All categories

2 years ago

Please help with this question

Mathematics

1 answer:

stiks02 [169]2 years ago

7 0

Answer:

I don't know

Step-by-step explanation:

You might be interested in

Suppose that is in standard position and the given point is on the terminal side of 0. Give the exact value of the indicated tri

Lilit [14]

Answer:

12/13

Step-by-step explanation:

The length of the segment from the origin to the terminal point is ...

r = √((-5)² +12²) = √169 = 13

The sine of the angle is the ratio of the y-coordinate to this distance

sin(θ) = y/r = 12/13

_____

Additional comment

The other trig functions are ...

cos(θ) = x/r = -5/13

tan(θ) = y/x = -12/5

This is a 2nd-quadrant angle, where the sine is positive, but the cosine and tangent are negative.

6 0

2 years ago

Connie and two of her friends completed a 10-mile bike relay. each friend road an equal part of the race. what distance did each

IrinaVladis [17]

Connie + 1 friend +2 friend = 10
3 people = 10 miles
3 people/3= 10 miles / 3
Each friend rode 3 1/3 miles

8 0

3 years ago

Read 2 more answers

120 beats per 45 seconds

fomenos

Answer: 120 beats per 45 seconds

Step-by-step explanation:

8 0

3 years ago

A Circle has a diameter of 9 cm. Find the Circumference. Give a Simplified Exact answer. Enter pi with answer. For example 4pi

lutik1710 [3]

Circumference = Diameter * $\pi$

So 9 $\pi$ is the answer

3 0

3 years ago

Read 2 more answers

PLEASE SOLVE AND CHECK. SHOW COMPLETE SOLUTION

Alex17521 [72]

Solution:

$\sqrt{4x + 13} = x + 2$

First square both sides.

$= > ( \sqrt{4x + 13} ) ^{2} = (x + 2) ^{2}$

Now, square root and square gets cancel out in the LHS. And in the RHS, apply the identity: (a + b)² = a² + 2ab + b².

$= > 4x + 13 = {(x)}^{2} + 2 \times x \times 2 + (2) ^{2} \\ = > 4x + 13 = {x}^{2} + 4x + 4$

Now, transpose 4x and 4 to LHS.

$= > 4x - 4x + 13 - 4 = {x}^{2} \\$

Now, do the addition and subtraction.

$= > {x}^{2} = 9 \\ = > x = \sqrt{9} \\ = > x = ±3$

Answer:

x = ± 3

Hope you could understand.

If you have any query, feel free to ask.

3 0

2 years ago