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jeka94
2 years ago
5

What is the volume of the Triangular prism? Volume = 1/2 LWH (Geometry)

Mathematics
1 answer:
Mariulka [41]2 years ago
3 0

Answer:

B

Step-by-step explanation:

You might be interested in
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
What is an expression that equals 24 that includes an exponent
statuscvo [17]
5∧2(exponent of 2) - 1 =24
7 0
2 years ago
I’m need this worked out step by step by tonight
Pie

9514 1404 393

Answer:

  -3 ≤ x ≤ 19/3

Step-by-step explanation:

This inequality can be resolved to a compound inequality:

  -7 ≤ (3x -5)/2 ≤ 7

Multiply all parts by 2.

  -14 ≤ 3x -5 ≤ 14

Add 5 to all parts.

  -9 ≤ 3x ≤ 19

Divide all parts by 3.

  -3 ≤ x ≤ 19/3

_____

<em>Additional comment</em>

If you subtract 7 from both sides of the given inequality, it becomes ...

  |(3x -5)/2| -7 ≤ 0

Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well.

__

For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...

  |f(x)| < (some number) . . . . or ≤

If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:

  |f(x)| > g(x)   ⇒   f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0

Any solutions to these inequalities must respect their domains.

8 0
3 years ago
If a triangle has one obtuse angle, then it is an obtuse triangle.
nalin [4]
No the sum of measures would exceed 180, therefore it would not be a obtuse triangle
6 0
3 years ago
On sunday 1/16 of the people who visited for senior citizens, 3/16 or infants, 3/8 were children, and 3/8 for adults. How many o
Trava [24]
1/16=1,003, so

senior= 1003
infants=3009
children=6018
adults=6018
3 0
3 years ago
Read 2 more answers
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