reserve

dedylja [7]3 years ago

4 0

Answer:

The answer is D. The federal Reserve is the central banking system of U.S.A

You might be interested in

64÷54 is the number of.... whar

ankoles [38]

Your answer is 1.18518519

8 0

3 years ago

Read 2 more answers

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

In the past ten years, the population of a city decreased from 90,000 to 75,000 Find the percent decrease.

vfiekz [6]

The answer is 31.5%.

8 0

3 years ago

Find two positive consecutive odd integers such that square of the smaller integer is 10 more than the larger integer

Otrada [13]

Let $2n+1$ be the smaller integer. The larger integer is then $2n+3$ , and we have

$(2n+1)^2=10+(2n+3)\implies4n^2+4n+1=2n+13$

$\implies4n^2+2n-12=0$

$\implies2n^2+n-6=0$

$\implies(2n-3)(n+2)=0$

$\implies 2n-3=0\text{ or }n+2=0$

$\implies n=\dfrac32\text{ or }n=-2$

We omit $n=-2$ , since $2(-2)+1=-3$ is negative.

Then for $n=\dfrac32$ we find $2\left(\dfrac32\right)+1=4$ , but this is not odd.

There are no consecutive odd integers that satisfy the given condition!

3 0

3 years ago

⚠️⚠️⚠️⚠️⚠️HELP HELP HELP HELP PLEASEEEE I

ser-zykov [4K]

Answer:

im stuck on that too man

Step-by-step explanation:

4 0

3 years ago