Answer:
Step-by-step explanation:
Slope of line A =
=
= 3
Slope of line B =
=
Slope of line C =
=
5). Slope of the hypotenuse of the right triangle =
=
=
Since slopes of line C and the hypotenuse are same, right triangle may lie on line C.
6). Slope of the hypotenuse =
= 3
Therefore, this triangle may lie on the line A.
7). Slope of hypotenuse =
=
Given triangle may lie on the line C.
8). Slope of hypotenuse =
=
Given triangle may lie on the line B.
9). Slope of hypotenuse =
=
Given triangle may lie on the line B.
10). Slope of hypotenuse =
= 3
Given triangle may lie on the line A.
Complete Question:
On a number line, the coordinates of X, Y, Z, and W are −8, −5, 4, and 6, respectively. Find the lengths of the two segments below. Then tell whether they are congruent. and
Answer:
They are not congruent
Step-by-step explanation:
Length of segment XY:
Coordinate of X = -8
Coordinate of Y = -5
= |-8 -(-5)| = |-8 + 5| = 3
Length of ZW:
Coordinate of Z = 4
Coordinate of W = 6
= |4 - 6| = 2
≠
, therefore, they are not congruent.
Answer:
The amount needed as a one-time deposit to earn $7,500 in 3 years is <em>$4388.17</em>
Step-by-step explanation:
<u>Basic Finance Formulas </u>
One of the most-used formulas to compute present and future values is
Where FV is the future value, PV is the present value, r is the interest rate and n is the number of periods. It's vital to keep in mind that r and n must be referred to the same compounded time, e.g. r is compounded monthly and n is expressed in months
The question requires to compute the PV needed as a one-time deposit to earn a future value of $7,500 in 3 years at a 1.5% rate compounded monthly.
FV=7,500
r=1.5%=0.015
n=3*12=36 months
We converted n to months because r is compounded monthly . The formula
must be managed to make PV isolated
Answer: The amount needed as a one-time deposit to earn $7,500 in 3 years is $4388.17
The actual value of sin(π/3) is (√3)/2 ≈ 0.86602540.
If the sine function is approximated by y=x (no error at x = 0), then the error at x=π/3 is ...
... x -sin(x) @ x=π/3
... π/3 -(√3)/2 ≈ 0.18117215 ≈ 0.1812
You know right away this is a bad approximation, because the approximate value is π/3 ≈ 1.04719755, a value greater than 1. The range of the sine function is [-1, 1] so there will be no values greater than 1.
___
If the sine function is approximated by y=(x+1-π/4)/√2 (no error at x=π/4), then the error at x=π/3 is ...
... (x+1-π/4)/√2 -sin(x) @ x=π/3
... (π/12 +1)/√2 -(√3)/2 ≈ 0.026201500 ≈ 0.02620
