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2 years ago

Find the equation of the circle with center at (3,-2) and radius of 3.

Mathematics

1 answer:

AURORKA [14]2 years ago

3 0

Answer:

The equation of the circle is (x - 3)² + (y + 2)² = 9

Step-by-step explanation:

The form of the equation of a circle is (x - h)² + (y - k)² = r², where

h is the x-coordinate center of the circle
k is the y-coordinate of the center of the circle

r is the radius of the circle

∵ The center of a circle is (3, -2)

∴ x-coordinate of the center is 3

∴ y-coordinate of the center is -2

∴ h = 3 and k = -2

∵ The radius of the circle is 3

∴ r = 3

→ Substitute them in the form of the equation above

∵ (x - 3)² + (y - -2)² = (3)²

∴ (x - 3)² + (y + 2)² = 9

∴ The equation of the circle is (x - 3)² + (y + 2)² = 9

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Step-by-step explanation:

the law of sines :

a/sin(A) = b/sin(B) = c/sin(C)

with the sides being opposite to the angles.

so,

14/sin(80) = x/sin(60)

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2 years ago

Can someone please help me with my maths question

DIA [1.3K]

Answer:

$a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}$

By expanding the expression, we get;

$\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}$

Collecting like terms gives;

$\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}$

(b) The given expression is presented as follows;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4$

Therefore, we get;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}$

Collecting like terms gives;

$x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )$

$x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

4 0

3 years ago

What is the constant of the expression? 5x−2y+2.4

Leno4ka [110]

The correct answer is the 2.4

The constant in an equation is the number(s) that don't have variables.

This is because variables can change the number they are attached to, making them not constant.

Hope this helps!

6 0

2 years ago

Read 2 more answers

Find the angle theta in radians. ( PLEASE SHOW STEPS )

yKpoI14uk [10]

Hello!

As you can see, we have a radius of 6. If we divide, this means that this is 2.5 radians. To convert radians to degrees we use the formula below.

$\frac{180}{ \pi }x$

First of all we divide 180 by pi.

180/ $\pi$ ≈57.3

Now we multiply by 2.5

57.3(2.5)=143.25°

Note that the angle we see is obtuse, or greater than 90°.

Therefore, ∠θ≈143.25°

Now we need to convert this back into radians. This can be represented by the equation below.

$\frac{ \pi }{180}x$

First we divide pi by 180 then multiply by our angle.

$\pi$ /180(143.25)≈2.5

Therefore, our angle theta is about 2.5 radians.

I hope this helps!

3 0

3 years ago

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Solve the equation.

lawyer [7]

Answer:

x=12
x=9

Step-by-step explanation:

1) Eliminate parentheses:

0.1x +18.8 = -4 +2x

22.8 = 1.9x . . . . . . . . . add 4 - 0.1x

12 = x . . . . . . . . . . . . . divide by 1.9

2) Eliminate parentheses:

-16 +4x = 0.8x +12.8

3.2x = 28.8 . . . . . . . . add 16 - 0.8x

x = 9 . . . . . . . . . . . . . .divide by 3.2

_____

Comments on the solutions

The expression we add in each case eliminates the constant on one side of the equation and the variable term on the other side. That leaves an equation of the form ...

variable term = constant

We choose to eliminate the smaller variable term (the one with the coefficient farthest to the left on the number line). Then the constant we eliminate is the on on the other side of the equation. This choice ensures that the remaining variable term has a positive coefficient, tending to reduce errors.

You can work these problems by methods that eliminate fractions. Here, the fractions are decimal values, so are not that difficult to deal with. In any event, it is good to be able to work with numbers in any form: fractions, decimals, integers. It can save some steps.

7 0

2 years ago