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Anastaziya [24]
3 years ago
7

Please help!

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

$644.19

Step-by-step explanation:

just took the quiz

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Limit as x approaches 0 of csc3x/cotx
jolli1 [7]
Hello Meggieh821, to find the lim as x approaches 0 we can check this by inserting a number that is close to 0 that is coming from the left and from the right.

For instance, we can find the lim by using the number -.00001 for x and solve
<span>csc(3x) / cot(x)
</span>csc(3*-.00001) / cot(-.00001) = .333333... = 1 /3

We also need to check coming from the right. We will use the number .00001 for x
csc(3x) / cot(x)
csc(3*.00001) / cot(.00001) = .333333... = 1 /3

So since we are getting 1/3 from the left and right we can say as x approaches 0 the limit is 1/3

<span>\lim_{x\to 0} \frac{csc(3x)}{cot(x)} = \frac{1}{3}</span>

6 0
3 years ago
Simplify the radical expression (-9√28a^2*1/3√63a)<br><br> show work
Dovator [93]
A ≥ 0

-\not9^3\sqrt{28a^2}*\frac{1}{\not3^1}\sqrt{63a}=\\\\ =-3a\sqrt{4*7}*\sqrt{9*7a}=\\\\=-3a*2\sqrt{7}*3\sqrt{7a}\\\\ =-18a\sqrt7*\sqrt 7*\sqrt a=\\\\=-18a*(\sqrt7)^2*\sqrt a =\\\\ =-18a*7*\sqrt a=\\\\ =-126a\sqrt a
8 0
3 years ago
Write the perimeter of the triangle in simplest form<br><br> can i please get help
Zepler [3.9K]
This is easy. You just add all the like terms
8 0
3 years ago
What is the volume of a cone when the radius is 6 and the height is 13
lidiya [134]

Answer:

<h2>489.84 units³</h2>

Step-by-step explanation:

Given,

Radius ( r ) = 6

Height ( h ) = 13

Volume of cone = ?

Now, Let's find the volume of cone:

= \pi \:  {r}^{2}  \frac{h}{3}

plug the values

= 3.14 \times  {6}^{2}  \times  \frac{13}{3}

Evaluate the power

= 3.14 \times 36 \times  \frac{13}{3}

Calculate

489.84 units³

Hope this helps..

Best regards!!

6 0
3 years ago
3. A rocket is launched vertically from the ground with an initial velocity of 64 ft / sec . ( a ) Write a quadratic function h
lys-0071 [83]

Answer:

See below.

Step-by-step explanation:

The rocket's  flight is controlled by its initial velocity and the acceleration due to gravity.

The equation of motion is h(t) = ut + 1.2 g t^2 where  u = initial velocity, g = acceleration due to gravity ( = - 32 ft s^-2) and t = the time.

(a) h(t) = 64t  - 1/2*32 t^2

h(t) = 64t - 16t^2.

(b) The graph will be a parabola which opens downwards  with a maximum at the point (2, 64) and x-intercepts at (0, 0) and (4, 0).

The y-axis is the height of the rocket and the x-axis  gives the time.

Maximum height = 64 feet, Time to maximum height = 2 seconds, and time in the air = 4 seconds.

6 0
3 years ago
Read 2 more answers
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