<span>4x² - 5x + 1 = 0
a = 4; b = - 5, c = 1
</span>Δ = b² - 4.a.c
Δ = (-5)² - 4.4.1
Δ = 25 - 16
Δ = 9
<span>
x = - b </span>± √Δ / 2.a
A. Two real solutions (x' = -1 and x'' = -1/4)
Answer:
The expression 1.02^(4+n) can be used because 1.02^4 * 1.02^2= 1.02^(4+n) AND this is option C.
Step-by-step explanation:
For 4 + N years = (1 + 2/100)^(4+N)
= 1.02 ^ (4+N)
Remember, that the number of years is increased by N and interest rate remain the same, and n is years, and which is now 4+ N.
-3x + 9 - 4 = x + 3 + 2x
-3x + 9 - 4 = 3x + 3
9 - 4 = 6x + 3
5 = 6x + 3
2 = 6x
1/3 = x
Answer:
75 mins or 1 hour 15 min :)
Step-by-step explanation:
9514 1404 393
Answer:
4a. ∠V≅∠Y
4b. TU ≅ WX
5. No; no applicable postulate
6. see below
Step-by-step explanation:
<h3>4.</h3>
a. When you use the ASA postulate, you are claiming you have shown two angles and the side between them to be congruent. Here, you're given side TV and angle T are congruent to their counterparts, sides WY and angle W. The angle at the other end of segment TV is angle V. Its counterpart is the other end of segment WY from angle W. In order to use ASA, we must show ...
∠V≅∠Y
__
b. When you use the SAS postulate, you are claiming you have shown two sides and the angle between them are congruent. The angle T is between sides TV and TU. The angle congruent to that, ∠W, is between sides WY and WX. Then the missing congruence that must be shown is ...
TU ≅ WX
__
<h3>5.</h3>
The marked congruences are for two sides and a non-included angle. There is no SSA postulate for proving congruence. (In fact, there are two different possible triangles that have the given dimensions. This can be seen in the fact that the given angle is opposite the shortest of the given sides.)
"No, we cannot prove they are congruent because none of the five postulates or theorems can be used."
__
<h3>6.</h3>
The first statement/reason is always the list of "given" statements.
1. ∠A≅∠D, AC≅DC . . . . given
2. . . . . vertical angles are congruent
3. . . . . ASA postulate
4. . . . . CPCTC
