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Snezhnost [94]
2 years ago
5

(10 points)List the angles in order from greatest to least. mpn​

Mathematics
1 answer:
Blizzard [7]2 years ago
3 0

Answer:

C

Step-by-step explanation:

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At the North campus of a performing arts school 30% of students are music majors at the South campus 80% of the students are mus
OLEGan [10]

Answer:

It should be option B

Step-by-step explanation:

4 0
3 years ago
Find the distance CA
katrin [286]

C' ( 0, 3)

A' ( 2 , 1)

C'A' = CA

C'A' = √(2^2 + 2^2) = √8 = 2√2

C'A' = CA = 2√2

3 0
3 years ago
Half-life of Po-210 is 140 days. If the initial mass of the sample is 5
Anna35 [415]

Answer:

0.625kg

Step-by-step explanation:

we are given half-life of PO-210 and the initial mass

we want to figure out the remaining mass <u>after</u><u> </u><u>4</u><u>2</u><u>0</u><u> </u><u>days</u><u> </u>

in order to solve so we can consider the half-life formula given by

\displaystyle f(t) = a {0.5}^{t/T}

where:

  • f(t) is the remaining quantity of a substance after time t has elapsed.
  • a is the initial quantity of this substance.
  • T is the half-life

since it halves every 140 days our T is 140 and t is 420. as the initial mass of the sample is 5 our a is 5

thus substitute:

\displaystyle f(420)=5\cdot{0.5}^{420/140}

reduce fraction:

\displaystyle f(420)=5\cdot{0.5}^{3}

By using calculator we acquire:

\displaystyle f(420)=0.625

hence, the remaining sample after 420 days is 0.625 kg

4 0
3 years ago
I need help on completing this
Allushta [10]
I don’t know I need points
7 0
3 years ago
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
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