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87 ? 8t + t3 ? 23 when t = 1? 88 73 57 59

7 Answers
87 ? 8t + t3 ? 23 when t = 1? 88 73 57 59
I

There are three possible scenarios here arising from typographical issue

(1) It's possible that you meant to write: 87 - 8t + t^3 - 23
When t = 1, AND t is an integer, a substitution results in 87 - 8*1 + 1^3 - 23 = 87 - 8 + 1 - 23 = 57
The Answer is 57

(2) It's also possible that you meant to write: 87 - 8t + t(3) - 23
When t = 1, AND t is an integer, a substitution results in 87 - 8*1 + 1*3 -23 = 87 - 8 + 3 - 23 = 59
The Answer is 59

(3) It's again possible that t is isn't an integer but a DIGIT hence a substitution of t into 87 - 8t + t3 - 23 = 87 - 81 + 13 - 23 = -4
The Answer is -4

Feel free to pose more confusing questions, we are here for you

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S

If you do it this way it is incorect

Put t=1, then the resulting value is:
87 - 8*1 + 1*1*1 - 23
= 87 - 8 +1 -23
= 57 <------- Answer!

Because it is not " T cubed" it is t(3) which is 1x3 being 3

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J

To solve this

87-8(1)+1(3)-23 *because t=1

87-8+3-23

87+3-8-23 *adding +ve and _ve integers

90-31 *add first and then sub.

59

so,answer is 59

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S

87 ? 8t + t3 ? 23 when t = 1?

87-8(1)+(1)3-23=?
87-8+3-23=?
87-8=79+3=82-23=59

Answer is 59

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Q

Put t=1, then the resulting value is:
87 - 8*1 + 1*1*1 - 23
= 87 - 8 +1 -23
= 57 <------- Answer!

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B

When t=1 than the equation is

87 - 8 + 3 - 23 = 59

So we get the answer is 59

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S

57 dude

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