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BartSMP [9]
3 years ago
10

Look at the images please I WILL MRK THE BRIANLEST

Mathematics
1 answer:
professor190 [17]3 years ago
8 0

Answer: V=234\ yd^3

Step-by-step explanation:

By definition, the volume of a rectangular prism can be calculated with the following formula:

V=lwh

Where "l" is the length, "w" is the width and "h" is the height of the rectangular prism.

In this case, you can identify that the length, the width and the height of this rectangular prism given in the exercise, are:

l=13\ yd\\\\w=6\ yd\\\\h=3\ yd

Then, knowing its dimensions, you can substitute them into the formula:

V=(13\ yd)(6\ yd)(3\ yd)

Finally, evaluating, you get that the volume of that rectangular prism is:

V=234\ yd^3

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Answer:

singe

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2 years ago
The town of Passaic had 58,753 people in 1990 and 71,509 people in 2014.
OLEGan [10]

Answer:

21.71% increase

Step-by-step explanation:

increase = Increase ÷ Original Number × 100

12756/58753*100

0.217112318*100

21.71% increase.

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2 years ago
What is the product of the polynomials below? (6x^2-3x-6)(4x^2+5x+4)​
Lunna [17]

Answer:

(6 - 5x(2))(x(4) - x(3))

(6 - 5x^2) (x^4 - x^3)

6x^4 - 6x^3 - 5x^6 + 5x^5

-5x^6 + 5x^5 + 6x^4 - 6x^3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
Harrizon [31]

Answer:

The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

f_z = -ye^{yz}

we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

3 0
3 years ago
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tiny-mole [99]

Answer:

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