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Ivan
3 years ago
7

(Photo)

Mathematics
2 answers:
gogolik [260]3 years ago
8 0
I think it's B. 168 units
Because I multiplied them all together and halved the answer
Nuetrik [128]3 years ago
4 0
The answer is B. 168 3(cube)
You welcome
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Can anyone help me with this.<br>Thank You.
Cerrena [4.2K]
I can not see the question it is blurry
4 0
3 years ago
I need project on thales theorem
Naily [24]
Dont under stand put pitire so we can help u
7 0
3 years ago
In the diagram below, if &lt; A = 97 °, &lt; B = 46 °, &lt; D = 97 ° and &lt; F = 37 °, we can say that
sdas [7]

Answer:

  (a)  The two triangles are similar by AA

Step-by-step explanation:

Similar triangles have congruent corresponding angles and proportional corresponding side lengths.

<h3>Third angle</h3>

The third angle in triangle ABC is found using the fact that the sum of angles is 180°.

  ∠C = 180° -∠A -∠B

  ∠C = 180° -97° -46° = 37°

Two of the angles, A and C, match the measures of two of the angles in triangle DEF. The matches are ...

  ∠A = ∠D = 97°

  ∠C = ∠F = 37°

The two triangles are similar by AA.

__

<em>Additional comment</em>

The similarity statement can be ΔABC ~ ΔDEF.

8 0
2 years ago
2000 people are selected randomly from a certain population and it is found that 389 people in the sample are over 6 feet tall.
Airida [17]

Answer: 0.195

Step-by-step explanation:

given data:

population = 2000 people.

people Who are over 6 feet tall = 389.

Solution:

the point estimate of people over 6feet tall

= no of people over 6 feet tall / total population size

= 389/2000

= 0.195

the point estimate of people over 6 feet tall is 0.195

8 0
3 years ago
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
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