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monitta
3 years ago
10

Plea look at the photo of Parallelogram properties.

Mathematics
1 answer:
Yuliya22 [10]3 years ago
3 0

Answer:

1. 45

2. 105

3. 95

4. 85

Step-by-step explanation:

For problems 1 and 4:

Consecutive angles are supplementary in a parallelogram, which means they equal 180.

180 - 135 = 45

180 - 95 = 85

For problems 2 and 3:

Opposite angles are congruent or equal each other in a parallelogram,

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Step-by-step explanation:

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3 years ago
Would you need to use the chain rule to find the derivative of this function?
Nutka1998 [239]

Answer:

TRUE. We need to use the chain rule to find the derivative of the given function.

Step-by-step explanation:

Chain rule to find the derivative,

We have to find the derivative of F(x)

If F(x) = f[g(x)]

Then F'(x) = f'[g(x)].g'(x)

Given function is,

y = \sqrt{2x+3}

Here g(x) = (2x + 3)

and f[g(x)] = \sqrt{2x+3}

\frac{dy}{dx}=\frac{d}{dx}(\sqrt{2x+3}).\frac{d}{dx} (2x+3)

y' = \frac{1}{2}(2x+3)^{(1-\frac{1}{2})}.(2)

   = (2x+3)^{-\frac{1}{2}}

y' = \frac{1}{\sqrt{2x+3}}

Therefore, it's true that we need to use the chain rule to find the derivative of the given function.

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
4 years ago
Vernon is having a fish fry. Catfish is on sale for 6.10 per pound. Vernon buys 2 3 5 pounds of catfish. How do you write 2 3 5
konstantin123 [22]

Answer:

76.8

Step-by-step explanation:

6 0
3 years ago
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