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3 years ago

Can you help me with this question?-8(-4x-1)-9x

1 answer:

4 0

First you have to multiply -8 to (-4x-1)
which is, -12x+8
then there is -9x you have to add like terms
-12x-9x+8
-21x+8 is your final answer

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The sum of two numbers is 19. Their difference is 65. What are the numbers?

Ludmilka [50]

X = 42 and y = - 23 are the two numbers

3 0

3 years ago

Which equation is rewritten in vertex form?

Zinaida [17]

(x + 3)^2 + (x + 4)^2
= x^2 + 6x + 9 + x^2 + 8x + 16
= 2x^2 + 14x + 25
= 2(x^2 + 7x) + 25
= 2[(x + 7/2)^2 - 49/4] + 25
= 2(x + 7/2)^2 - 98/4 + 25
= 2(x + 7/2)^2 + 1/2

Its B

7 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

Which of the following fractions is in simplest form? 2/10 3/18 5/65 10/37

AlexFokin [52]

The answer is 10/37 because you cannot reduce it anymore

4 0

3 years ago

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3 years ago