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2 years ago

In the accompanying diagram, point P(0.6, −0.8) is on unit circle O. What is the value of θ, to the nearest degree?

Mathematics

2 answers:

MakcuM [25]2 years ago

8 0

Answer:

if I answer ..will you mark as BRAINLIEST

Triss [41]2 years ago

7 0

ANSWER:

To find the size of the angle when you know the cos, use cos-1(0.324).

On hand-held calculators use (2nd) (cos) , enter 0.324 to get the answer: 71.1°.

On this calculator, use (2nd) (acos) and then enter the value.

“acos” means “arc cos”, which is the inverse cosine.

Similarly, find sin-1(0.123) and tan-1(0.783), using (2nd) (asin) and (2nd) (atan).

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Please include equation and solution thanks

ehidna [41]

John is 31 and Jenny is 26

7 0

3 years ago

A quadratic equation has a

11111nata11111 [884]

Answer:

$\boxed {\boxed {\sf 2 \ real \ solutions }}$

Step-by-step explanation:

The quadratic formula is:

${x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}$

The discriminant of a quadratic is just the expression under the square root, or $b^2-4ac$ . This can tell us the number of solutions a quadratic has.

If the discriminant is:

Positive = 2 real solutions
Equal to Zero = 1 real double/repeated solution
Negative = 0 real solutions, but 2 imaginary solutions

Our quadratic equation has a discriminant of 5, which is positive. Therefore, it has 2 real solutions.

6 0

2 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

2 years ago

I need help with this math problem.

DiKsa [7]

Answer:Multiply 410,000 with 26:)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Try to slove this question for me please

cestrela7 [59]

C. s is your answer.

The sum of 2 angles on a straight line (supplementary angles) is 180°. Vertical angles (opposites where a line intersects) are equal.

The upper left and lower right angles where p crosses the other lines (q, r, t) are 105°.

180 - 105 = 75°

So the upper right and lower left angles are 75°. The lower left angle where p intersects s is 85°, so it is not parallel to the others.

Please let me know if you have any questions.

6 0

3 years ago