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3 years ago

There of the ABCD are given. fine the coordinates of point D. A(3, 6) B(6,3) D(x, y)

Mathematics

2 answers:

Evgen [1.6K]3 years ago

4 0

Answer:

A. (3,2) . I recommend using desmos to find the coordinates

shutvik [7]3 years ago

4 0

9514 1404 393

Answer:

A. (3, 2)

Step-by-step explanation:

The midpoints of the diagonals are the same point, so you have ...

X = (A +C)/2 = (B +D)/2

A +C = B +D . . . . . . . multiply by 2

A +C -B = D . . . . . . . subtract B

The coordinates of D are ...

D = (3, 6) +(6, 3) -(6, 7) = (3+6-6, 6+3-7)

D = (3, 2)

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Answer:

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3 years ago

What is x equal to in the problem 10+3(x+2)=36? Please show your work.

scZoUnD [109]

Answer:

x = 6 2/3

Step-by-step explanation:

10+3(x+2)=36

10+3x+6=36 Solve inside the parenthesis.

16+3x=36 Combine like terms.

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3 years ago

Read 2 more answers

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madam [21]

A. x= 2 y=4

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5 0

3 years ago

Given that the diameter of Circle A is 6 cm , and the radius of Circle B is 18 cm , what can be concluded about the two circles?

ELEN [110]

Answer:

The radius of circle B is 6 times greater than the radius of circle A

The area of circle B is 36 times greater than the area of circle A

Step-by-step explanation:

we have

<em>Circle A</em>

$D=6\ cm$

The radius of circle A is

$r=6/2=3\ cm$ -----> the radius is half the diameter

<em>Circle B</em>

$r=18\ cm$

Compare the radius of both circles

$3\ cm< 18\ cm$

$18=6(3)$

The radius of circle B is six times greater than the radius of circle A

Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared

All circles are similar

In this problem the scale factor is 6

$6^{2}=36$

therefore

The area of circle B is 36 times greater than the area of circle A

8 0

3 years ago

You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The o

Aleksandr [31]

Answer:

Therefore the circumference of the circle is $=\frac{20\pi}{4+\pi}$

Step-by-step explanation:

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

$\Rightarrow s=\frac{10-\pi r}{2}$

The area of the circle is = πr²

The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

$A=\pi r^2+ (\frac{10-\pi r}{2})^2$

$\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2$

$\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}$

$\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25$

For maximum or minimum $\frac{dA}{dr}=0$

Differentiating with respect to r

$\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi$

Again differentiating with respect to r

$\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}$ > 0

For maximum or minimum

$\frac{dA}{dr}=0$

$\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0$

$\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}$

$\Rightarrow r=\frac{10}{4+\pi}$

$\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0$

Therefore at $r=\frac{10}{4+\pi}$ , A is minimum.

Therefore the circumference of the circle is

$=2 \pi \frac{10}{4+\pi}$

$=\frac{20\pi}{4+\pi}$

4 0

3 years ago