All categories

2 years ago

Which of the following must be true?

Mathematics

1 answer:

galben [10]2 years ago

6 0

Answer:

answer is B because in the property of parallelogram it says ad is parallel to bc

You might be interested in

6x=12 with explanation plz

lord [1]

Answer:

X = 2

Step-by-step explanation:

6(2) = 12

6 X 2 = 12

4 0

3 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

Henry buys a large boat for the summer, however he cannot pay the full amount of $32,000 at

Anna007 [38]

Answer:

Monthly payments=$418.14

Total amount will be=down payment + 48×$418.14

$14000+$20070.84=$34070.84

Step-by-step explanation:

Loan payment per month=Amount to pay÷discount factor

Mathematically P=A÷D

where D is the discount factor calculated using the formula;

$\frac{(1+i)^n-1}{i(1+i)^n}$

where i=periodic interest rate=annual rate divided by number of payment periods

A is the amount to pay after downpayment

P is the loan monthly payment amount

n=number of periodic payments=payments per year times number of years

⇒In this question you find the discount factor then divide the amount remaining to pay with the discount factor to get monthly payments

Given;

Cost of boat=$32000

Down payment=$14000

Loan to pay=$32000-$14000=$18000

Annual rate=5.5%=i=5.5%÷12=0.458%⇒0.00458

Periodic payments, n=4×12=48

Finding the discount factor D;

$D=\frac{(1+i)^n-1}{i(1+i)^n} \\\\\\D=\frac{(1+0.00458)^{48} -1}{0.00458(1+0.00458)^{48} } \\\\\\D=\frac{1.2455-1}{0.005703} \\\\\\D=\frac{0.2455}{0.005703} =43.05$