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3 years ago

20) Simplify: 3ab(4b-ab) + 5b2(a2 + 2)

Mathematics

2 answers:

shutvik [7]3 years ago

6 0

2a^2 b^2 + 12ab^2 + 10b^2

raketka [301]3 years ago

6 0

Answer:

$12ab^2+2a^2b^2+10b^2$

Step-by-step explanation:

$3ab(4b-ab) + 5b^2(a^2 + 2)\\\\\mathrm{Expand}\:3ab\left(4b-ab\right):\quad 12ab^2-3a^2b^2\\\\=12ab^2-3a^2b^2+5b^2\left(a^2+2\right)\\\\\mathrm{Expand}\:5b^2\left(a^2+2\right):\quad 5a^2b^2+10b^2\\\\=12ab^2-3a^2b^2+5a^2b^2+10b^2\\\\\mathrm{Add\:similar\:elements:}\:-3a^2b^2+5a^2b^2=2a^2b^2\\\\=12ab^2+2a^2b^2+10b^2$

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30 points<br> can somebody please help me<br> question below

Inessa05 [86]

Find the total number of dogs by adding each day:

Total dogs = 23

Percent of dogs: Divide the number of dogs by the total number of patients:

= 23 / 50 = 0.46 x 100 = 46%

Use the given formula to find the standard error:

Standard error = √(0.46*(1-0.46)/50) = 0.07

90%: Find Z value for 90% ( 1.645) multiply by SE:

1.645 x 0.07 = 0.115 = 0.12

Now add and subtract that value from the Percent from above:

46 + 12 = 58%

46-12 = 34%

Answer (34%, 58%)

95%: Find Z value for 95% ( 1.96) multiply by SE:

1.96 x 0.07 = 0.137 = 0.14

Now add and subtract that value from the Percent from above:

46 + 14 = 60%

46-14 = 32%

Answer (32%, 60%)

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3 years ago

What is the median of 30,33,38,40,51,57,81,91?

enot [183]

Answer: ${ \left[\begin{array}{ccc}45.5\\\\\end{array}\right] }{}\\\\\end{array}\right] } ----$

Step-by-step explanation: Hey, Basically we are going to arrange the data in an ascending order and the median is the middle value. If the number of values is an even number, the median will be the average of the two middle numbers.

Hope this helps!

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2 years ago

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.