Answer:
Given: A triangle ABC and a line DE parallel to BC.
To prove: A line parallel to one side of a triangle divides the other two sides proportionally.
Proof: Consider ΔABC and DE be the line parallel to Bc, then from ΔABC and ΔADE, we have
∠A=∠A (Common)
∠ADE=∠ABC (Corresponding angles)
Thus, by AA similarity, ΔABC is similar to ΔADE, therefore
AB/AD= AC/AE
⇒AD+DB/AD = AE+EC/AE
⇒1+DB/AD = 1+ EC/AE
⇒DB/AD = EC/AE
Therefore, a line parallel to one side of a triangle divides the other two sides proportionally.
⇒Therefore Proved
Hope this helps!!!
Answer:
This is late, but for anyone else that needs it, it's B. 4x-5
For roots of -2, 5, and 7.
x = -2, x = 5, and x = 7
x = -2 x = 5 x = 7
(x + 2) = 0 (x - 5) = 0 (x - 7) = 0
The polynomial of least degree would be:
(x -2)(x - 5)(x - 7) = 0
(x -2)(x -5) = x(x - 5) - 2(x -5)
= x² - 5x - 2x + 10
= x² - 7x + 10
(x² - 7x + 10)(x -7)
x(x² - 7x + 10) - 7(x² - 7x + 10)
x³ - 7x² + 10x - 7x² + 49x - 70
x³ - 7x² - 7x² + 10x + 49x - 70
x³ - 14x² + 59x - 70
The least is x³ - 14x² + 59x - 70
Here, regrouping is basically carrying.
64+43 shown vertically would be:
64
+43
-------
107
4+3 is 7, so seven is in the ones place, but that's not the point.
60+40 is 100, so you regroup by carrying the one to the hundreds place.
Hope I helped!
9514 1404 393
Answer:
6
Step-by-step explanation:
The value of f(x) for x=0 is f(0). The first line of the table tells you that value is 6.
f(0) = 6