Please help ASAP!<br> 1. 83.40 divided by 12<br> 2. 53.2 divided by 4

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

Subtract<br> (3x2 + 2x – 9) - (4x2 - 6x + 3)<br> Enter your answer, in standard form in the box.

ss7ja [257]

Answer:

2

Step-by-step explanation:

7 0

2 years ago

Family Video charges $10 for

Irina18 [472]

Answer:

4 movies, $20

Step-by-step explanation:

Given data

Family Video

charges $10 for a monthly membership and

$2.50 per movie

let the number of movies be x

and the charge for x movies be y

y= 10+2.5x----------1

Babe's Movies

charges a $12

membership and $2 per

movie.

let the number of movies be x

and the charge for x movies be y

y= 12+2x----------2

equate 1 and 2

10+2.5x= 12+2x

collect like terms

12-10= 2.5x-2x

2= 0.5x

x= 2/0.5

x= 4 movies

put x= 4 in equation 1 or 2

y= 12+2x----------2

y= 12+2*4

y= 12+8

y= $20

4 0

3 years ago

Which of the following numbers are perfect squares?

professor190 [17]

Answer:

B and D

Step-by-step explanation:

I think i am right

3 0

2 years ago

Read 2 more answers

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago

Please help ASAP! 1. 83.40 divided by 12 2. 53.2 divided by 4