The trapezoid did a reflect transformation
1 answer:
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Answer:
Hence proved △ABE∼△CBF.
Step-by-step explanation:
Given,
ABCD is a parallelogram.
BF ⊥ CD and
BE ⊥ AD
To Prove : △ABE∼△CBF
We have drawn the diagram for your reference.
Proof:
Since ABCD is a parallelogram,
So according to the property of parallelogram opposite angles are equal in measure.
⇒1
And given that BF ⊥ CD and BE ⊥ AD.
So we can say that;
⇒2
Now In △ABE and △CBF
∠A = ∠C (from 1)
∠E = ∠F (from 2)
So by A.A. similarity postulate;
△ABE∼△CBF
Micah was asked to add the following expressions:
First, he combined like terms in the numerator and kept the common denominator
First step is correct. He added the like terms in the numerator, because the denominators are same.
So he got ,
In the next step, he cannot cancel out x^2 from the top and bottom . Because x-4 and 3x+2 are added with x^2
If we have x^2 is multiplied with other terms at the top and bottom , then we can cancel out x^2.
So Micah added the expression incorrectly. Final answer is not correct.
Answer:
[1]
Given:
Remove the parenthesis, we get;
Like terms are the those terms with same variable and powers.
Combine like terms;
To write this polynomial in standard form, you write starting with the term with the highest degree, or exponent(i.e ), and then in decreasing order .
Standard form:
To, classify a polynomial by degree, you just look at the highest exponent, or degree.
Since, 3 is the highest degree (), it is a cubic.
Now, classify a polynomial by the number of terms, count how many terms are in the polynomial( )
Number of terms: 4 (so this is polynomial)
[2]
Similarly,
for
Remove the parenthesis, we get;
Combine like terms; we have
Standard form:
Degree of the polynomial is, 2
Number of terms: 3 ( so, this is trinomial)