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2 years ago

ITS 25 POINTS PLEASE ANSWER

Mathematics

1 answer:

lord [1]2 years ago

3 0

Answer:

17 rows

Step-by-step explanation:

We know that f(x) = 2x² - 10x models the total number of spaces and x models the number of rows.

Since we are given the total number of spaces of 408. We let f(x) = 408

2x² - 10x = 408

Now we solve as a quadratic, subtract 408 to both sides

2x² - 10x - 408 = 0

Divide 2 on both sides

x² - 5x - 204 = 0

We look for terms that multiply to -204 and add to -5.

Those terms are -17 and 12

x² - 17x + 12x - 204 = 0

x(x-17) + 12(x-17)

(x+12)(x-17)

x = -12 x = 17

We can not have -12 rows but we can have 17 rows.

17 rows is the answer.

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Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = yzi + 4xzj + ex

natima [27]

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The result of the integral is 81π

Step-by-step explanation:

We can use Stoke's Theorem to evaluate the given integral, thus we can write first the theorem:

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S$

Finding the curl of F.

Given $F(x,y,z) = < yz, 4xz, e^{xy} >$ we have:

$curl \vec F =\left|\begin{array}{ccc} \hat i &\hat j&\hat k\\ \cfrac{\partial}{\partial x}& \cfrac{\partial}{\partial y}&\cfrac{\partial}{\partial z}\\yz&4xz&e^{xy}\end{array}\right|$

Working with the determinant we get

$curl \vec F = \left( \cfrac{\partial}{\partial y}e^{xy}-\cfrac{\partial}{\partial z}4xz\right) \hat i -\left(\cfrac{\partial}{\partial x}e^{xy}-\cfrac{\partial}{\partial z}yz \right) \hat j + \left(\cfrac{\partial}{\partial x} 4xz-\cfrac{\partial}{\partial y}yz \right) \hat k$

Working with the partial derivatives

$curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(4z-z\right) \hat k\\curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k$

Integrating using Stokes' Theorem

Now that we have the curl we can proceed integrating

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S$

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot \hat n dS$

where the normal to the circle is just $\hat n= \hat k$ since the normal is perpendicular to it, so we get

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S \left(\left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k\right) \cdot \hat k dS$