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3 years ago

In circle T, A. 3 B. 4 C. 6 D. 7

Mathematics

2 answers:

Maksim231197 [3]3 years ago

5 0

Answer:

Im confused

Step-by-step explanation:

You didn't show us what were supposed to be answering

ArbitrLikvidat [17]3 years ago

4 0

I’m confused ?
Where is the chart ?

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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

{[14-2×3]-[18÷(12-3)]}+{42-11×3+[(65-9×5)+3]}

natali 33 [55]

Use Bidmas, solve all the inner brackets (), with first division, then multiplication then addition then subtraction

5 0

3 years ago

Divide. −36/−8 Enter your answer as a mixed number, in simplified form, in the box.

Pavel [41]

Answer:

= 4 whole 1/2

Step-by-step explanation:

Given that:

= −36/−8

"-" signs will be cancelled out with each other so

= 36/8

By reducing to lowest term

= 9/2

When writing into mixed form:

quotient = 4, remainder = 1, divisor = 2 so:

= 4 1/2

i hope it will help you!

8 0

3 years ago

The differfence between the product of seven and a number and two times the number

olga2289 [7]

Answer:

7(x-10)

Step-by-step explanation:

7×(x-10)

you ×7by whatever x-10 is

4 0

2 years ago

If the acute angles in a right triangle are in the ratio of 5:4 find the measure of the smallest angle

deff fn [24]

Let's create this right triangle. We will call it triangle ABC with angle B as the right angle, angle A being the larger of the angles A and C. If the ratio is 5:4, then it is A:C. 5+4 = 9. The measure of the right angle, angle ABC = 90, and 90 divided by 9 is 10. So angle A is 5 * 10 which is 50 degrees, and angle C is 4 * 10 which is 40 degrees. The smallest angle, angle C, is 40 degrees.

6 0

3 years ago