Answer:
1)
(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12
Instead of multiplying 3 and 4 the student added them which gave them the answer of 7 when the correct answer is 12.
2)
a) x^2 + x - 2
b) x^2 - 7x + 10
c) 4x^2 - 1
d) x^2 + 10x + 25
Step-by-step explanation:
a. (x - 1)(x + 2)
x(x) - 1(x) + 2(x) - 1(2)
x^2 - 1x + 2x - 2
x^2 + x - 2
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b. (x - 5)(x - 2)
x(x) - 5(x) - 2(x) - 5(-2)
x^2 - 5x - 2x + 10
x^2 - 7x + 10
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c. (2x + 1)(2x - 1)
2x(2x) + 1(2x) - 1(2x) + 1(-1)
4x^2 + 2x - 2x - 1
4x^2 - 1
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d. (x + 5)^2
(x + 5)(x + 5)
x(x) + 5(x) + 5(x) + 5(5)
x^2 + 5x + 5x + 25
x^2 + 10x + 25
Answer:
ok this is easy all it is 16-7=9 so a=9 so all it is 7x9=72 then its 72 divided by 2 so your answer is 36
If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.
Answer:
Step-by-step explanation:
Given is a function as 
Equating to 0 we have equation
If the function f(x) has x intercepts then the solutions are real
Let us use remainder theorem and change of signs rule
f(0) = 1>0
f(-1) = -1+3+1=3
f(-2) = -32+6+1<0
This implies there is a real root between -1 and -2.
f(1) = -1
Since f(0) and f(1) have different signs, there exists a real root between 0 and 1.
f(2) = 32-6+1>0
Since f(1) and f(2) have different signs there exists a real root between 1 and 2.
Thus there are definitely three real solutions as
one between -1 and 0, one between 0 and 1, and third between 1 and 2.
