Factorise : x^2+x-72 Step by Step

(x + 9)(x - 8)

Step-by-step explanation:

f(x) = x^2 + x - 72

     = x^2 - 8x + 9x - 72

     = (x^2 - 8x) + (9x - 72)

     = x(x - 8) + 9(x - 8)

     = (x + 9)(x - 8)

              x² + x - 72 = (x + 9)(x - 8)

Step-by-step explanation:

or:

     x² + x - 72 =

= x² + 9x - 8x - 72 =

= x(x + 9) - 8(x + 9) =

= (x + 9)(x - 8)

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.