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Katyanochek1 [597]
3 years ago
13

Please help asap pleaseeeeeeeeee ASAPPPPPPPPPP

Mathematics
1 answer:
zheka24 [161]3 years ago
7 0

DUDE THESE WERE SO EASY I DID THIS IN 4TH GRADE YOYR TOO YOUNG OK OK

Answer: do L x W x H = a

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The sum of three consecutive odd integers is -381
AVprozaik [17]
Given that the numbers are consecutive:
let the first number be x, the second number will be x+1, the third number be x+2
thus the sum of the numbers will be:
x+(x+1)+(x+2)=-381
3x+3=-381
3x=-381-3
3x=-384
x=-128
thus the first number is -128, the second number is -127 and the third number is -126
4 0
3 years ago
Find Measure of angle 1<br><br> Options: 60<br> 50<br> 40<br> 30
Fofino [41]

Answer:

Its equal so its 40

Step-by-step explanation:

prove me wrong

8 0
3 years ago
Jan is trying to increase her water intake to meet recommended guidelines. She currently consumes 2 (24 oz. each) bottles of wat
castortr0y [4]

Answer:

Extra 4 cups to make up 11 cups a day

Step-by-step explanation:

Here we have that Jan obtains 15% of her fluid needs from food, therefore, she gets the remaining 85 % from drinking water

If the recommended daily intake = 2.7 liters  or 91.3 oz

And she already consumes 48 oz of water and  16 oz of milk

Note milk is equivalent to 0.87 water therefore total intake = 48 + 16*0.87= 61.92 oz

She needs to take additional (91.3 - 61.92) = 29.4 oz or 3.68 ≈ 4 cups extra to meet her recommended daily dosage.

8 0
4 years ago
How many positive integers $N$ from 1 to 5000 satisfy both congruences, $N\equiv 5\pmod{12}$ and $N\equiv 11\pmod{13}$?
True [87]
Use the Chinese remainder theorem. Suppose we set N=5\cdot13+11\cdot12. Then clearly taken modulo 12, the second term vanishes, and incidentally 5\cdot13\equiv65\equiv5\pmod{12}; taken modulo 13, the first term vanishes, but the second term leaves a remainder of 2. To counter this, we can multiply the second term by the inverse of 12 modulo 13, which is 12 since 12^2\equiv144\equiv11\cdot13+1\equiv1\pmod{13}.

So, we found that N=5\cdot13+11\cdot12^2=1649, but the least positive solution is 1649\equiv89\pmod{\underbrace{156}_{12\cdot 13}}, and in general we can have N=89+156k for any integer k.

Now, since 5000=32\cdot156+8, or 4992=32\cdot156, we know that there are 32 possible integers N that satisfy the congruences.
7 0
3 years ago
Do the products 40 times 500 and 40 times 600 have the same number of zeros
alexandr1967 [171]
Yes they do have the same number of zeros
3 0
3 years ago
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