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3 years ago

Im giving brainliest

Mathematics

2 answers:

Alla [95]3 years ago

8 0

Answer:

80 is the measure because if aob is 40 then doe is 40 and dof is 120 minus the doe that 40 then you are going to get the mesuare which is 80

polet [3.4K]3 years ago

7 0

Answer:

BOC is 80

COD is 60

Step-by-step explanation:

hope this helps

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Equation: A number decreased by 5 is equal to 10 (seperate eqaution) The sum of a number and 12 is the same as twice the number.

kirza4 [7]

Answer:

1. 15

2. 12

Explanation:

1. n-5=10

n=10+5

n=15

2. n+12=2n

12=2n-n

12=n

3 0

2 years ago

Please help me find the value of x

Aleonysh [2.5K]

Answer:

Step-by-step explanation:

$(tangent)^{2}$ = AC × CB

$12^{2}$ = ( x + 9 ) 9

144 = 9x + 81

9x = 144 - 81

9x = 63

x = 63 ÷ 9

x = 7

3 0

2 years ago

Find the quotient of 85 and 4find the quotient of 85 and4

nignag [31]

Quotient is 21

Step-by-step explanation:

Step 1:

Divide 85 by 4. Using long division method, find the quotient.

21

4|85

-------

- 4

--------

4 0

3 years ago

what is 900+100 I'm stuck 100 points for the people who can answer i know its 1000 still 100 points go listen to (joey trap) bar

charle [14.2K]

Answer:

1000

Step-by-step explanation:

x = 1, y = -1, and z = 2

5 0

2 years ago

Read 2 more answers

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago