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lesya692 [45]
3 years ago
15

What is 43.78 in word form

Mathematics
2 answers:
maria [59]3 years ago
4 0
Forty three point seventy eight.
balu736 [363]3 years ago
3 0
Forty three and seventy eight thousands
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I’ll give u brainliest!!<br> what are the areas for shape 1,2 and 3
JulsSmile [24]

Answer:

I think it is

1: 22.37

2:16.81

3:83.16

Step-by-step explanation:

3 0
3 years ago
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-5a+9x=-27<br> Solve for x. <br> X=
Tresset [83]

Answer:

x= 5a -3/9

Step-by-step explanation:

-5a+9x=-27

9x=5a-27

x= 5a-27÷9

x= 5a - 27×1/9

x= 5a -3/9

Hope this helps

Keep Smiling :)

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4 years ago
50 POINTS!!!! If you were on a rocket ship for a month that sat in between the moon and earth, how might moon phases be differen
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Answer:I would sayif the planet earth was making a shadown on it. when u on other side

Step-by-step explanation:

That is the best i got

5 0
3 years ago
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A light source is located over the center of a circular table of diameter 4 feet. (See picture below) Find the height h of the l
Alex
Very nice to have an accompanied image!Illumination is proportional to the intensity of the source, inversely proportional to the distance squared, and to the sine of angle alpha.so that we can writeI(h)=K*sin(alpha)/s^2 ................(0)where K is a constant proportional to the light source, and a function of other factors.
Also, radius of the table is 4'/2=2', therefore, using Pythagoras theorem,s^2=h^2+2^2 ...........(1), and consequently,sin(alpha)=h/s=h/sqrt(h^2+2^2)..............(2)
Substitute (1) and (2) in (0), we can writeI(h)=K*(h/sqrt(h^2+4))/(h^2+4)=Kh/(h^2+4)^(3/2)
To get a maximum value of I, we equate the derivative of I (wrt alpha) to 0, orI'(h)=0or, after a few algebraic manipulations, I'(h)=K/(h^2+4)^(3/2)-(3*h^2*K)/(h^2+4)^(5/2)=K*sqrt(h^2+4)(2h^2-4)/(h^2+4)^3We see that I'(h)=0 if 2h^2-4=0, giving h=sqrt(4/2)=sqrt(2) feet above the table.
We know that I(h) is a minimum if h=0 (flat on the table) or h=infinity (very, very far away), so instinctively h=sqrt(2) must be a maximum.Mathematically, we can derive I'(h) to get I"(h) and check that I"(sqrt(2)) is negative (for a maximum).  If you wish, you could go ahead and find that I"(h)=(sqrt(h^2+4)*(6*h^3-36*h))/(h^2+4)^4, and find that the numerator equals -83.1K which is negative (denominator is always positive).
An alternative to showing that it is a maximum is to check the value of I(h) in the vicinity of h=sqrt(2), say I(sqrt(2) +/- 0.01)we findI(sqrt(2)-0.01)=0.0962218KI(sqrt(2))     =0.0962250K   (maximum)I(sqrt(2)+0.01)=0.0962218KIt is not mathematically rigorous, but it is reassuring, without all the tedious work.
3 0
3 years ago
Find the zeros of the function f(x) = 1/3 x^2 + 72
Alexus [3.1K]
To find the zeros of an equation equate the equation to zero

Please with this question I don’t really get it looking at how u hv written it
Is the x^2 a numerator or a denominator
I don’t really get how u wrote ur question

6 0
3 years ago
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