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EleoNora [17]
3 years ago
6

Find the volume Find the volume

Mathematics
2 answers:
ValentinkaMS [17]3 years ago
5 0
The answer should be v=360
baherus [9]3 years ago
4 0

v=360

formula

v =  \frac{1}{2}(bl)h

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Help me with precal pleaseee
yuradex [85]

Answer:  Choice C

\left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right]

==================================================

Explanation:

Let's look at the function y = sec(x) first, which is the secant function.

Recall that secant is 1 over cosine, so sec(x) = 1/cos(x)

We can't divide by zero, so cos(x) = 0 can't be allowed. If x = pi/2, then cos(pi/2) = 0 will happen. So we must exclude pi/2 from the domain of sec(x).

If we look at the interval from 0 to pi, then the domain of sec(x) is 0 \le x < \frac{\pi}{2} \ \ \text{ and } \ \ \frac{\pi}{2} < x \le \pi

we can condense that into the interval notation \left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right]

Note the use of curved parenthesis to exclude the endpoint; while the square bracket includes the endpoint.

So effectively we just poked at hole at x = pi/2 to kick that out of the domain. I'm only focusing on the interval from 0 to pi so that secant is one to one on this interval. That way we can apply the inverse. When we apply the inverse, the domain and range swap places. So the range of arcsecant, or \sec^{-1}(x) is going to also be \left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right]

8 0
3 years ago
6.
NISA [10]

Answer:

6

Step-by-step explanation:

5 0
3 years ago
Triangle DEF has vertices D = (-4,-4), E = (-2,-4), and F = (-3,-1). 5! 3 2 1 -7 6 5 4 3 2 2 3 4 5 6 10 5 3. What is the area of
maxonik [38]

Answer:

Area = 3

Step-by-step explanation:

Given

D = (-4,-4)

E = (-2,-4)

F = (-3,-1)

Solving (a) and (b):

See attachment for plot and labelled vertices

Solving (c): The area

This is calculated using:

Area = \frac{1}{2}|x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2+x_3y_1 - x_1y_3|

Where:

D = (-4,-4) -- (x_1,y_1)

E = (-2,-4)  -- (x_2,y_2)

F = (-3,-1) -- (x_3,y_3)

This gives:

Area = \frac{1}{2}|(-4*-4) - (-4*-2) + (-2 *-1) - (-3*-4)+(-3*-4) - (-4*-1)|

Area = \frac{1}{2}|16 - 8 +2- 12+12 - 4|

Area = \frac{1}{2}|6|

This gives

Area = \frac{1}{2} * 6

Area = 3

7 0
3 years ago
A line segment is part of a line, true or false?
11Alexandr11 [23.1K]

Answer:

geometry ?

Step-by-step explanation:

If so I gotchu this is my subject

7 0
3 years ago
Read 2 more answers
If h(x) = 5x − 3 and j(x) = −2x, solve h[j(2)] and select the correct answer below.
Sati [7]
Sub 2 for x in j(x) and evalueate
then sub that result for x in h(x)

j(2)=-2(2)
j(2)=-4

now sub-4 for  x in h(x)

h(-4)=5(-4)-3
h(-4)=-20-3
h(-4)=-23

answer is -23
8 0
3 years ago
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