All categories

3 years ago

The circumference of a fountain is 63 feet. What is the approximate diameter?

Mathematics

1 answer:

Alik [6]3 years ago

3 0

Answer:

Approximantly 20

Step-by-step explanation:

Circumference = pi times diameter

63/pi = diameter

Answer: 20.0535228

You might be interested in

In the expression -3×+4y-2 what are the variables

SSSSS [86.1K]

Answer:

x and y are the variables

7 0

2 years ago

Allisa [31]

Answer:

Triangle PRT is similar to triangle SRQ

Step-by-step explanation:

AA similarity Postulate :If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

one angle is already given : angle RPT congruent to angle RSQ

The other is : angle TRP congruent to angle SRQ (both triangles share the same angle)

Triangle PRT is similar to triangle SRQ (AA postulate)

3 0

3 years ago

Elena starts her new job working at an office supply store. She works for 6 hours and makes a total of $54.

docker41 [41]

Answer:

I believe the answer would be A.

Step-by-step explanation:

A says 6 x h is 54.

^ ^ ^

important!

6 is the number of hours.

h is how much money made per hour.

54 is the total after working 6 hours.

4 0

3 years ago

Help me 15 points plz fast

yaroslaw [1]

Answer:

The two numbers marked on the number line are -5 and 7.

Two inequalities to compare the marked numbers on the number line:

-5 < 7 or 7 > -5

The written form of the two inequalities are:

-5 < 7 : Negative five is less than positive seven.

7 > -5 : Positive seven is greater than negative five.

6 0

2 years ago

Read 2 more answers

Let II be the tangent plane to the graph of f(x, y) = 8 – 2x^2 – 3y^2 at the point (1, 2,-6). Let S, x² + y^2 + z = 4 be another

stealth61 [152]

Let $F(x,y,z)=f(x,y)-z$ . The tangent plane to $f(x,y)$ at (1, 2, -6) has equation

$\nabla F(1,2,-6)\cdot(x-1,y-2,z+6)=0$

We have

$\nabla F(x,y,z)=(-4x,-6y,-1)\implies\nabla F(1,2,-6)=(-4,-12,-1)$

Then the tangent plane has equation

$(-4,-12,-1)\cdot(x-1,y-2,z+6)=0\implies -4(x-1)-12(y-2)-(z+6)=0\implies 4x+12y+z=22$

Let $g(x,y)=4-x^2-y^2$ , and $G(x,y,z)=g(x,y)-z$ . The tangent plane to $S$ at a point $(a,b,c)$ is

$\nabla G(a,b,c)\cdot(x-a,y-b,z-c)=0$

We have

$\nabla G(x,y,z)=(-2x,-2y,-1)\implies \nabla G(a,b,c)=(-2a,-2b,-1)$

so that this plane has equation

$(-2a,-2b,-1)\cdot(x-a,y-b,z-c)=0\implies2ax+2by+z=2a^2+2b^2+c$

In order for this plane to be parallel to the previous plane, we need to have

$\begin{cases}2a=4\\2b=12\end{cases}\implies a=2,b=6\implies g(a,b)=c=-36$

so the point we're looking for is (2, 6, -36).

6 0

3 years ago