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2 years ago

Irrigation is important because it

Mathematics

1 answer:

olasank [31]2 years ago

8 0

Answer:

helps to cultivate superior crops with the water supply as per need of the crops. Ultimately it helps in economic development. Irrigation water improves water conditions in the soil, increases the water content of plant fibers, dissolves nutrients & makes them available to plants.

Step-by-step explanation:

You might be interested in

Properties of 2-Dimensional Figures 1-7

Basile [38]

Answer:

1. PNT

2.RNM

3.QNM

4.UYZ

5. None

6.VYZ

7.none

Step-by-step explanation:

suplametery means when you add to angles together it will equal 180

while complementery is when to angels are added together to make 90 degrees

4 0

2 years ago

the radius of the curvature is 10ft and the height of the segment is 2ft whatthe length of the chord ? a.10ft, b.15, c.7.5 ,d.12

arsen [322]

Radius of curvature = 10'
Diameter of circle = 2 * 10' = 20'
height of segment = 2'
Let semi-chord be x.
Then by the intersecting chord property,
2*(20-2)=x^2
x^2=36
x=sqrt(36)=6
=>
chord=2x=2*6=12'

3 0

2 years ago

How do i solve this

MAXImum [283]

Answer:

x = 0, x = 3

Step-by-step explanation:

Given

- 4x² = - 12x ( add 12x to both sides )

- 4x² + 12x = 0 ← factor out - 4x from each term

- 4x(x - 3) = 0

Equate each factor to zero and solve for x

- 4x = 0 ⇒ x = 0

x - 3 = 0 ⇒ x = 3

5 0

3 years ago

Please help w this! Its a calculus question! look at the picture for the problem,

neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines $x=\pm3\sqrt2$ and $y=\pm3\sqrt2$ .

Let $R$ be the region bounded by the line $x=3\sqrt2$ and the circle $x^2+y^2=36$ (the rightmost blue region). The right side of the circle can be expressed in terms of $x$ as a function of $y$ :

$x^2+y^2=36\implies x=\sqrt{36-y^2}$

Then the area of this circular segment is

$\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy$

Substitute $y=6\sin t$ , so that $\mathrm dy=6\cos t\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18$

Then the area of the entire blue region is 4 times this, a total of $\boxed{36\pi-72}$ .

Alternatively, you can compute the area of $R$ in polar coordinates. The line $x=3\sqrt2$ becomes $r=3\sqrt2\sec\theta$ , while the circle is given by $r=6$ . The two curves intersect at $\theta=\pm\dfrac\pi4$ , so that

$\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18$

so again the total area would be $36\pi-72$ .

Or you can omit using calculus altogether and rely on some basic geometric facts. The region $R$ is a circular segment subtended by a central angle of $\dfrac\pi2$ radians. Then its area is