Tan (0) is equal to 0
Sin (0) = 0
Cos (0) = 1
Tan = Sin/Cos
That means that Tan (0) = 0/1 or 0
Answer:
x = 8
Step-by-step explanation:
you have similar triangles here
so 2 is to 4 as 4 is to x
2/4 = 4/x
x = 8
<span><span> y2(q-4)-c(q-4)</span> </span>Final result :<span> (q - 4) • (y2 - c)
</span>
Step by step solution :<span>Step 1 :</span><span>Equation at the end of step 1 :</span><span><span> ((y2) • (q - 4)) - c • (q - 4)
</span><span> Step 2 :</span></span><span>Equation at the end of step 2 :</span><span> y2 • (q - 4) - c • (q - 4)
</span><span>Step 3 :</span>Pulling out like terms :
<span> 3.1 </span> Pull out q-4
After pulling out, we are left with :
(q-4) • (<span> y2</span> * 1 +( c * (-1) ))
Trying to factor as a Difference of Squares :
<span> 3.2 </span> Factoring: <span> y2-c</span>
Theory : A difference of two perfect squares, <span> A2 - B2 </span>can be factored into <span> (A+B) • (A-B)
</span>Proof :<span> (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 <span>- AB + AB </span>- B2 =
<span> A2 - B2</span>
</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication.
Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.
Check : <span> y2 </span>is the square of <span> y1 </span>
Check :<span> <span> c1 </span> is not a square !!
</span>Ruling : Binomial can not be factored as the difference of two perfect squares
Final result :<span> (q - 4) • (y2 - c)
</span><span>
</span>
Given: Triangle JKL is dilated from the origin at a scale factor of 10 to create triangle J′K′L′.
Required: Correct option that complete the statement.
Explanation:
Now, triangle is dilated from the origin at a scale factor of 10.
So, the sides will become larger, but angles will remain the same.
Hence, the triangles will be similar as sides will be in proption (SSS rule).
So, The triangles are similar because their corresponding sides are proptional and their corresponding angles are congurent.
Final Answer: Second group is correct answer.
I can't see them all the way
