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Vera_Pavlovna [14]
3 years ago
14

Solve for m m m Please help!!!

Mathematics
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

m∠ TRS = 60° , m∠ SRW = 120°

Step-by-step explanation:

First, find x

∠TRS = ∠VRW (vertically opposite angles are equal)

x + 40° = 3x

x - 3x = -40

-2x = -40

x = -40/-2

x = 20

m∠ TRS = 60°  [x + 40 = 20+40 = 60]

m∠ SRW + m∠ TRS = 180° (linear pair)

m∠ SRW + 60° = 180°

m∠ SRW = 180° - 60°

m∠ SRW = 120°

hope this helps you

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Step-by-step explanation:

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  • 9c=5f-160

  • -5f=-160-9c

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2 years ago
A 20 foot extension ladder liens against a building if the base of the latter is placed 6 feet from the base of the building wha
goldenfox [79]

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3 years ago
1. Ben had 50 baseball cards in each of 25 boxes. How many baseball cards did he have altogether?​
Romashka [77]
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8 0
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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

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3 years ago
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nalin [4]

Answer:

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Step-by-step explanation:

use PEMDAS=

2+6=12

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48/24=2

2-1= 1

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