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Sphinxa [80]
4 years ago
8

Find the quotient 45a2-5a3/5a​

Mathematics
1 answer:
NISA [10]4 years ago
4 0

Answer:

9a-a^2 is the required quotient.

Step-by-step explanation:

\frac{45 {a}^{2}  - 5 {a}^{3} }{5a}  \\  =  \frac{5a(9a - {a}^{2} )}{5a}  \\  = 9a -  {a}^{2}

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The area of a square is 4x^2 + 24x + 36. What expression
ss7ja [257]

Answer:

2x + 6

Step-by-step explanation:

The area of a square is the square of the side length:

A = s²

We know that the area of the square is 4x² + 24x + 36.

4 is 2², and 36 is 6².  24 is double 2 times 6.

Therefore, this is a perfect square trinomial:

A = 4x² + 24x + 36

A = (2x + 6)²

Therefore:

s = 2x + 6

7 0
4 years ago
I will make you brainlliest whoever answers this right and will give you points Please help me i am having a hard time
frutty [35]

14 + (-8) / 2 x -3

Follow order of operations: divide and multiply from left to right:

-8/2 = -4 x -3 = 12

Now you have 14 + 12

The answer is 26

4 0
3 years ago
Read 2 more answers
What is the interquartile range
elena-14-01-66 [18.8K]
The answer is six because you have to subtract the upper and lower quartile which are 18 and 16 in this case
3 0
3 years ago
X / 6 equals -7 solve using the multiplication principle
garik1379 [7]

Answer:

x = -42 is your answer

Step-by-step explanation:

x/6 = -7

Isolate the x. Note the equal sign, what you do to one side, you do to the other.

Multiply 6 to both sides

(x/6)(6) = (-7)(6)

x = (-7)(6)

Simplify. Multiply 6 with -7

x = (-7)(6)

x = -42

<em>~So shines a good deed in a weary world</em>

8 0
3 years ago
Read 2 more answers
Prove that one of every three consecutive positive integer is divisible by 3
Inessa05 [86]
Hello : 
all n in N ; n(n+1)(n+2) = 3a    a in  N  or : <span>≡ 0 (mod 3)
1 ) n </span><span>≡ 0 ( mod 3)...(1)

     n+1 </span>≡ 1 ( mod 3)...(2)
      n+2 ≡ 2 ( mod 3)...(3)
by (1), (2), (3)  : n(n+1)(n+2) ≡ 0×1×2   ( mod 3)   : ≡ 0 (mod 3)
2) n ≡ 1 ( mod 3)...(1)

     n+1 ≡ 2 ( mod 3)...(2)
      n+2 ≡ 3 ( mod 3)...(3)
by (1), (2), (3)  : n(n+1)(n+2) ≡ 1×2 × 3  ( mod 3)   : ≡ 0 (mod 3) , 6≡ 0 (mod)
 3) n  ≡ 2 ( mod 3)...(1)

     n+1 ≡ 3 ( mod 3)...(2)
      n+2 ≡ 4 ( mod 3)...(3)
by (1), (2), (3)  : n(n+1)(n+2) ≡ 2×3 × 4  ( mod 3)   : ≡ 0 (mod 3) , 24≡ 0      (mod3)
5 0
3 years ago
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