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El matadora 24/02/2019 at 07:29Very Easy
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Quân Nguyễn 29/05/2019 at 14:52post for what? don't need to solve these? =))
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Bùi Đức Lôc 31/05/2017 at 13:47Chúng ta có: 3aa1 chia hết cho 9
<=> (3 + a + a + 1) chia hết cho 9
<=> (2a + 4) chia hết cho 9
<=> 2 (a + 2) chia hết cho 9
<=> a + 2 chia hết cho 9
<=> a = 7
Trả lời 3771
Summer Clouds selected this answer. -
→இے๖ۣۜBoy™†hïếuGîa 31/05/2017 at 16:44We have : 3aa1 is divisible by 9
<=> (3 + a + a + 1) is divisible by 9
<=> (2a + 4) is divisible by 9
<=>2(a + 2) is divisible by 9
<=> a + 2 is divisible by 9
<=> a = 7
Answer 3771
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NVH_HVN 31/05/2017 at 12:17We have : 3aa1 is divisible by 9
<=> (3 + a + a + 1) is divisible by 9
<=> (2a + 4) is divisible by 9
<=>2(a + 2) is divisible by 9
<=> a + 2 is divisible by 9
<=> a = 7
Answer 3771
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Searching4You 27/07/2017 at 09:29The answer is 901.
From 1 to 9 we have sum : 45.
From 10 to 19 we have : 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 +...+ 1 + 9 we have sum 55.
Next from 20 to 29 we have sum 65.
................. 30 to 39 we have sum 75.
................. 40 to 49 we have sum 85.
................. 50 to 59 we have sum 95.
................. 60 to 69 we have sum 105.
................. 70 to 79 we have sum 115.
................. 80 to 89 we have sum 125.
................. 90 to 99 we have sum 135.
And the last 100 we have sum 1 + 0 + 0 = 1.
So the sum she obtained is : 65 + 75 + 85 + 95 +105 +115 +125 +135 + 1 = 901.
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FA KAKALOTS 08/02/2018 at 22:01The answer is 901.
From 1 to 9 we have sum : 45.
From 10 to 19 we have : 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 +...+ 1 + 9 we have sum 55.
Next from 20 to 29 we have sum 65.
................. 30 to 39 we have sum 75.
................. 40 to 49 we have sum 85.
................. 50 to 59 we have sum 95.
................. 60 to 69 we have sum 105.
................. 70 to 79 we have sum 115.
................. 80 to 89 we have sum 125.
................. 90 to 99 we have sum 135.
And the last 100 we have sum 1 + 0 + 0 = 1.
So the sum she obtained is : 65 + 75 + 85 + 95 +105 +115 +125 +135 + 1 = 901.
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From 1 to 9 the sum of the individual digits are: \(1+2+3+4+5+6+7+8+9=45\)
From 10 to 19: the tens digit all have the same digit, the ones digit have the rules from 1 to 9.
So as the same with 20-99.
The number 100 has the individual digits sum: \(1+0+0=1\)
She has obtained the sum: \(\left[\left(1+2+3+4+5+6+7+8+9\right).10\right]+\left[\left(1+2+3+4+5+6+7+8+9\right).10\right]+1\)
\(=450+450+1=901\)
So she obtained the sum: \(901\)
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If each number is doubled, their sum is doubled and their mean is doubled and equal to : 9 x 2 = 18
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Nicolas has a bird, so he has a parrot
Manon has a furry animal, so he has a cat or a dog. Since he doesn't cats, he has a dog
Fabienne has a 4-legged creature, so he has a cat or a dog. Since Manon has a dog, he has a cat
So, Julien has a goldfish
The statement A is not true
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a) \(x^5+x^4+1=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
b) \(x^8+x^4+1=x^8+2x^4+1-x^4=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)=\left(x^4-x^2+1\right)\left(x^4+2x^2+1-x^2\right)\)
\(=\left(x^4-x^2+1\right)\left[\left(x^2+1\right)^2-x^2\right]\)
\(=\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
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The smallest prime number has 2 digits is 11
The greatest prime number has 2 digits is 97
So their product is:
11.97 = 1067
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Draw \(BE\perp CD\), then ABED is a rectangle since it has 3 right angles. So, DE = AB = 4 units and :
\(AD=BE=\sqrt{BC^2-EC^2}=\sqrt{15^2-\left(16-4\right)^2}=9\)(units)
The volume of the frustum is :
\(\dfrac{1}{3}\pi\left(AB^2+CD^2+AB.CD\right).AD=\dfrac{1}{3}\pi\left(4^2+16^2+4.16\right).9=1008\pi\) (units3)
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! Help!-
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Nguyen Huu Ai Linh 09/12/2017 at 10:22Answer: 16048523266853
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Nguyen Huu Ai Linh 09/12/2017 at 10:20
50% corret all!
