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hjlf
3 years ago
10

What is the value of x ÷ y + z if x = 32, y = 4, and z = 7? PLEASE HELP!

Mathematics
1 answer:
Zepler [3.9K]3 years ago
7 0
The answer is 15 my dear boi, just plug in the numbers
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Use the following figure to answer the question. if line t is perpendicular to both line l and m then 1 and 2 are both right ang
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3 years ago
The sum of two numbers is -6. The difference is 16. Find the two numbers.
Bess [88]

Answer:

-5\:\mathrm{and}\: 11

Step-by-step explanation:

We can write the following system of equations:

\begin{cases}x+y=6\\x-y=16\\\end{cases}

Adding both equations together, we isolate x and get:

2x=22,\: x=11

Plugging x=11 in any of the equations, we can solve for y:

11+y=6,\: y=-5

Verify that the solution pair (11, -5) works \checkmark

Therefore, the two numbers are \fbox{$-5\:\mathrm{and}\: 11$}.

7 0
3 years ago
Where does the 1 come from in (1.0325)
NemiM [27]

Answer: *1*.0325

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3 0
2 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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