The Area of the Rectangle

Mathematics

1 answer:

pshichka [43]3 years ago

3 0

Answer:

450 I believe is correct

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Find the lateral area of the right prism with height 0.3 dm, if the base of the prism is Isosceles trapezoid with bases of 3 cm

KIM [24]

The lateral area of that prism consists of 4 rectangles.
h = 0.3 dm = 30 cm.
The lateral area:
A = ( a + b + c + c ) · h
A = ( 5 + 3 + 4 + 4 ) · 30
A = 16 · 30 = 480 cm²
Answer: The lateral area of the prism is 480 cm².

8 0

4 years ago

Convert r =-72/12+6 sin theta to Cartesian form.

PolarNik [594]

For converting polar to cartesian form we know
$r = \sqrt{x^2 + y^2}$
$x = rcos \theta , y = rsin \theta$

Given equation is

$r = \frac{-72}{12} + 6sin \theta$

We can simplify it as
$r = -6 + 6sin \theta$

Now we can write it as
$r^2 = -6r + 6rsin \theta$

Now we can use
$r^2 = x^ 2+ y^2 , rsin \theta = y , r = \sqrt{x^2 + y^2}$
So equation we can write it as
$x^2 + y^2 = -6 \sqrt{x^2 + y^2} + 6y$
$x^2 + y^2 - 6y = -6 \sqrt{x^2 + y^2}$
On squaring both sides
$(x^2 + y^2 - 6y)^2 = (-6 \sqrt{x^2 + y^2})^2$
$x^4 + y^4 + 36y^2 +2x^2y^2 -12y^3 -12x^2y = 36(x^2 + y^2)$
$x^4 + y^4 +36y^2 + 2x^2y^2 -12y^3 -12x^2y = 36x^2 + 36y^2$
$x^4 + y^4 + 36y^2 +2x^2y^2 -12y^3-12x^2y - 36x^2 - 36y^2 = 0$
$x^4 + y^4 -12y^3 + 2x^2y^2 - 12x^2y - 36x^2 = 0$