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Step2247 [10]
2 years ago
12

And simplify this too ​

Mathematics
2 answers:
Andrew [12]2 years ago
6 0

Answer:

-5x^4

Step-by-step explanation:

Cube root of -125 is -5

Cube root of n^12 is n^4

gavmur [86]2 years ago
5 0

Answer:

3 x (-125n^12)^1/2

Step-by-step explanation:

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Maximum value. F=4x+3y; x+3y<=6; x+y<=4
levacccp [35]

Answer:

6

Step-by-step explanation:

hope it helps

8 0
3 years ago
The quotient of j and 8" can be expressed as:
jek_recluse [69]
The quotient is the result of dividing one number by another.
Quotient = dividend / divisor
For example:
7 : 5 = 7/5 ( 7/5 is the quotient )
Quotient = j : 8 = j / 8
Answer:
A ) j / 8 
6 0
3 years ago
Consider M, N, and P. collinear points on MP.
Kobotan [32]

Answer:

See explanation

Step-by-step explanation:

There are three possible cases:

1. Point N lies between M and P, then MN + NP = MP. Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{NP}-\dfrac{MN}{MN+NP}=\dfrac{MN(MN+NP)-MN\cdot NP}{NP(MN+NP)}=\\ \\=\dfrac{MN^2+MN\cdot NP-MN\cdot NP}{NP(MN+NP)}=\dfrac{MN^2}{NP(MN+NP)}

2. Point N lies to the right from point P, then MP + PN = MN.  Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MP+PN}{NP}-\dfrac{MP+PN}{MP}=\dfrac{MP}{NP}+1-1-\dfrac{NP}{MP}=\dfrac{MP^2-NP^2}{NP\cdot MP}

3. Point N lies to the left from point M, then NM + MP = NP. Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{MN+MP}-\dfrac{MN}{MP}=\dfrac{MN\cdot MP-MN(MN+MP)}{MP(MN+MP)}=\\ \\=\dfrac{MN\cdot MP-MN^2-MN\cdot MP}{MP(MN+MP)}=\dfrac{-MN^2}{MP(MN+MP)}

3 0
2 years ago
Sheldon wants to access the money in this checking account. Which of the following will work?
bulgar [2K]
D. he can access it by taking it out of his account
7 0
2 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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