Answer:
In the first account was invested
at 3%
In the second account was invested
at 5%
Step-by-step explanation:
we know that
The simple interest formula is equal to
where
I is the Interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
in this problem we have
First account
substitute in the formula above
Second account
substitute in the formula above
Remember that
The interest is equal to
so
Adds the interest of both accounts
therefore
In the first account was invested
at 3%
In the second account was invested
at 5%
The measure of ∠BAF is 54°.
Solution:
DF and CE are intersecting lines.
m∠EAF = 72° and AB bisects ∠CAF.
∠EAF and ∠DAC are vertically opposite angles.
Vertical angle theorem:
<em>If two lines are intersecting, then vertically opposite angles are congruent.</em>
∠DAC ≅ ∠EAF
m∠DAC = 72°
<em>Sum of the adjacent angles in a straight line = 180°</em>
m∠DAE + m∠EAF = 180°
m∠DAE + 72° = 180°
Subtract 72° from both sides.
m∠DAE = 108°
∠CAF and ∠DAE are vertically opposite angles.
⇒ m∠CAF = m∠DAE
⇒ m∠CAF = 108°
AB bisects ∠CAF means ∠CAB = ∠BAF
m∠CAB + m∠BAF = 108°
m∠BAF + m∠BAF = 108°
2 m∠BAF = 108°
Divide by 2 on both sides, we get
m∠BAF = 54°
Hence the measure of ∠BAF is 54°.
Answer:
The first and second iteration of Newton's Method are 3 and
.
Step-by-step explanation:
The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form
based on the following formula:
![x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}](https://tex.z-dn.net/?f=x_%7Bi%2B1%7D%20%3D%20x_%7Bi%7D%20-%5Cfrac%7Bf%28x_%7Bi%7D%29%7D%7Bf%27%28x_%7Bi%7D%29%7D)
Where:
- i-th Approximation, dimensionless.
- (i+1)-th Approximation, dimensionless.
- Function evaluated at i-th Approximation, dimensionless.
- First derivative evaluated at (i+1)-th Approximation, dimensionless.
Let be
and
, the resultant expression is:
![x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}](https://tex.z-dn.net/?f=x_%7Bi%2B1%7D%20%3D%20x_%7Bi%7D%20-%5Cfrac%7Bx_%7Bi%7D%5E%7B2%7D-8%7D%7B2%5Ccdot%20x_%7Bi%7D%7D)
First iteration: (
)
![x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%202-%5Cfrac%7B2%5E%7B2%7D-8%7D%7B2%5Ccdot%20%282%29%7D)
![x_{2} = 2 + \frac{4}{4}](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%202%20%2B%20%5Cfrac%7B4%7D%7B4%7D)
![x_{2} = 3](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%203)
Second iteration: (
)
![x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}](https://tex.z-dn.net/?f=x_%7B3%7D%20%3D%203-%5Cfrac%7B3%5E%7B2%7D-8%7D%7B2%5Ccdot%20%283%29%7D)
![x_{3} = 2 - \frac{1}{6}](https://tex.z-dn.net/?f=x_%7B3%7D%20%3D%202%20-%20%5Cfrac%7B1%7D%7B6%7D)
![x_{3} = \frac{11}{6}](https://tex.z-dn.net/?f=x_%7B3%7D%20%3D%20%5Cfrac%7B11%7D%7B6%7D)
I think isn't anyone of these alternatives because if we put number 15 in the place of y it is x-y=30. X-15=30
X=30+15=45.
X=45