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jasenka [17]
3 years ago
14

It says to find the value of x is this right?

Mathematics
2 answers:
dalvyx [7]3 years ago
8 0
The answer is x equals 42.
ss7ja [257]3 years ago
6 0

Yes, because

3x-2+55+85 = 180

138+x = 180

x = 42

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Yolanda is making a banner for a school pep rally. She cuts fabric in the shape of a parallelogram. The angle at the bottom-left
Roman55 [17]

Answer: x=100

Step-by-step explanation:

In parallelograms the opposite angles are equal. As the bottom left angle is 80° the opposite angle (top right), is also 80°. All 4 angles in a quadrilateral must add up to 360. Because you know 2 angles added together are 160° the other two angles must equal 200°. As opposite angles are equal you know 2x=200, divide by 2 x=100

8 0
2 years ago
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Carlos earns $3.50 a day on his paper route. He runs
loris [4]

Answer:

their are 4 weeks in a month so 4*7=

28

so its 1/28 which is equals to (1 divided by 28)

3.57%

Hope This Helps!!!

6 0
2 years ago
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What is the equation of the line that passes through the point (6, 8) and has a slope 1/2
Korvikt [17]

Answer:

y=1/2x+5

Step-by-step explanation:

Plug it into the formula for a point and a slope.

(y-(y(subscript 1)))=(m)(x-(x(subscript 1)))

Y-8=1/2(x-6)

Then multiply the 1/2 to the x and -6. After you do that, move everything so that it has y by itself on the left. Your ending should be y=1/2x+5.

7 0
2 years ago
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

So the scalar potential function is

\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

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

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:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\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
2 years ago
Please Help!
victus00 [196]
The GCF is 2
Pull out 2 from each number:
2(2n + 5)
Hope this helps!
7 0
3 years ago
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