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VashaNatasha [74]
3 years ago
10

Which of the following statements are not true about parallelograms?

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
3 0

9514 1404 393

Answer:

  1, 2

Step-by-step explanation:

1. The area of a parallelogram is the product of length and width. The given statement is FALSE.

2. The angles formed by the diagonals are linear pairs, so supplementary. The given statement is FALSE.

3. TRUE

4. TRUE

5. TRUE

__

Statements 1 and 2 are not true.

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Which statement best explains the relationship between
Nataly [62]

Answer:

The best statement which explains the relationship between lines AB and CD is "They are parallel because their slopes are equal" ⇒ A

Step-by-step explanation:

  • Parallel lines have equal slopes and different y-intercepts
  • The rule of the slope of a line passes through points (x1, y1) and (x2, y2) is m = \frac{y2-y1}{x2-x1}

In the given figure

∵ The blue line passes through points A and B

∵ A = (-4, -2) and B = (4, 4)

∴ x1 = -4 and y1 = -2

∴ x2 = 4 and y2 = 4

→ Substitute them in the rule of the slope

∵ m(AB) = \frac{4--2}{4--4} = \frac{4+2}{4+4} = \frac{6}{8} = \frac{3}{4}

∴ The slope of line AB is  \frac{3}{4}

∵ The green line passes through points C and D

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→ Substitute them in the rule of the slope

∵ m(CD) = \frac{0--3}{4-0} = \frac{0+3}{4} = \frac{3}{4}

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∵ The slope of line AB = the slope of line CD

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The best statement which explains the relationship between lines AB and CD is "They are parallel because their slopes are equal"

7 0
3 years ago
Integrate xe^(-x^2).
Gekata [30.6K]
You can easily integrate it using a simple substitution:

\large\begin{array}{l} \mathsf{\displaystyle\int\!x\,e^{-x^2}\,dx}\\\\ =\mathsf{\displaystyle\int\!\left(-\frac{1}{2}\right)\cdot (-2)x\,e^{-x^2}\,dx}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\int\!\,e^{-x^2}\cdot (-2x)\,dx\qquad\quad(i)}\\\\ \end{array}


\large\begin{array}{l} \textsf{Let}\\\\ \mathsf{-x^2=u,}\quad\textsf{then}\quad\mathsf{-2x\,dx=du}\\\\\\ \textsf{so (i) becomes}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\int\!e^u\,du}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\cdot e^u+C}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\cdot e^{-2x}+C}\\\\\\ \boxed{\begin{array}{c} \mathsf{\displaystyle\int\!x\,e^{-x^2}\,dx=-\,\frac{1}{2}\,e^{-x^2}+C} \end{array}}\qquad\checkmark \end{array}


If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2159730


\large\textsf{I hope it helps. :-)}


Tags: <em>integrate substitution indefinite integral exponential composite calculus</em>

7 0
4 years ago
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