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3 years ago

Find the center and the radius of the circle. (x - 5)2 + (y + 1)2 = 9

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

5 0

Answer:

The centre's coordinates are: (5; -1)

Radius is the square of 9:

$\sqrt{9} = 3 \: and \: - 3 \\ it \: has \: to \: be \: positive \: therefore \: it \: is 3$

Answer: Radius=3; the cente's coordinates are(5; -1)

Good luck!

Intelligent Muslim,

From Uzbekistan.

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An eraser is in the shape of a triangular prism. Its dimensions are shown in the diagram. What is surface area of the

Colt1911 [192]

Answer:

6000 square millimeters

Step-by-step explanation:

Side length of the slanting side can be calculated using the Pythagorean theorem,

L = sqrt(30^2+40^2) = 50

sufrace area of a triangular prism

= surface area of the end sections + the surface area of the three side faces

= 2*(30*40/2) + 40*(30+40+50)

= 1200 + 4800

= 6000 mm^2

4 0

3 years ago

Read 2 more answers

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .

For triangle DEF:

The length of segment DF is to be found,
The length of segment EF is 9,
The sine of angle E is $\sin{64\textdegree}}$ , and
The sine of angle D is $\sin{39\textdegree}$ .

Apply the law of sine:

$\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}$

$\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9$ .

7 0

3 years ago

AB and BC are tangents to P. what is the value of x? this is really so confusing

Westkost [7]

1. Angle PAB is 90 degrees, as it is formed from the tanget to the circle at A, and the radius drawn to A.

2. AB=BC, because tangents drawn to a circle from the same point are equal.

3. PB is Common, so by the side-side-side congruence postulate, triangles ABP and CBP are congruent.

4. So measure of m(BPA)=x/2 and m(ABP)=73/2.

5. $\frac{x}{2}+90+ \frac{73}{2}=180$

$\frac{x+73}{2}=90$

$x+73=180$ , x= 107 degrees.