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3 years ago

I need help with this!

Mathematics

1 answer:

natulia [17]3 years ago

4 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

12. The first attachment has the rate of change on each interval. (It is listed at the end of the interval.) It is computed by dividing the difference in length by the corresponding difference in months. The units of rate of change are inches per month.

The baby was growing fastest in the first 3 months.

13. The first attachment has the rate of change on each interval. It is computed by dividing the difference in height by the corresponding difference in seconds. The units of rate of change are meters per second. A negative number indicates the elevator is going down.

The second attachment shows a graph of the distance versus time.

_____

I find it convenient to let a spreadsheet do repetitive calculations. I can enter the formula once and copy it where needed. I don't have to worry about the accuracy of the arithmetic, and the result can be formatted to suit the problem requirements. (The basics of most spreadsheet programs can be learned in a few minutes.)

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Find an (rational) approximation of the following values with the

Dmitry_Shevchenko [17]

The right answer for this one is c

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3 years ago

Determine the zeroes of 10x2−5=35

OLEGan [10]

Answer:

x = +2, x = -2

Step-by-step explanation:

The equation to solve in this problem is

$10x^2-5=35$

The first step we do is to subtract 35 on both sides of the equation, so we get:

$10x^2-5-35=0\\10x^2-40=0$

Now we simplify the equation by dividing both terms by 10:

$\frac{10x^2-40}{10}=0\\x^2-4=0$

Now we observe that the term on the left is the difference between two squares, so it can be rewritten using the property:

$a^2-b^2=(a+b)(a-b)$

Where here,

a = x

b = 2

So we can rewrite the equation as:

$x^2-4=0\\(x+2)(x-2)=0$

And this equation is zero when either one of the two factors is zero, so the two solutions are:

$x+2=0\rightarrow x=-2\\x-2=0 \rightarrow x=+2$

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3 years ago

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th

Lena [83]

F(t) = (t^-1) - t
Taking derivative:
f'(t) = d/dt(t^-1) - t) = d/dt(t^-1) - d/dt(t) 
f'(t) = -1(t^-2) -1
= -t^-2 - 1 
f'(t) = v(t),
It can be also written as :
-1/t^2 - 1 

So v(3) = -1/(3^2) - 1 = -1/9 - 1 = -10/9

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3 years ago

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The answer should be B

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(y-8)(y-8)(y-8) How do I multiply OR find the special product of this problem ?

photoshop1234 [79]

Answer:

2x^3-23y^2+72-128

Step-by-step explanation:

its like when you just do two

do the first 2 first

(y-8)(y-8)

this is (y^2 -8y+y^2+16) simplify (2x^2-7y+16)

then add the other bracket

(2x^2-7y+16) (y-8)

2x^3-16x^2-7y^2+56y+16y-128 simplify

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3 years ago