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2 years ago

What is the area, in square feet, of the shaded part of the rectangle below?

Mathematics

1 answer:

likoan [24]2 years ago

8 0

<h3>Answer: 208</h3>

===============================================

Work Shown:

A = area of the rectangle

A = length*width

A = 16*17

A = 272 sq ft

B = area of triangle

B = base*height/2

B = 16*8/2 ..... note that 17-9 = 8

B = 16*4

B = 64 sq ft

C = area of shaded trapezoid

C = A - B

C = 272 - 64

C = 208 sq ft

--------------------

Alternative Method:

Apply the trapezoid area formula

b1 = 17 and b2 = 9 are the parallel bases

h = 16 is the height perpendicular to either base

A = h*(b1+b2)/2

A = 16*(17+9)/2

A = 16*(26)/2

A = 16*13

A = 208 sq ft

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Which expression can be used to find the surface area of the following square pyramid?

telo118 [61]

Answer:

Step-by-step explanation:

The slant height of one side of this pyramid is 5, and the base of this side is 4. Thus, the area of one slant side is (1/2)(5)(4) = 10 units^2.

There are 4 such sides. Thus, the total slant surface area is 4(10 units^2), or 40 units^2.

If you also want to include the base area, the total would be

40 units^2 + 16 units^2 = 56 units^2.

8 0

3 years ago

26 × (– 48) + (– 48) × (–36).

Rina8888 [55]

Answer:

480

Step-by-step explanation:

(26*-48)+(-48*-36)

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6 0

1 year ago

What is the measure of angle RST? Help!!

erik [133]

Answer:

Step-by-step explanation:

Angle Formed by Two Chords = 1/2(sum of Intercepted Arcs)

RST = 1/2 ( 47+77)

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4 0

3 years ago

Read 2 more answers

Factor the four-term polynomial ab + ac - 4b - 4c

Troyanec [42]

A b + a c - 4 b - 4 c = a ( b + c ) - 4 ( b + c ) =
= (b + c) (a - 4)

5 0

2 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

4 0

3 years ago