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nata0808 [166]
3 years ago
12

It is a six-digit

Mathematics
1 answer:
KonstantinChe [14]3 years ago
8 0
Ok, the first clue is it has six digit and the second clue is it’s a whole number!
So we know it lies between 99999 and 1000000
3rd clue tells it has only 3 different digits and 4th clue tells us each are used twice!
Moving on, 5th clue says none of its digits are even! 6th speaks none are divisible be 3
So the possibilities for digits are 1, 5, 7
And it’s greater than 600000, then the 1st digit must be 7! It is divisible be 5, so last digit must be 5!
7th clue states that It’s tenth digit is same as hundred-thousand! Means the tenth digit is 7
Let’s see what we got!
{7xxx75}
Clue no 8 as you can see says that it’s thousands digit is same as unit digit
So the number now is {7x5x75}
9th clue says it’s hundreds digit is different from tens digit meaning the hundreds digit is either 1 or 7 and we used 7 two times, so it’s 1 and clue 10 says it’s ten thousands digit is 1 so the number that’s playing hide ‘n seek or most probably riddle game is 715175!
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if in one min they make 1/12 then  in 12  min they  paint all the wall

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3 years ago
What is the answer to (-3x) + x
Dafna11 [192]
I'm pretty sure it's -4x. 
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Use your calculator to select the best answer below:
bixtya [17]

Answer:

-2\sqrt{a}

Step-by-step explanation:

\lim_{x \to a} \frac{a-x}{\sqrt{x} -\sqrt{a} } \\ =-\lim_{x \to a} \frac{x-a}{\sqrt{x} -\sqrt{a} }  \\=-\lim_{x \to a} \frac{(\sqrt{x} -\sqrt{a})(\sqrt{x} +\sqrt{a})}{\sqrt{x} -\sqrt{a} }  \\=- \lim_{x \to a} \sqrt{x} +\sqrt{a}\\=-2\sqrt{a}

6 0
3 years ago
A rectangle field is 350 meters long and 300 meters wide. What is the area of the field in square kilometers
balu736 [363]

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4 0
3 years ago
How do I solve for the missing lengths?
Advocard [28]

Answer:

Part 4) ER=3\ units

Part 5) DF=9\sqrt{10}\ units

Part 6) DE=30\ units

Step-by-step explanation:

Part 4) Find ER

we know that

In the right triangle ERF

Applying the Pythagorean Theorem

EF^2=ER^2+RF^2

substitute the given values

(3\sqrt{10})^2=ER^2+9^2

solve for ER

ER^2=(3\sqrt{10})^2-9^2

ER^2=90-81\\ER^2=9\\ER=3\ units

Part 5) Find DF

we know that

In the right triangle DRF

Applying the Pythagorean Theorem

DF^2=DR^2+RF^2

substitute the given values

DF^2=27^2+9^2

DF^2=810\\DF=\sqrt{810}\ units

simplify

DF=9\sqrt{10}\ units

Part 6) Find DE

we know that

DE=DR+RE ----> by segment addition postulate

we have

DR=27\ units\\RE=ER=3\ units

substitute

DE=27+3=30\ units

3 0
3 years ago
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