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2 years ago

PLEASE PLEASE HELP ME ASAP

Mathematics

1 answer:

Vika [28.1K]2 years ago

4 0

Answer: DC=30.5

Step-by-step explanation:

Use trigonometry to solve this question

sin=O/H
cos=A/H
tan=O/A

----------------------------------------------------------

First Step: find BD

- AB relative to 54° is [hypotenues]

- BD relative to 54° is [opposite]

- Therefore, we need to use O/H, which is Sin

- sin(54°)=O/20=20·sin(54°)=16.2

Second Step: Find DC

- DC relative to 28° is [adjacent]

- BD relative to 28° is [opposite]

- Therefore, we need to use O/A, which is tan

- tan(28°)=16.2/A=16.2/tan(28°)=30.5

Hope this helps!! :)

Please let me know if you have any questions

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Will is buying a house for $185,000. He is financing $165,000 and obtained a 30-year, fixed-rate mortgage with a 6.725% interest

guajiro [1.7K]

<h3>Answer: $1067.45</h3>

=================================================

Work Shown:

L = 165000 = loan amount or amount financed

r = 0.06725 = annual interest rate in decimal form

i = r/12 = 0.06725/12 = 0.005604167

i = 0.005604167 = approximate monthly interest rate in decimal form

n = number of months = 30*12 = 360 months

P = unknown monthly payment

--------------

Apply the monthly payment formula

P = (L*i)/( 1-(1+i)^(-n) )

P = (165000*0.005604167)/(1-(1+0.005604167)^(-360))

P = 1067.44636311118

P = 1067.45

6 0

2 years ago

1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given

neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

$x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}$

2) Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: $x^2-2x-4=0$

$x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}$

As for

$x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\$

3) Rewriting and calculating its derivative. Remember to do it, in radians.

$5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1$

$x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}$

For the second root, let's try -1.5

$x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\$

For x=-3.9, last root.

$x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\$

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

$x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}$

$f(x)=x^6-x^4+3x^3-2x$

$\mathbf{f'(x)=6x^5-4x^3+9x^2-2}$

$\mathbf{f''(x)=30x^4-12x^2+18x}$

For -1.2

$x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx \mathbf{-1.29322}\\$

For x=0.4

$x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx \mathbf{0.50785}\\$

and for x=-0.4

$x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\$

These roots (in bold) are the critical numbers

3 0

2 years ago

Luna flew from Boston to Orlando with a stop in Atlanta to switch planes. The flight from Boston to Atlanta was 3 hours and 30 m

sertanlavr [38]

Answer:

1:15 minuites

Step-by-step explanation:

30 minuites

8 0

2 years ago

Read 2 more answers

Select the graph that shows the solution set for the following system.<br> x + y < 2<br> x>2

kompoz [17]

Answer:

Answer is in the attachment.

Step-by-step explanation:

To graph x>2 consider first x=2. x=2 is a vertical line and if you want to graph x>2 you need to shade to the right of the vertical line.

To graph x+y<2, I will solve for y first.

x+y<2

Subtract x on both sides:

y<-x+2

Consider the equation y=-x+2. This is an equation with y-intercept 2 and slope -1 or -1/1. So the line you have in that picture looks good for y=-x+2. Now going back to consider y<-x+2 means we want to shade below the line because we had y<.

Now where you see both shadings will be intersection of the shadings and will actually by your answer to system of inequalities you have. In my picture it is where you have both blue and pink.

I have a graph in the picture that shows the solution.

Also both of your lines will be solid because your question in the picture shows they both have equal signs along with those inequality signs.

Just in case my one graph was confusing, I put a second attachment with just the solution to the system.

5 0

2 years ago

Leo is planning a trip with his friends by renting a bus. He found that the bus cost $75.50 for 5 people if he is planning to go

kolezko [41]

Step-by-step explanation:

75.50 x 13

=151 is the anser for 10 people you need to figure out how much it costs per person to find your complete answer and don't forget get that your adding "Leo " into the eqation

8 0

2 years ago