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spayn [35]
3 years ago
9

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
Lunna [17]3 years ago
6 0

Answer:

6^11

Step-by-step explanation:

you have 6 and 6 so it stays 6 and then you add the exponents on this problem which gives u 11

Norma-Jean [14]3 years ago
5 0

Answer:

6^11

Step-by-step explanation:

In this case the base remains the same but you add exponents. Try it on your calculator

6^5 x 6^6 = 362797056

6^11 = same as above

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Widget Wonders produces widgets. They
navik [9.2K]
Find the total cost of producing 5 widgets. Widget Wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $15.50 to produce 3 widgets, $23.50 to produce 7 widgets, and $56 to produce 12 widgets.
OK...so we have
a(7)^2 + b(7) + c = 23.50 → 49a + 7b + c = 23.50 (1)
a(3)^2 + b(3) + c = 15.50 → 9a + 3b + c = 15.50 subtracting the second equation from the first, we have
40a + 4b = 8 → 10a + b = 2 (2)
Also
a(12)^2 + b(12) + c = 56 → 144a + 12b + c = 56 and subtracting (1) from this gives us
95a + 5b = 32.50
And using(2) we have
95a + 5b = 32.50 (3)
10a + b = 2.00 multiplying the second equation by -5 and adding this to (3) ,we have
45a = 22.50 divide both sides by 45 and a = 1/2 and using (2) to find b, we have
10(1/2) + b = 2
5 + b = 2 b = -3
And we can use 9a + 3b + c = 15.50 to find "c"
9(1/2) + 3(-3) + c = 15.50
9/2 - 9 + c = 15.50
-4.5 + c = 15.50
c = 20
So our function is
c(x) = (1/2)x^2 - (3)x + 20
And the cost to produce 5 widgets is = $17.50
7 0
3 years ago
If square ABCD with side length of 7 units is reflected over the line y=x, what is length A'B'?
Art [367]

Answer:

  7

Step-by-step explanation:

Reflection is a rigid transformation, so the side length does not change.

  A'B' = 7

5 0
3 years ago
The area of an equilateral triangle with a 6 inch apothem
skad [1K]
Answer is 187.06 and the work is in the attached picture.

6 0
3 years ago
The circumference is __
Anestetic [448]

Answer:

<em>As </em><em>we </em><em>know </em><em>that </em><em>there </em><em>is </em><em>a </em><em>radius </em><em>which </em><em>is </em><em>2 </em><em>cm</em>

<em>so </em>

<em>circumference </em><em>=</em><em> </em><em>2</em><em> </em><em>π</em><em> </em><em>r</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>2</em><em> </em><em>*</em><em> </em><em>2</em><em>2</em><em>/</em><em>7</em><em> </em><em>*</em><em> </em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>1</em><em>2</em><em>.</em><em>5</em><em>7</em><em>c</em><em>m</em>

<em>it's </em><em>circumference </em><em>is </em><em>1</em><em>2</em><em>.</em><em>5</em><em>7</em><em> </em><em>cm</em>

3 0
3 years ago
Read 2 more answers
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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