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deff fn [24]
3 years ago
6

Which expression is equivalent to 10x - 5 + 3x - 2

Mathematics
2 answers:
Goryan [66]3 years ago
8 0
13x-7 I would love some points
Serga [27]3 years ago
3 0

Answer:

13x-7

Step-by-step explanation:

add 10x and 3x.

13x-5-2

subtract 2 from -5

13x-7

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Mrs. Baker uses 7.5 cups of sugar to make 168 cookies. To the nearest hundredth, what is the unit rate in cups of sugar per cook
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Answer:0.04

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The tree diagram represents an experiment consisting two trials
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0.35

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How many liters give u 3.1KL
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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
Reconocimiento de suceciones aritmeticas y geometricas
Monica [59]

Answer:

el producto es 4 o 3 por que dice hallar el producto de CUATRO termino estonses 3×4= 12

Step-by-step explanation:

si necesita ayuda o necesita responder sus preguntas en cuestión de segundos o minutos, hágamelo saber y lo ayudaré :)

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