2 years ago

What's the solution to –9x + 4 < 40?

Mathematics

1 answer:

olganol [36]2 years ago

6 0

C it’s very easy if you want me to explain I will

You might be interested in

I need to solve this question asap

jeyben [28]

Answer:

im sorry I just need to us ur page to right on

7 0

1 year ago

Simplify the fraction below. −63/99

vivado [14]

Answer:

-7/11

Step-by-step explanation:

3 0

2 years ago

250% of a number is 60. What is the number

katrin [286]

60 / 2.50 = 24

the number is 24

7 0

2 years ago

Read 2 more answers

Find the values of x that satisfy

mr_godi [17]

Setup 2 problems
2x - 7 < 15   and    2x - 7 > -15
2x < 22                  2x > -8
x < 11                    x > -4

Or you can write it -4 < x < 11

7 0

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

8 0

2 years ago