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mart [117]
3 years ago
9

Justine answered 68 questions correctly on a 80-question test.

Mathematics
1 answer:
cestrela7 [59]3 years ago
5 0
For A the answer is 85% correct. Then 15% incorrect. I got the number correct by taking 68 and dividing by 80. To get the number incorrect I took the 15 other questions (80-68) and did 12 divided by 80 to get the 15% incorrect.
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Which of the following is a factor of 3x^3+18x^2+27x? <br> A. 9x<br> B. x^3<br> C. x+3<br> D. x-3
Tju [1.3M]
3x³ + 18x² + 27x = 3x(x² + 6x + 9) = 3x(x+3)²

Answer: <span>C. x+3</span>
3 0
3 years ago
Find a number if: 3/7 of it is 2 1/4
Sergeeva-Olga [200]

Answer:

x=5 1/4

Step-by-step explanation:

2 1/4= \frac{9}{4}---> \frac{9}{4} divided by 3 to find 1/7x----> \frac{3}{4}=1/7x---> x=21/4---> x=5 1/4

8 0
3 years ago
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What is the answer to 15 - 5x??​
anzhelika [568]

Answer:

x=3

Step-by-step explanation:

6 0
3 years ago
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The sum of a rational number and an even integer is rational. A) Always True B) Sometimes True C) Usually True D) Never True
SOVA2 [1]
I think that the sum will always be a rational number
let's prove that


<span>any rational number can be represented as a/b where a and b are integers and b≠0

</span>and an integer is the counting numbers plus their negatives and 0
so like -4,-3,-2,-1,0,1,2,3,4....

<span>so, 2 rational numbers can be represented as

</span>a/b and c/d (where a,b,c,d are all integers and b≠0 and d≠0)

their sum is
a/b+c/d=
ad/bd+bc/bd=
(ad+bc)/bd

1. the numerator and denominator will be integers
2. that the denominator does not equal 0

alright
1.
we started with that they are all integers
ab+bc=?
if we multiply any 2 integers, we get an integer
<span>like 3*4=12 or -3*4=-12 or -3*-4=12, etc.
</span>even 0*4=0, that's an integer
the sum of any 2 integers is an integer
like 4+3=7, 3+(-4)=-1, 3+0=3, etc.
so we have established that the numerator is an integer

now the denominator
that is just a product of 2 integers so it is an integer


<span>2. we originally defined that b≠0 and d≠0 so we're good

</span>therefore, the sum of any 2 rational numbers will always be a rational number <span>is the correct answer.</span>
6 0
3 years ago
Difference in the following polynomials (6x^3 - 2x^2 + 4)<br> -(2x^3 + 4x^2 - 5)
Colt1911 [192]

\bold{[ \ Answer \ ]}

\boxed{\bold{4x^3-6x^2+9}}

\bold{[ \ Explanation \ ]}

  • \bold{Find \ Difference: \ \left(6x^3\:-\:2x^2\:+\:4\right)-\left(2x^3\:+\:4x^2\:-\:5\right)}

\bold{-----------------------}

  • \bold{Remove \ Parenthesis}

\bold{6x^3-2x^2+4-\left(2x^3+4x^2-5\right)}

  • \bold{Remove \ Parenthesis \ / \ Simplify}

\bold{-\left(2x^3+4x^2-5\right): \ -2x^3-4x^2+5}

  • \bold{Rewrite}

\bold{6x^3-2x^2+4-2x^3-4x^2+5}

  • \bold{[Simplify \ 6x^3-2x^2+4-2x^3-4x^2+5] \ 4x^3-6x^2+9}

\bold{4x^3-6x^2+9}

\boxed{\bold{[] \ Eclipsed \ []}}

6 0
3 years ago
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