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Naily [24]
3 years ago
9

A digital picture is made up of many small dots called pixels. Rayma has a picture that is 40 pixels wide and wants to enlarge i

t. Every time she increases the size, the width of the picture doubles. Write a formula that gives the width, w, of the picture after doubling it n times.
Mathematics
1 answer:
Xelga [282]3 years ago
4 0
The answer to your question is w =40(2)n [n is in exponent form I just couldn’t type it that way ]
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AleksandrR [38]

Answer:b2254

Step-by-step explanation:

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|2/3-3/2|3^2-17/3 a)11/16 b)11/6 c)79/6 d)83/6
vfiekz [6]

Answer:  The correct answer is:  [B]:  

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                  →   " \frac{11}{6} " .

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Explanation:  

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Let's rewrite this problem as:  

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Simplify:

 "  | \frac{2}{3} - \frac{3}{2} | * 3^{2} - \frac{17}{3} |  " ;

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3 years ago
Solve the equation 5.5(x-3)=17+10.5
sveticcg [70]

Answer:

x= 8

Step-by-step explanation:

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8 0
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In the figure, PQ is parallel to RS. The legth of RP is 4 cm; the length of PT is 16 cm; the length of QT is 20 cm. What is the
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Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=x^2+y

\dfrac{\partial f}{\partial y}=y^2+x

\dfrac{\partial f}{\partial z}=ze^z

Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

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\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

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and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

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\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
3 years ago
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