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3 years ago

Can you show that this quadrilateral is a parallelogram?

Mathematics

1 answer:

larisa [96]3 years ago

5 0

The diagonals intersect perpendicularly, two consecutive angled are supplementary, and that opposite angles are equal

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Let g(x)=Intragal from 0 to x f(t) dt, where r is the function whos graph is shown.

leonid [27]

$\displaystyle g(x) = \int_0^x f(t) \, dt$

then g(x) gives the signed area under f(x) over a given interval starting at 0.

In particular,

$\displaystyle g(0) = \int_0^0 f(t) \, dt = 0$

since the integral of any function over a single point is zero;

$\displaystyle g(4) = \int_0^4 f(t) \, dt = 8$

since the area under f(x) over the interval [0, 4] is a right triangle with length and height 4, hence area 1/2 • 4 • 4 = 8;

$\displaystyle g(8) = \int_0^8 f(t) \, dt = 0$

since the area over [4, 8] is the same as the area over [0, 4], but on the opposite side of the t-axis;

$\displaystyle g(12) = \int_0^{12} f(t) \, dt = -8$

since the area over [8, 12] is the same as over [4, 8], but doesn't get canceled;

$\displaystyle g(16) = \int_0^{16} f(t) \, dt = 0$

since the area over [12, 16] is the same as over [0, 4], and all together these four triangle areas cancel to zero;

$\displaystyle g(20) = \int_0^{20} f(t) \, dt = 24$

since the area over [16, 20] is a trapezoid with "bases" 4 and 8, and "height" 4, hence area (4 + 8)/2 • 4 = 24;

$\displaystyle g(24) = \int_0^{24} f(t) \, dt = 64$

since the area over [20, 24] is yet another trapezoid, but with bases 8 and 12, and height 4, hence area (8 + 12)/2 • 4 = 40, which we add to the previous area.

5 0

3 years ago

8. f(x) = 2x + 2<br> what’s the solve for x ?

Aleksandr-060686 [28]

Answer:

X=-1

Step-by-step explanation:

0=2x+2

subtract 2 out

-2 = 2x

divide 2 on both sides x = -1

5 0

3 years ago

[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation

irina1246 [14]

Answer:

$y=-\frac{158}{579}$

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

$A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]$

And the vector B is formed with the solution of each equation of the system: $b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]$

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called $A_{2}$ .

$A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]$

The value of y using Cramer's rule is:

$y=\frac{det(A_{2}) }{det(A)}$

Find the value of the determinant of each matrix, and divide:

$y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}$