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4 years ago

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Lucas plants 20% of his garden with cucumber plants. The cucumber plants cover an area of 24m square. What is the total area of

Lucas's garden?

1 answer:

enyata [817]4 years ago

7 0

Answer:

<em>120m square.</em>

Step-by-step explanation:

If on average 20% of the garden is equal to 24m square, 24m * 4 = 80%, which is 96.

96msq + 24msq = 120msq

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The plane x + y + 2z = 12 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on the ellipse that are nearest t

Soloha48 [4]

The distance between a point $(x,y,z)$ and the origin is $\sqrt{x^2+y^2+z^2}$ . But since $(\sqrt{f(x)})'=\frac{f'(x)}{2\sqrt{f(x)}}$ , both $f(x)$ and $\sqrt{f(x)}$ have the same critical points, so we can consider instead the squared distance, $x^2+y^2+z^2$ .

We're looking for the extrema of $x^2+y^2+z^2$ subject to $x+y+2z=12$ and $z=x^2+y^2$ . The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(x+y+2z-12)+\mu(z-x^2-y^2)$

with critical points where the partial derivatives vanish:

$L_x=2x+\lambda-2\mu x=0\implies\lambda=2x(\mu-1)$

$L_y=2y+\lambda-2\mu y=0\implies\lambda=2y(\mu-1)$

$L_z=2z+2\lambda+\mu=0$

$L_\lambda=x+y+2z-12=0$

$L_\mu=z-x^2-y^2=0$

From the first two equations, it follows that $x=y$ .

Then in the last two equations,

$x+y+2z-12=0\implies x+z=6$

$z-x^2-y^2=0\implies z=2x^2$

$\implies x+2x^2=6\implies2x^2+x-6=(2x-3)(x+2)=0\implies x=\dfrac32\text{ or }x=-2$

If $x=\frac32$ , then $z=6-\frac32=\frac92$ ; if $x=-2$ , then $z=8$ .

So there are two critical points, $\left(\frac32,\frac32,\frac92\right)$ and $(-2,-2,8)$ .

Let $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ . We have a minimum distance of $f\left(\frac32,\frac32,\frac92\right)=\boxed{\frac{3\sqrt{11}}2}$ and maximum distance of $f(-2,-2,8)=\boxed{6\sqrt2}$ .

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4 years ago

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Lynna [10]

Definition of acute angle

6 0

4 years ago

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MrMuchimi

Answer:

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Step-by-step explanation:

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3 years ago

8. Solve the equation<br>x² - 1 = 48

crimeas [40]

Answer:

x^2-1=48

x^2=48+1

x^2=49

x= $\sqrt{49}$

x=7

Step-by-step explanation:

5 0

3 years ago

Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle, or a parallelogram. The

vodomira [7]

Answer:

2) ABCD is a RECTANGLE.

Perimeter = 30 units

Step-by-step explanation:

Here, the given points of the quadrilateral are:

A(6, –4), B(11, –4), C(11, 6), D(6, 6)

Now, if P(a,b) and Q(c,d) are any two given points, then the distance between them is given by DISTANCE FORMULA as:

$PQ = \sqrt{(c-a)^2 + (d-b)^2$

Use this to find the length of all 4 sides:

AB with coordinates A(6, –4), B(11, –4) is given as:

$AB = \sqrt{(11-6)^2 + (-4-(-4))^2} = \sqrt{(5)^2 + 0} = 5$

⇒ AB = 5 units

BC with coordinates B(11, –4), C(11, 6) is given as:

$BC = \sqrt{(11-11)^2 + (6-(-4))^2} = \sqrt{(0)^2 + 10^2} = 10$

⇒ BC = 10 units

CD with coordinates C(11, 6), D(6, 6) is given as:

$CD = \sqrt{(11-6)^2 + (6-6)^2} = \sqrt{(5)^2 + 0} = 5$

⇒ CD = 5 units

AD with coordinates A(6, –4),D(6, 6) is given as:

$AD = \sqrt{(6-6)^2 + (-4-(-4))^2} = \sqrt{(0)^2 + 10^2} = 10$

⇒ AD = 10 units

Now, here AB = CD = 5 units, AD = BC = 10 units

Also, in a RECTANGLE, OPPOSITE SIDES ARE EQUAL IN LENGTH.

Hence, ABCD is a RECTANGLE.

Perimeter = AB + BC+ CD + AD = 5 + 10 + 5 + 10 = 30 units

8 0

3 years ago