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katrin [286]
3 years ago
12

What polynomial identity should be used to prove that 16^2=(10+6)^2? (A. Difference of Cubes; B. Difference of Squares; C. Squar

e of Binomial; D. Sum of Cubes)
Mathematics
2 answers:
Marysya12 [62]3 years ago
7 0

Answer:

C. Square of Binomial

Step-by-step explanation:

To prove the identity you should use square of Binomial that states the following:

(a+b)^{2}=a^{2}+2ab+b^{2}

Lets prove it, so first take the equation to solve:

(10+6)^{2}

Then square the first term:

(10+6)^{2}=10^{2}

Then multiply by 2 the first and second terms:

(10+6)^{2}=10^{2}+2(10)(6)

Finally square the second term:

(10+6)^{2}=10^{2}+2(10)(6)+6^{2}

Solve the values:

(10+6)^{2}=100+2(10)(6)+6^{2}

(10+6)^{2}=100+120+6^{2}

(10+6)^{2}=100+120+36

(10+6)^{2}=256

And prove the polynomial identity:

16^{2}=256

Anna71 [15]3 years ago
3 0
Not difference of cubes since it is 2nd degree
not difference of squares since it is plus
not sum of cubes because 2nd degreee

answer is square of binomial (why do we even need this property, oh well)


C
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Answer:

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Step-by-step explanation:

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3 years ago
WILL GIVE BRAINLIEST
vesna_86 [32]

The simplified forms of the given expressions are

a. 32x

b. 0

c. -26t + 14x

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<h3>Simplifying an expression </h3>

From the question, we are to simplify the given expressions by adding or subtracting

a. (10x) + (22x)

= 32x

b. (-6x) - (-6x)

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c. (-13t) + (14x) - (+13t)

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Collect like terms

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= 3t

Hence, the simplified forms of the given expressions are

a. 32x

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d. 3t

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7 0
2 years ago
Eight values have a mean of 7. a ninth value is added. the new mean is 9. what is the new value
My name is Ann [436]

Answer:

25 because the 8 values must times with 7 and will get 56. when the ninth value added , let the ninth value as y , 56+y , the new mean is 9 . so it will be like this , 56+y over 9 times with 9

4 0
3 years ago
Given the trinomial 2x2 − 5x + 3, what is the value of the Discriminant and what can be said about the solutions.
valina [46]
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3 years ago
Evaluate the algebraic expression below for the given value.
MrRa [10]

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