Let the fist integer be x, the second is x+20
the product of the numbers is:
x(x+20)
the sum of the numbers is:
x+x+20=2x+20
the sum of the above operations will give us:
2x+20+x^2+20x=95
x^2+22x+20=95
this can be written as quadratic to be:
x^2+22x-75=0
solving the above we get:
x=3 and x=-25
but since the integers should be positive, then x=3
the second number is x+20=3+20=23
hence the numbers are:
3 and 23
The positive square roots of the number 151,321 according to the task content can be determined by means of division as; 389.
<h3>What are the square roots of 151,321 by means of division method?</h3>
It follows from.the task content above that the number given is; 151,321 whose positive square roots is to be determined.
Upon testing different integers as divisor on the number 151,321; it is concluded that the only positive integer by which 151,321 can be divided to result in a whole is; 389.
Hence, the positive square root of the number 151,321 is; 389.
Consequently, it can be concluded that the positive square root of the number, 151,321 as in the task content is; 389 which is itself a prime number as it is only divisible by 1 and itself.
Read more on square root of numbers;
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Answer:
3p³ + 2p² – 3p – 11
Step-by-step explanation:
From the question given above, the following data were obtained:
Side 1 (S₁) = –1(p + 5)
Side 2 (S₂) = 2(p² – 3)
Side 3 (S₃) = 3p³ – 2p
Perimeter (P) =?
The perimeter of the triangle can be obtained as follow
P = S₁ + S₂ + S₃
P = –1(p + 5) + 2(p² – 3) + 3p³ – 2p
Clear bracket
P = –p – 5 + 2p² – 6 + 3p³ – 2p
Rearrange
P = 3p³ + 2p² – 2p – p – 6 – 5
P = 3p³ + 2p² – 3p – 11
Therefore, the perimeter of the triangle is 3p³ + 2p² – 3p – 11
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Answer:
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Step-by-step explanation:
