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KengaRu [80]
3 years ago
7

Is 8397 divisible by 9 ? Explain your answer

Mathematics
2 answers:
Ierofanga [76]3 years ago
8 0

Answer:

Yes

Step-by-step explanation:

9 can go into 8397 evenly without any remainders.

zaharov [31]3 years ago
3 0

Answer:

8397/9= 933

8397 is divisible by 9 if you divide 8397 by 9 and the quotient is an integer. The easiest way to determine if 8397 is divisible by 9 is to calculate the sum of the digits in 8397. If the sum of the digits is divisible by 9, then 8397 is divisible by 9.

8+3+9+7= 27

27 is divisible by 9

Step-by-step explanation:

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Answer: C {3, 4}

Step-by-step explanation:

<em>Solve the inequality:</em>

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7x > 20 - 6

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x > 2

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7 0
3 years ago
Graph Y equals -5/6x -4
svetlana [45]

Step-by-step explanation:

y = -5/6x - 4

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4 years ago
Reciprocal of. 0×7/11​
Tpy6a [65]

Answer:

it doesn't exist

Step-by-step explanation:

the expression 0×7/11​ is equivalent to 0. 1/0 isn't possible, so its reciprocal doesn't exist.

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3 years ago
I need help with &lt;1 please and thanks ​
Afina-wow [57]

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6 0
3 years ago
Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 7x − 5x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a
Margaret [11]

Answer:

Therefore the solution is = k{-7,5,1} where k ∈R

Step-by-step explanation:

Given that,

f₁(x) =x

f₂(x)= x²

f₃(x)= 7x - 5x²

Also,

g(x) = c₁f₁(x)+c₂f₂(x)+c₃f₃(x)

Putting the values of f₁(x), f₂(x) and f₃(x).

g(x) = c₁.x+c₂x²+c₃(7x-5x²)

Given condition that g(x)= 0

∴ c₁.x+c₂x²+c₃(7x-5x²)=0

⇒(c₁+7c₃)x +(c₂-5c₃)x² = 0

Comparing the coefficients of x and x²

∴c₁+7c₃=0               and       c₂-5c₃ =0

\Rightarrow c_1 =-7c_3                     \Rightarrow c_2=5c_3

Let c₃= k   [k∈R]  

Then c₁ = -7k   and   c₂=5k

Therefore the solution is = { c₁,c₂,c₃}  

                                           = {-7k, 5k, k}

                                            =k{-7,5,1}

5 0
3 years ago
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