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3 years ago

Add the two expressions. 3y + 5 and y + 24 Enter your answer in the box.

Mathematics

2 answers:

nadezda [96]3 years ago

8 0

4y+29 is the answer to the qustion

Sedaia [141]3 years ago

6 0

Answer:

4y+29

Step-by-step explanation:

hello, I can help you with this question.

to add two or more algebraic expressions you must add the similar terms, it is, variables ( the same letter or symbol) with the same exponent( 1 in this case)

Step 1

put the terms in orden and then add similar terms

+ 3y+ 5

y + 24

________________

(3y+y)+(5+24)=

4y +29

so,3y+5+y+24=4y+29

I really hope it helps.

Have a great day.

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A piece of pie costs .10 and a cup of coffee costs .5. How much will 40 pieces of pie and 20 cups of coffee cost?

dybincka [34]

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3 years ago

1c) What is 25% of 40? 8.5

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3 years ago

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t),

pishuonlain [190]

Answer:

S(3)=22

Step-by-step explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

$\dfrac{dS}{dt}=k(30-S(t))\\ \dfrac{dS}{dt}+kS(t)=30k\\$The integrating factor: e^{\int k dt}=e^{kt}\\$Multiply all through by the integrating factor\\ \dfrac{dS}{dt}e^{kt}+kS(t)e^{kt}=30ke^{kt}$

$(Se^{kt})'=30ke^{kt} dt\\$Integrate both sides\\ Se^{kt}=\dfrac{30ke^{kt}}{k}+C$ (C a constant of integration)\\Se^{kt}=30e^{kt}+C\\$Divide both sides by e^{kt}\\S(t)=30+Ce^{-kt}$

When t=0, S(t)=15

$15=30+Ce^{-k*0}\\C=15-30\\C=-15$

When t = 2, S(t)=20

$20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\$Take the natural log of both sides$\\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}$

Therefore:

$S(t)=30-15e^{\frac{\ln(2/3)}{2}t}\\$When t=3$\\S(t)=30-15e^{\frac{\ln(2/3)}{2} \times 3}\\S(3)=21.8 \approx 22$

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3 years ago