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lorasvet [3.4K]
4 years ago
14

(2x+8)^2+(X)^2= 2704 solve for x

Mathematics
1 answer:
bonufazy [111]4 years ago
3 0
Answer:

x=25.69046516

(round as you please)
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Jonathan Broke what is he doing buying cheap stuff

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Algebra 2!..!!!! please help
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We need to multiply the factors (x - 2) and (x^2 + 9x + 10) to see if the product is the original expression given.

(x - 2)(x^2 + 9x + 10)

x^3 + 9x^2 + 10x - 2x^2 - 18x - 20

x^3 + 7x^2 - 8x - 20.

Since the product just found is not the original expression given, Jimmy is wrong.

I will complete the work on paper by synthetic division and post my answer.

After using synthetic division, the correct quotient is x^2 + 9x + 20.
5 0
4 years ago
For the following geometric sequence find the explicit formula. {12, -6, 3, ...}
Furkat [3]
X divide by -2

this is your formula

x/ (-2)

5 0
3 years ago
Knowing that P = X x i / 100, C = main, e = Rate. Calculate the rate for the formula: A principal of R $ 105,000.00 obtained a p
Brut [27]

Answer:

Interest expressed as a percent of the principal will be referred to as an ... earned on an investment of P at the annual interest rate r for a period of t ... (c) monthly, (​d) weekly, (e) daily, and (f) continuously at a nominal annual ... and substitute x = 3−2y in for x in second equation and find 4y−3(3−2y) = 1.

Step-by-step explanation: sorry if that doesn't help :(

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.

5 0
3 years ago
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