Answer: (3x + 11y)^2
Demonstration:
The polynomial is a perfect square trinomial, because:
1) √ [9x^2] = 3x
2) √121y^2] = 11y
3) 66xy = 2 *(3x)(11y)
Then it is factored as a square binomial, being the factored expression:
[ 3x + 11y]^2
Now you can verify working backwar, i.e expanding the parenthesis.
Remember that the expansion of a square binomial is:
- square of the first term => (3x)^2 = 9x^2
- double product of first term times second term =>2 (3x)(11y) = 66xy
- square of the second term => (11y)^2 = 121y^2
=> [3x + 11y]^2 = 9x^2 + 66xy + 121y^2, which is the original polynomial.
Answer:
64 teachers
Step-by-step explanation:
We can use ratios to solve
14 students 896 students
----------------- = -------------
1 teacher x teachers
Using cross products
14 * x = 896*1
Divide each side by 14
14x/14 = 896/14
x =64
<h3>
Answer: c = 7/4</h3>
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Work Shown:
Compute the function value at the endpoints
With a = -5 and b = 4, we have
So,
Use algebra to solve for c
Answer:
See explanation
Step-by-step explanation:
Triangles ΔABC and ΔBAD are congruent. So,
- AB ≅ BA;
- AC ≅ BD;
- BC ≅ AD;
- ∠ABC ≅ ∠BAD;
- ∠BCA ≅ ∠ADB;
- ∠CAB ≅ ∠DBA.
Consider triangles AEC and BED. In these triangles,
- AC ≅ BD;
- ∠EAC ≅ ∠EBD (because ∠CBA ≅ ∠BAD);
- ∠AEC ≅ ∠BED (as vertical angles).
So, ΔAEC ≅ ΔBED. Thus,
AE ≅ EB.
This means that segment CD bisects segment AD.