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3 years ago

The area of a circle is equal to the area of a square measuring 5 centimeters on each side find the radius of the circle

Mathematics

1 answer:

Gre4nikov [31]3 years ago

4 0

The area os the square is 25 cm sq so the radius of the circle should be about 2.82 cm

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x2 + y2 − 4x + 12y − 20 = 0 (x − 6)2 + (y − 4)2 = 56 x2 + y2 + 6x − 8y − 10 = 0 (x − 2)2 + (y + 6)2 = 60 3x2 + 3y2 + 12x + 18y −

snow_tiger [21]

For this case, what we must do is fill squares in all the expressions until we find the correct result.
We have then:

x2 + y2 − 4x + 12y − 20 = 0 x2 + y2 − 4x + 12y = 20
x2 − 4x + y2 + 12y = 20
x2 − 4x + (12/2)^2 + y2 + 12y + (-4/2)^2 = 20 + (12/2)^2 + (-4/2)^2
x2 − 4x + (6)^2 + y2 + 12y + (-2)^2 = 20 + (6)^2 + (-2)^2
x2 − 4x + 36 + y2 + 12y + 4 = 20 + 36 + 4
(x − 2)2 + (y + 6)2 = 60

3x2 + 3y2 + 12x + 18y − 15 = 0
x2 + y2 + 4x + 6y − 5 = 0
x2 + y2 + 4x + 6y = 5
x2 + 4x + (4/2)^2 + y2 + 6y + (6/2)^2 = 5 + (4/2)^2 + (6/2)^2
x2 + 4x + (2)^2 + y2 + 6y + (3)^2 = 5 + (2)^2 + (3)^2
x2 + 4x + 4 + y2 + 6y + 9 = 5 + 4 + 9
(x + 2)2 + (y + 3)2 = 18

2x2 + 2y2 − 24x − 16y − 8 = 0
x2 + y2 − 12x − 8y − 4 = 0
x2 + y2 − 12x − 8y = 4
x2 − 12x + (-12/2)^2 + y2 − 8y + (-8/2)^2 = 4 + (-12/2)^2 + (-8/2)^2
x2 − 12x + (-6)^2 + y2 − 8y + (-4)^2 = 4 + (-6)^2 + (-4)^2
x2 − 12x + 36 + y2 − 8y + 16 = 4 + 36 + 16
(x − 6)2 + (y − 4)2 = 56

x2 + y2 + 2x − 12y − 9 = 0
x2 + y2 + 2x - 12y = 9
x2 + 2x + y2 - 12y = 9
x2 + 2x + (2/2)^2 + y2 - 12y + (-12/2)^2 = 9 + (2/2)^2 + (-12/2)^2
x2 + 2x + (1)^2 + y2 - 12y + (-6)^2 = 9 + (1)^2 + (-6)^2
x2 + 2x + 1 + y2 - 12y + 36 = 9 + 1 + 36
(x + 1)2 + (y − 6)2 = 46

7 0

4 years ago

Given right triangle QRS, what is the value of sin(30°)? StartFraction StartRoot 3 EndRoot Over 3 EndFraction One-half StartFrac

baherus [9]

Answer:

Option 2: $\sin(30)=\frac{1}{2}$

Step-by-step explanation:

Given:

From the triangle shown below;

A triangle QRS with angle QRS = 90°, ∠QSR = 30°.

Side QR = 5, SQ = 10 and RS = 5√3

Now, we know from trigonometric ratio that,

$\sin (A) = \frac{Opposite\ side}{Hypotenuse}$

Here, opposite side of angle QSR is QR and Hypotenuse is the side opposite angle QRS which is SQ. Therefore,

$\sin(\angle QSR)=\dfrac{QR}{SQ}\\\\\\\sin(30)=\dfrac{5}{10}\\\\\\\sin(30)=\dfrac{5}{2\times 5}=\dfrac{1}{2}$

Therefore, the value of sine of 30° is one-half. So, second option is correct.

6 0

3 years ago

Read 2 more answers

Paula got $10.50 from her grandparents on her birthday. She earned $4.45 by babysitting. How much money is this in all? c HAPPY

Aloiza [94]

We have the following:

To calculate the total money, you have to add the amounts that Paula won, as follows

$\begin{gathered} T=10.5+4.45 \\ T=14.95 \end{gathered}$

Which means that in total they are $ 14.95

8 0

1 year ago

X^2 -3x + 0 = 0 Factor and solve

Luda [366]

So if you subtract the 0 out from the left side, the equation is now $x^{2}-3x=0$ . Since a x is common in both of the terms on the left side, you can factor it out, now making the equation look like $x(x-3)=0$ . When factoring, each of the separate things (x and (x-3)), must equal zero while the other one doesn't matter. So in order for the equation to be 0, the only potential options for x is 0 and 3. Therefore x=0 and x=3 are the two solutions

3 0

3 years ago

Read 2 more answers

Brainliest!!

Andreyy89

answer of all tasks ( 1 to 6) is attached below in 4 images

6 0

4 years ago