All categories

3 years ago

A) Why are all parallelograms not rectangles?

1 answer:

8 0

a) Rectangles always have 4 right angles (90 degrees), but parallelograms do not always have 4 right angles rather they can, but do not have to. Rectangles have to.

b) A square is a type of rectangle, but a rectangle is not a type of square. Both rectangles & squares have 4 right angles and are always quadrilaterals, but squares always have all sides that have the same length, but rectangles do not have to have their sides all equivalent lengths.

c) In the case of a parallelogram, the opposite sides are equal whereas in a rhombus all four sides are equal.

You might be interested in

Find the length of x

zaharov [31]

Answer:

The answer is 2

Step-by-step explanation:

The hypotenuses of the triangles are different. One is 4 and the other is 5.

Lets use ratios to find the relationship of the hypotenuse and the base.

2.5:5

Turn this into a fraction

2.5/5

Now divide

2.5/5 = 0.5

The base of the triangle is half the length of the hypotenuse.

So 4 is x’s hypotenuse so:

4 x 0.5 = 2

X = 2

6 0

2 years ago

Calcule a medida da altura da cachoeira do Aracá, da torre Eiffel e da torre CN,

daser333 [38]

Answer:

Step-by-step explanation:

6 0

1 year ago

If f(x) = 3 x-2 what is f(5.9)

Korvikt [17]

$3(5.9) = 17.7 \: \: \: \: 17.7 - 2 = 15.7$

3 0

2 years ago

Read 2 more answers

Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

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

And the normal vector is given by

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

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

6 0

2 years ago

Solve the equation A = 2LW + 2LH + 2WH for H.

Fed [463]

H = A - 2LW / 2 (L+W)

6 0

3 years ago