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Volgvan
3 years ago
9

What is the product? 2x+8 x2 + 3x-4 x(x-4)(x-1) 20x+4) 2 (x+4)(x-4) 2x(x-1) 2x(x+4)​

Mathematics
2 answers:
guajiro [1.7K]3 years ago
7 0

=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2

step by step

(2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x(x+4)

=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x+4)

=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x)+((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(4)

=−640x10+3840x9+4544x8−58904x7+91128x6−40608x5+128x4+512x3−2560x9+15360x8+18176x7−235616x6+364512x5−162432x4+512x3+2048x2

=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2

Oduvanchick [21]3 years ago
4 0
The product is the same as the answer
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aleksley [76]
The answer would be -0.5.
4 0
3 years ago
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Invasive species often experience exponential population growth when introduced into a new environment. Zebra mussels are an inv
Oliga [24]

Answer:

If ‘a’ is the initial population of the Zebra mussels, then every six months, the population of Zebra mussels quadruples, i.e. it becomes 4a. In t years, i.e. in 2t intervals of 6 months each, the population of Zebra mussels can be computed with the help of a geometric series a, ar, ar2, ar3, …. arn-1 ( the series being finite with n terms) where a is the 1st term , r is the common ratio and n is the number of terms in the series..

Here, in 2t years, there will be 2t + 1 terms in the geometric series including the initial term. Thus in the above series, a = 10, r = 4 and n = 2t + 1. Then the population of the Zebra mussels after 2t years is the (2t +1)st term of the geometric series I.e. 10* ( 42t)                  

In 15 months, t = 15/12 = 1.25. Then the population of Zebra mussels after 15 months will be 10*(42.5 ) = 10 * 25 = 320

If after t years, the population of the Zebra mussels become 1 million , then we have

1000000 = 10 * (42t) or, 100000= 42t or,105 = 42t    Taking logarithms of both sides, we have 5 log 10 = 2t log 4 or, t = (5log10)/(2log4) = 5/1.20 years or (5/1.20) * 12 months = 50 months, i.e. 4 years and 2 months.

Step-by-step explanation:

8 0
3 years ago
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Please help me this is due in 20 min
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Answer:

7.50 + 14P < 60; 24 slices

Step-by-step explanation:

8 0
3 years ago
A pretzel maker was interested in knowing the number of pretzels in each bag is sold. The results of the research are shown in t
Vladimir [108]

Answer:

23 pretzels

Step-by-step explanation:

The range value is obtained by taken the difference between the maximum and minimum values ;

The range = maximum - minimum

From the box and whisker plot attached ; the maximum value = 68

Minimum value = 45

Hence, the range in the number of pretzels :

68 - 45 = 23

4 0
3 years ago
How do I solve these problems?
lesya [120]

Domain:\\D:x > 0\\\\\ln x=5.6+\ln(7.5)\ \ \ \ |-\ln(7.5)\\\\\ln x-\ln7.5=5.6\ \ \ |Use\ \log x-\log y=\log\dfrac{x}{y}\\\\\ln\dfrac{x}{7.5}=5.6\ \ \ \ |use\ \log_ab=c\iff a^c=b\\\\\dfrac{x}{7.5}=e^{5.6}\ \ \ \ |\cdot7.5\\\\\boxed{x=7.5e^{5.6}}\in D

\log x=5.6-\log7.5\ \ \ \ |+\log7.5\\\\\log x+\log7.5=5.6\ \ \ \ |use\ \log x+\log y=\log (xy)\\\\\log(7.5x)=5.6\ \ \ \ |use\ \log_ab=c\iff a^c=b\\\\7.5x=10^{5.6}\ \ \ \ |:7.5\\\\\boxed{x=\dfrac{10^{5.6}}{7.5}}\in D

5 0
3 years ago
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