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cluponka [151]
2 years ago
5

What is the measure of angle ACB

Mathematics
1 answer:
zubka84 [21]2 years ago
7 0
Answer:

Angle ACB: 28
Arc AB: 56

Explanation:
- To find AB, the angle of BOA is a central angle. And its according arc is AB. The angle is always the same as the arc as long if its a central angle.

- Angle ACB is an inscribed angle. Its according arc is AB. Whenever its an inscribed angle, its half of its arc.
56 / 2 = 28
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Translate and solve using proportions: What number is 30% of 60
balandron [24]

Answer: 18

Step-by-step explanation:

The question is asked to know 30% of 60, we get

60 x 30% = 60 x 0.3 = 18

3 0
3 years ago
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Write an equation that represents the cost of tickets from Ticket System.
Akimi4 [234]
Is there any other information behind this problem?
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4 years ago
What is the value of x in the equetion 3/4(1/4x+8)-(1/2+2)-3/8(4-x)-1/4x?​
ikadub [295]
3/4*1/4= 3/16 or 0.1875 (you times 3 by 1 and 4 by 4)
3/4*8= 6
first bracket = 3/16+ 6
1/2+2= 2 1/2 or 2.5
second bracket = -2.5
-3/8*4= -1.5
-3/8*x= -3/8x or -0.375x
third bracket= -0.375x
so

(0.1875+6) - (2.5) - (0.375x) - (0.25x)
collect like terms
6.2 - 2.5 = 3.7
0.375x - 0.25x = 0.125x
3.7-0.125x is your answer in decimal form

(3/16+6/1) - (2 1/2) - (3/8x-1/4x)

(6 3/16 - 2 1/2 )

(3 11/16)

3/8x-1/4x= 1/8x

final answer = 3 11/16 - 1/8x
-
6 0
3 years ago
How many solutions exist for the given equation?
lesya [120]

Answer:

one

Step-by-step explanation: you cant simplify the equation so that both sides can be divided to get the same equation, if you graph it theres not a parabola so the only thing left is one.

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The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find
9966 [12]

Answer:

<em>The particular integral of given differential equation</em>

<em>                  </em>y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))<em></em>

<em> General solution of given differential equation</em>

<em>      </em>y = y_{c} + y_{p}<em></em>

<em>  </em>Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))<em></em>

<em></em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given Differential equation  y'' − 5 y' + 4 y = x

Given equation in operator form

        D²y - 5 Dy +  4 y = x

⇒     ( D² - 5 D +  4 ) y =x

⇒    f(D) y = Q

where  f(D) = D² - 5 D +  4 and Q(x) = x

<em>The auxiliary equation  f(m) =0</em>

<em>           m²-5 m + 4 =0</em>

         m² - 4 m - m + 4 =0

        m ( m -4 ) -1 ( m-4) =0

         (m - 1) =0   and ( m-4) =0

        <em> m = 1 and m =4</em>

<em>The complementary function </em>

<em></em>Y_{c} = C_{1} e^{x} + C_{2} e^{4x}<em></em>

<u><em>Step(ii)</em></u>:-

<u><em>particular integral</em></u>

<em>Particular integral</em>

<em>     </em>y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2}  - 5 D +  4} X<em></em>

<em>taking common '4' </em>

<em>                          </em>= \frac{1}{4(1 +  (\frac{D^{2}  - 5 D}{4} ))} X<em></em>

<em>                         </em>

<em>                           </em>=\frac{1}{4}  (1 + (\frac{D^{2} -5D}{4})^{-1} )} X<em></em>

<em>applying binomial expression</em>

<em>      ( 1 + x )⁻¹    = 1 - x + x² - x³ +.....       </em>

<em>                          </em>=\frac{1}{4}  (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X<em></em>

<em>Now simplifying and we will use notation D = </em>\frac{dy}{dx}<em></em>

<em>                        </em>=\frac{1}{4}  (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}<em></em>

<em>Higher degree terms are neglected</em>

<em>                     </em>=\frac{1}{4}  (x - (\frac{ -5 D}{4}) x)<em></em>

<em>The particular integral of given differential equation</em>

<em>                  </em>y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))<em></em>

<u><em>Final answer</em></u><em>:-</em>

<em>          General solution of given differential equation</em>

<em>      </em>y = y_{c} + y_{p}<em></em>

<em>  </em>Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))<em></em>

<em></em>

<em></em>

<em>         </em>

<em> </em>

     

4 0
3 years ago
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