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3 years ago

Write the ratios for tan D

Mathematics

1 answer:

Maurinko [17]3 years ago

5 0

Answer:

3/5 3:5 3 to 5

1/4 1 to 4 1:4

Step-by-step explanation:

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I WILL GIVE BRAINLIEST!!!! A teacher is grading the final exam. He notices that the mean test score is 61, and the standard devi

marin [14]

Answer:

153.586 close to 154 students

Step-by-step explanation:

mean=61

standard deviation =10

sample 450 students , x between 61 and 71

(x-mean)/deviation for score 61 and 71

61-61/10=0

71-61/10=1

the z-scores is between 0 and 1 ( use calculator)

the percentage is 0.3413 close to 34%

number of students that score between 61 and 71=

450*0.3413= 153.586 close to 154 students

3 0

3 years ago

Solve by Elimination. (10 Points)

Vilka [71]

Make y the subject:

y = 6 - x

Now, use Elimination.

0 = 3x - 3

Divide both sides by 3:

0 = x - 1

So x = 1.

Substitute back into the equation:

y = 6 - (1)

So y = 5.

Hence, the answers are x = 1, y = 5.

3 0

3 years ago

How many solutions does this SOE have?

4vir4ik [10]

Answer:

1 solution

Step-by-step explanation:

The solution is the intersection point, which is at (3, -2)

5 0

3 years ago

The height (feet) of an object moving vertically is given by s= (-16t^2)+(208t)+(156), where t is in seconds.

OverLord2011 [107]

Answer:

The object's velocity at t = 7 is -16 ft/s

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

Step-by-step explanation:

From the information given, the equation of motion of the object is

$s(t)= -16t^2+208t+156$

For any equation of motion s(t), the instantaneous velocity at time t is given by

$v(t)=\frac{ds}{dt}$

(a) To find the object's velocity at t = 7, you must:

$v(t)=\frac{d}{dt}(-16t^2+208t+156)\\\\v(t)=-\frac{d}{dt}\left(16t^2\right)+\frac{d}{dt}\left(208t\right)+\frac{d}{dt}\left(156\right)\\\\v(t)=-32t+208+0\\\\v(t)=-32t+208$

Next, we evaluate when t = 7

$v(7)=-32(7)+208=-224+208\\\\v(7)=-16$

The object's velocity at t = 7 is -16 ft/s

(b) <em>To find the maximum of a function we always use the derivative of the function and we set it equal zero to find the</em><em> critical points.</em>

To find the maximum height and when it occurs, we set the velocity function equal to 0 and solve for t.

$-32t+208=0\\\\-32t+208-208=0-208\\\\-32t=-208\\\\\frac{-32t}{-32}=\frac{-208}{-32}\\\\t=\frac{13}{2}$

Next, we substitute this value into the equation of motion to find the maximum height

$s(\frac{13}{2})= -16(\frac{13}{2})^2+208(\frac{13}{2})+156\\\\s(\frac{13}{2})=-676+1352+156)=832$

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

6 0

3 years ago

Find the integral of 9^(6x)

vredina [299]

Answer:

(1/ 6 ln (9) ) x 9^6x + C

Step-by-step explanation:

3 0

3 years ago