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3 years ago

Find the value of a. A. 58 B. 130 C. 86 D. 65

Mathematics

1 answer:

tatiyna3 years ago

5 0

Answer:

$C. \ \ \ 86$ °

Step-by-step explanation:

1. Approach

In order to solve this problem, one must first find a relationship between arc (a) and arc (c). This can be done using the congruent arcs cut congruent segments theorem. After doing so, one can then use the secants interior angle to find the precise measurement of arc (a).

2. Arc (a) and arc (c)

A secant is a line or line segment that intersects a circle in two places. The congruent segments cut congruent arcs theorem states that when two secants are congruent, meaning the part of the secant that is within the circle is congruent to another part of a secant that is within that same circle, the arcs surrounding the congruent secants are congruent. Applying this theorem to the given situation, one can state the following:

$a = c$

3. Finding the degree measure of arc (a),

The secants interior angle theorem states that when two secants intersect inside of a circle, the measure of any of the angles formed is equal to half of the sum of the arcs surrounding the angles. One can apply this here by stating the following:

$86=\frac{a+c}{2}$

Substitute,

$86=\frac{a+c}{2}$

$86=\frac{a+a}{2}$

Simplify,

$86=\frac{a+a}{2}$

$86=\frac{2a}{2}$

$86=a$

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9x-6+11x what would the answer be

frez [133]

Answer:

20x-6

Step-by-step explanation:

5 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

Which number(s) below belong to the solution set of the equation? Check all that apply.x + 12 = 13A.14B.1C.13D.312E.0F.25

Darya [45]

X + 12 = 13
x = 13 - 12
x = 1

The only solution is x = 1
letter B

4 0

3 years ago

PLEASE HELP! WILL GIVE BRAINLIEST!!!!!

Sophie [7]

The 8 percent commission must be at least $1000.

8% = $1000

1% = $125

100% = $12500

Hence, Jim needs to sell $12500 dollars worth of widgets.

6 0

3 years ago

Help help help! 25 points

Paraphin [41]

Step-by-step explanation:

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8 0

2 years ago

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