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3 years ago

Yet again I still get confused

Mathematics

1 answer:

ryzh [129]3 years ago

6 0

The left side is a half circle.
The area of a circle is PI x r^2

The diameter of the circle would be the height of the rectangle, given as 10 feet.
The radius is 1/2 the diameter, so 5 feet.

The area of half the circle is (3.14 x 5^2) / 2 = 39.25 square feet.

The area of the rectangle is 20 x 10 = 200 square feet.

Total area = 200 + 39.25 = 239.25 square feet.

You might be interested in

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

6 0

2 years ago

What is the center of the circle that you can circumscribe about a triangle with vertices A(2, 6), B(2, 0), and C(10, 0)?

denis23 [38]

The three points A,B,C are all points on this circle.
Each point is then equal distance from the center, that distance being the radius of the circle.
Using the distance formula, we can find the center of the circle (x,y):
$d^2 = (x-x_0)^2 + (y-y_0)^2$
Plugging in points A and B into distance formula, then setting them equal to each other gives:
$(x-2)^2+(y-6)^2 = (x-2)^2 + y^2$
Right away we can cancel out the x terms leaving:
$(y-6)^2 = y^2$
Expand Left side and Solve for y:
$y^2 -12y +36 = y^2$
$y = 3$
Plug in points B and C as before:
$(x-2)^2 + y^2 = (x-10)^2 + y^2$
Here we can cancel the y-terms.
Expand and solve for x:
$x^2 -4x+4 = x^2 -20x+100$
$16x = 96$
$x = 6$
Therefore the center of the circle is the point (6,3)

6 0

3 years ago

Write (1/3i)-(-6+2/3i) as a complex number in standard form

Eddi Din [679]

Answer:

$6+\frac{i}{3}$