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2 years ago

The length of a rectangle is 4 cm less than twice its width. The perimeter of the

Mathematics

1 answer:

myrzilka [38]2 years ago

4 0

Step-by-step explanation:

Let width be x

according to question

length = 2x-4

perimeter = 58

now,

perimeter = 2(length + width)

58 = 2(x+2x-4)

58 = 6x-8

6x = 66

x = 11

putting value of x in length

2x-4

= 2 X 11 -5

= 18

hence the dimensions of the rectangle is 18 and 11

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Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.

rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed as:

$y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}$

Thus, the area of the surface is as follows:

$\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA$

$= \iint_D \sqrt{x^2 +y^2 +1 } \ dA$

$= \int^{2 \pi}_{0} \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr \ d \theta$

$=2 \pi \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr$

$= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0$

$= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg]$

$= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]$

3 0

2 years ago

A cylinder and a cone have the same volume. The cylinder has radius x and height y. The cone has radius 2x. Find the height of t

weqwewe [10]

V(cylinder)=πR²H
Radius of the cylinder R=x, height of the cylinder H=y.
We can write for the cylinder
V(cylinder)=πx²y

V(cone) =(1/3)πr²h
Radius of the cone r=2x.
We can write for the cone
V(cone)= (1/3)π(2x)²h=(1/3)π *4*x²h

V(cylinder) =V(cone)
πx²y=(1/3)π *4*x²h
y=(4/3)*h
h=(3/4)*y

7 0

3 years ago

X + 3/2 = 3/4 (x - 1/2)

Iteru [2.4K]

this would be your answer, <em>x = 1 - 15/2.</em>

Thanks,

8 0

3 years ago

Please please please please help

Solnce55 [7]

Answer:

31.40