When you multiply 8•26 you get 208
Answer:
A.
by the SAS postulate.
Step-by-step explanation:
We have been two triangles. We are asked to determine the theorem by which both triangles could be proven congruent.
We can see that side DF of triangle DEF is equal to side AC of triangle ABC.
We can also see that side BC of triangle ABC is equal to side EF of triangle DEF.
The including angle between sides AC and BC of triangle ABC is equal to the including angle between sides DF and EF of triangle DEF.
Since both triangles have two sides and their included angles equal, therefore, triangle ABC is congruent to triangle DEF by SAS (Side-Angle-Side) congruence and option A is the correct choice.
10
b: 5+5
c: 5+5
thats thw andeer
The correct answer is 2y(3+4x)
Answer:
See Below.
Step-by-step explanation:
We want to verify the equation:

We can convert sec(α) to 1 / cos(α):

Multiply both layers of the first fraction by cos(α):

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

Simplify:

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

Subtract:

Distribute:

Rewrite:

Split:

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

Factor:

Cancel:

Divide the second fraction by cos(α):

Hence proven.