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poizon [28]
3 years ago
6

The 9’s in 9,905,482

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
7 0

Answer:Your question is very detailed but I’ll try anyways, there are two 9,s in that number and 9 million is one of them and the other is 900,000

Step-by-step explanation:

If this is wrong you can report it to remove I’m sure,

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Steven fill 3 pages in a stamp album. this is one sixth of the pages in the album. how many pages are there in steven stamp albu
Arturiano [62]
3 pages x 6 times=18 pages
7 0
3 years ago
How do I solve this problem because I don’t understand
Citrus2011 [14]
You are using a^2+b^2=c^2
since c is the hypotenuse

#1 the equation would be 16^2+b^2=34^2
16^2= 256
34^2= 1156
1156-256= 900
b^2=900 (now find square root of 900)
b= 30

You are now trying to find the hypotenuse or c in this so it is simpler

#2 equation - 4^2+2^2=c^2
16+4=c^2
20=c^2
c= 4.27



4 0
3 years ago
Or which equation would x = 4 be a solution?
ElenaW [278]

2x+7=22\ \ \ \ |-7\\2x=15\ \ \ \ \ |:2\\x=7.4\neq4\\----------------------\\6x\div8=3\ \ \ \ |\cdot8\\6x=24\ \ \ \ |:6\\x=4\\-------------------\\8-3x=20\ \ \ \ |-8\\-3x=12\ \ \ \ |:(-3)\\x=-4\neq4\\---------------------\\2x+8=4\ \ \ \ \ |-8\\2x=-4\ \ \ \ \ |:2\\x=-2\neq4\\\\Answer:\ 6x\div8=3

Other method.

Enter the value of x for each equation and check the equality of the pages of the equation

2x+7=22\\L=2(4)+7=8+7=15;\qquad R=22\qquad L\neq R\\\\6x\div8=3\\L=6(4)\div8=24\div8=3;\qquad R=3;\qquad L=R\\\\8-3x=20\\L=8-3(4)=8-12=-4;\qquad R=20;\qquad L\neq R\\\\2x+8=4\\L=2(4)+8=8+8=16;\qquad R=4;\qquad L\neq R

6 0
3 years ago
Which statement is TRUE about
igor_vitrenko [27]
I am bad at Common Core (I was good before, but not anymore) so I am saying none of the above
6 0
4 years ago
Using separation of variables technique, solve the following differential equation with initial condition y'= e sinx and y(pi) =
inessss [21]

\begin{aligned}&y'=e^y\sin x\\&\dfrac{dy}{dx}=e^y\sin x\\&\dfrac{dy}{e^y}=\sin x\, dx\\&\int \dfrac{dy}{e^y}=\int \sin x\, dx\\&-e^{-y}=-\cos x+C\\&e^{-y}=\cos x+C\\\end

e^{-0}=\cos \pi +C\\1=1+C\\C=0\\\\\boxed{\boxed{e^{-y}=\cos x}}

3 0
2 years ago
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