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3 years ago

If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. *

Mathematics

2 answers:

damaskus [11]3 years ago

6 0

Answer:

Yes

Step-by-step explanation:

Yes , if both pairs of opposite sides of a quadrilateral are congruent then the quadrilateral is a parallelogram . The parallelogram has opposite sides equal and parallel . Diagonals bisect each other . The sum of alternate angles is 180° and the opposite angles are equal .

Gennadij [26K]3 years ago

5 0

Answer:

true

Step-by-step explanation:

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3 years ago

When I triple my number and add five I get twenty six what is my number

fenix001 [56]

26-5=21

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3 years ago

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-4x-2y=-12 <br>4x+8y=-24

sweet [91]

Answer:

x = 6

y = -6

Step-by-step explanation:

By adding both equations :-

=》-4x -2y + 4x + 8y = -12 + (-24)

=》-4x + 4x + 8y - 2y = -12 - 24

=》6y = -36

=》y = -36 ÷ 6

=》y = -6

putting the value of y in equation 2

=》4x + 8y = -24

=》4x + (8 × -6) = -24

=》4x - 48= -24

=》4x = 48 - 24

=》x = 24 / 4

=》x = 6

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3 years ago

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The sum of a rational number and an irrational number equals.

VashaNatasha [74]

Answer:

I think D. but I'm not sure

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3 years ago

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interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$