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Zarrin [17]
3 years ago
6

Find the least common denominator for these two rational expressions,

Mathematics
2 answers:
kap26 [50]3 years ago
5 0

Answer:

(y + 2)²(y + 4)

Step-by-step explanation:

Factor the denominators of both fractions

y² + 4y + 4 = (y + 2)²

y² + 6y + 8 = (y + 2)(y + 4)

The least common denominator is (y + 2)²(y + 4)

Artyom0805 [142]3 years ago
3 0

Answer:

Step-by-step explanation:

y²+4y+4=(y+2)² =(y+2)(y+2)

y²+6y+8 = (y+2)(y+4)

the least common denominator for these

two rational expressions is : (y+2)²(y+4)

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PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS!!<br><br> Simplify.
NNADVOKAT [17]

Answer:  \frac{2x(5x + 1)}{4x - 1}

<u>Step-by-step explanation:</u>

\frac{4x^{3}-12x^{2}}{4x^{2}+7x - 2} ÷ \frac{2x^{2}-6x}{5x^{2}+11x + 2}

= \frac{4x^{2}(x - 3)}{(4x - 1)(x + 2)} ÷ \frac{2x(x - 3)}{(5x + 1)(x + 2)}

= \frac{4x^{2}(x - 3)}{(4x - 1)(x + 2)} x \frac{(5x + 1)(x + 2)}{2x(x - 3)}

= \frac{2x(5x + 1)}{4x - 1}

8 0
3 years ago
What is the solution of |2x| = –4?
DENIUS [597]
-2x=-4 or x=2
2x =-4 or x=-2
7 0
3 years ago
Let f(x)=18/1+3^e-0.1x What is the point of maximum growth rate for the logistic function f(x)? Show all work. Round your answer
xz_007 [3.2K]

Answer:

(0, 4.5)

Step-by-step explanation:

*The equation can be put into Desmos, to find the point, but the work to prove it is here*

f(x)=c/1+Ae^-Bx

Y=C

C=18      A=3      -B=-0.1

*Replace x with 0 in the equation, so you know 0 is the x value, and it leads you to the y value*

f(0)=18/1+3e^-o.1(0)

= 18/1+3e^0

=18/1+3(1)

=18/1+3

=18/4

=4.5

x=0    y=4.5

Maximum growth rate = (x,y) --> (0, 4.5)

Hope this helps:))!!

5 0
3 years ago
If 3(x+2)=5(x-8) what is the value of x+2?<br><br>A. 23<br>B.25<br>C.40<br>D.46
lorasvet [3.4K]

3*x

3*2

3x+6

5*x

5*-8

5x-40

Story short the answer is 25

3 0
3 years ago
Read 2 more answers
What is 1/4-1/6 equals to
klemol [59]

Answer:

3/16

Step-by-step explanation:

1. you convert 1/4 into 16ths and you get 4/16

2. 4/16 - 1/16= 3/16

7 0
3 years ago
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