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mixer [17]
3 years ago
8

Can some one friend me plz

Mathematics
1 answer:
Alona [7]3 years ago
4 0

Answer: I didn't understand

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What is the value of x?<br><br> 16<br> 50<br> 130<br> 164<br> Please hurry !!!
Dmitry_Shevchenko [17]
C) 130 i hope this helps good luck 
6 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B9x%7D%5E%7B2%7D%20-%20%7B%28x%7D%5E%7B2%7D%20-%204%29%20%7B%7D%5E%7B2%7D%20
Alenkinab [10]

Answer:

{ x^2+3x-4}

Step-by-step explanation:

Factor top and bottom.

The numerator is a difference of two squares, and the denominator is a quadratic.

\frac{ {9x}^{2} - {(x}^{2} - 4)^{2} }{4 + 3x - {x}^{2} }

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

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

If x does not equal -1 and does not equal 4, we can cancel the common factors in italics to give

= { (x-1)(x+4)}

= { x^2+3x-4}

3 0
3 years ago
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Two days after a biology experiment there was 49 million bacteria present. Four days after the experiment there was 600.25 milli
nataly862011 [7]

Answer:

y=Ae^(1.25t)

Step-by-step explanation:

From the expression  y=Ae^kt

After two days of the experiment, y = 49 million, t=2

After four days of the experiment, y= 600.25 million, t=4

A is the amount of bacteria present at time zero  and t is the time after the experiment (in days)

At t=2 and y =49

 49=Ae^2k…………….. (1)

At t=4 and y = 600.25

600.25=Ae^4k………… (2)

Divide equation (2) by equation (1)

600.25/49=(Ae^4k)/(Ae^2k )

12.25=e^2k

Take natural log of both sides

ln⁡(12.25) =2k

2.505⁡ =2k

k=1.25

The exponential equation that models this situation is y=Ae^(1.25t)

8 0
3 years ago
Answer the circle question please.
makkiz [27]

\frac{11}{-6}

3 0
3 years ago
Can someone please help me out?
djyliett [7]

Answer:

Correct answer is

165 {c}^{6}  {b}^{6}

7 0
3 years ago
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