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3 years ago

5

Duc has a trapezoid shaped section of his yard fenced off for his pets. The lengths of the parallel sides of the trapezoid are 8

ft and 14 ft. The height of the trapezoid section is 4 ft.
What is the area of the trapezoid section of the yard?

2 answers:

posledela3 years ago

7 0

The area is equal to 44 ft.

Hope this answer helps!

notka56 [123]3 years ago

6 0

Answer:

LOOK THE RIGHT ANWER IS DOWN BELLOW

The answer is 44 ft.

I hope this helps

dont forget to hit that thanks button

Step-by-step explanation:

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If AB=12 cm and BC =13 cm then AC = 25 cm

Maru [420]

Answer:

AB=12

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2 years ago

Please help me I really need it

Art [367]

Answer:

Angle C

Step-by-step explanation:

By applying sine rule in the given triangle,

$\frac{\text{sinB}}{CD}=\frac{\text{sinC}}{BD}=\frac{\text{sinD}}{BC}$

$\frac{\text{sinB}}{\text{sinC}}=\frac{\text{CD}}{\text{BD}}$

Therefore, sides of the given triangle will be in the same ratio as the sine of the angles.

Since, BD > BC > CD

Therefore, Opposite angles of these sides will be in the same ratio.

∠C > ∠D > ∠B

Largest angle of the triangle is angle C.

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3 years ago

What is the y-intercept of the line shown?

lys-0071 [83]

Simple.....

always remember the y-intercept is where your line hits the y-axis... or where it hits the vertical axis...

As we can see this line hits the y-axis at (0,3)-->> y-intercept=3

Thus, your answer.

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3 years ago

Read 2 more answers

Can you help Plzz..<br> Explain the process for find the product of two integers????

Katyanochek1 [597]

The process for finding the product of two integers is by multiplying them .

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4 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

3 years ago