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3 years ago

Hello. Someone help me solve all these. Will mark as brainliest. Thanks

Mathematics

1 answer:

iris [78.8K]3 years ago

4 0

Answer:

number 19 is letter B.hence express P(x) as the product of it's linear factors

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Mae ling earns a weekly salary of $320 plus a 8.0% commission on sales at a gift shop. How much would she make in a work week if

Alex787 [66]

Step-by-step explanation:

4300$ - 100%

x$ - 8%

(4300 × 8) ÷ 100 = 344$

344 + 320 = 664$

answer: 664$ :))))

8 0

3 years ago

Find the holes<br> <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx%2B3%7D%7B%28x%2B1%29%28x-2%29%7D" id="TexFormula1" t

olga55 [171]

Answer:

Step-by-step explanation:

The "holes" are the x-es where function doesn't exist.

one must not divide by 0, so given function doesn't exis if the denominator is equal 0

The product {(x+1)×(x-2)} is equal 0 if any of factors {(x+1),(x-2)} is 0, so

x + 1 = 0 or x - 2 = 0

x = - 1 or x = 2

8 0

3 years ago

Hello can u help plz?

I am Lyosha [343]

Answer:

slope would be 3 or 3/1

Step-by-step explanation:

I think i'm not totally sure it could be -3 or -3/1

7 0

3 years ago

98 POINTSSSSSS!!!!!!!!!

ddd [48]

Answer:

the answer is b -2u +w

or -2w - u

8 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