Ask question

Ask question

All categories

3 years ago

9

The diameter of a circular swimming pool is 8.5 m. What is the approximate circumference of the pool? Round to the nearest tenth

. Use 3.14 for π

2 answers:

Anna007 [38]3 years ago

6 0

The circumference would be 26.7 M.

Nikolay [14]3 years ago

3 0

here,

d=8.5m

c=π×d

=3.14*8.5

=26.67

=26.7m

the circumference of the pool is 26.7 m.

You might be interested in

Describe the transformation that maps the pre-image (AA) to the image (AA).

asambeis [7]

Answer:

it is reflection on y-axis

8 0

3 years ago

If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?

kondaur [170]

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Here, $x_1=-8, y_1=-6, x_2=-4, y_2=-14$

$$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)$

$$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)$

$$P(x, y) =(-6, -10)$

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Here, $x_1=-6, y_1=-10, x_2=-8, y_2=-6$

$r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}$

$r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}$

$r=\sqrt{(-2)^{2}+(4)^{2}}$

$r=\sqrt{20}$ units

The standard form of the equation of a circle is

$(x-a)^{2}+(y-b)^{2}=r^{2}$ , where (a, b) are center and r is the radius.

Here, center = (–6, –10) and $r=\sqrt{20}$

$(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}$

$(x+6)^{2}+(y+10)^{2}=20$

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

4 0

3 years ago

Simplify the expression <br><br> c+2+c+c+4

professor190 [17]

Answer:

2+4+ccc

Step-by-step explanation:

7 0

2 years ago

pychu [463]

Answer:

68.8 sq. inches.

Step-by-step explanation:

The triangular surfaces have sides 2 in by 3 in by 4 in.

So, half perimeter, s = $\frac{2 + 3 + 4}{2} = 4.5$

So, the area of the triangular face is $\sqrt{s(s - 4)(s - 2)(s - 3)} = \sqrt{4.5 \times 0.5 \times 2.5 \times 1.5} = 2.9$ sq. inches.

Now, the total surface area of the right triangular prism will be (2.9 × 2) + 7(2 + 3 + 4) = 68.8 sq. inches. (Answer)

4 0

2 years ago

Find a point on the ellipsoid x2+y2+4z2=36x2+y2+4z2=36 where the tangent plane is perpendicular to the line with parametric equa

Anvisha [2.4K]

Answer:

A point on the ellipsoid is (-4,2,2) or (4,-2,-2)

Step-by-step explanation:

Given equation of ellipsoid f(x,y,z) : $x^2+y^2+4z^2=36$

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

$\nabla f(x,y,z)=\\\nabla f(x,y,z)=$

So, The directions vectors=(-4,2,8)

Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

$\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)$

So, $2x=-4\lambda$

$\Rightarrow x=-2\lambda$

$2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda$

Substitute the value of x , y and z in the ellipsoid equation

$(2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2$

With $\lambda = 2$

x=-2(2)=-4

y=2

z=2

With $\lambda =- 2$

x=-2(-2)=4

y=-2

z=-2

Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)

7 0

3 years ago