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Furkat [3]
3 years ago
9

How does the value of the 8 in 589,310 compare to the value of the 8 in 598,301?

Mathematics
1 answer:
zloy xaker [14]3 years ago
6 0
589, 310

The 8 value is in the 10,000s place.

598, 301

The 8 is in the 1000s place.

So the difference of 10,000 over 1000 is 10

So your answer is B
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Answer:

Step-by-step explanation:

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If a kid is 145 cm and he owns an eraser that is 6.8 cm long, 1.5 cm wide, and 1.5 cm tall. If a scientist shoots him with an en
never [62]
To answer this you’d have to use an 2 step equation 145x= 1000
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3 years ago
Anybody help me to solve this question. ​
Mumz [18]

Answer:

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP

Step-by-step explanation:

Given that (b-c)^2, (c-a)^2 , (a-b)^2 are in AP

To prove: \dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are in AP

From given as we know if p , q, r are in AP then 2q= p+r.

2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab

Now

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc

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Hence proved

6 0
3 years ago
(<img src="https://tex.z-dn.net/?f=%28%5Csqrt%7B6%7D%20%2B%5Csqrt%7B3%7D%29%28%5Csqrt%7B2%7D-2%29" id="TexFormula1" title="(\sqr
Marta_Voda [28]

Answer:

-2.453

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sqrt(6) * sqrt(2) = 3.461

sqrt(6) * -2 = -4.899

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sum of numbers equals

-2.453

7 0
3 years ago
HELP ALGEBRA IS KILLING ME IMAGE IS BELOW
djyliett [7]

Answer:

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1425 ------ 45600

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82992000/1425 = x

58240 = x

6 0
3 years ago
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