Which products result in a difference of squares or a perfect square trinomial? Check all that apply.

Mathematics

2 answers:

SVEN [57.7K]4 years ago

3 0

The following are the answers for these:

(5x + 3)(5x – 3) ----> Difference of squares

(7x + 4)(7x + 4) ----> Perfect square trinomial

(2x + 1)(x + 2) ----> Neither

(4x – 6)(x + 8) ----> Neither

(x – 9)(x – 9)-----> Perfect square trinomial

(–3x – 6)(–3x + 6) -----> Difference of squares.

The easiest way to tell these is that if the parenthesis are exactly the same it will be a perfect square trinomial. If the parenthesis are the same except for the sign in between the terms, it is a difference of square. If it has neither of these, then it is neither.

kvv77 [185]4 years ago

3 0

Answer:

Your answers are:

Step-by-step explanation:

A) (5x + 3)(5x – 3)

B) (7x + 4)(7x + 4)

E) (x – 9)(x – 9)

F) (–3x – 6)(–3x + 6)

I just did the assignment, and I got a 100%. I swear, this is right!

I hope this helps!

- sincerelynini

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Paul [167]

Answer:

Right angle triangle goes to 15m, 8m, 17m

Step-by-step explanation:

Square root of 15^2 + 8^2 is 17

(The use of a^2 + b^2 = c^2 - Pythagoras's Theorem)

So, 10m ,11m ,15m isn't a right angle triangle

Hope this helps!

6 0

2 years ago

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times sm

sergey [27]

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel

<h3><u>Solution:</u></h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

$\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}$