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3 years ago

Select all the rates that are unit rates.

Mathematics

1 answer:

tensa zangetsu [6.8K]3 years ago

5 0

Pretty sure it’s 2 and 4. A unit rate is usually something to one 1. So 1 would be in the denominator.

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Digiron [165]

Pretty sure it is nine. I hope I helped

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3 years ago

Isabella tossed a coin and spun a spinner that is divided into 3 equal sections. She did this 50 times. The results are shown in

Hatshy [7]

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Step-by-step explanation:

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3 years ago

Find the solution to the differential equation dB/dt+4B=20 with B(1)=30

natita [175]

Answer:

The solution of the differential equation is $B=5+25e^{-4t+4}$

Step-by-step explanation:

The differential equation $\frac{dB}{dt}+4B=20$ is a first order separable ordinary differential equation (ODE). We know this because a separable first-order ODE has the form:

$y'(t)=g(t)\cdot h(y)$

where g(t) and h(y) are given functions.

We can rewrite our differential equation in the form of a first-order separable ODE in this way:

$\frac{dB}{dt}+4B=20\\\frac{dB}{dt}=20-4B\\\frac{dB}{dt}=4(5-B)\\\frac{1}{5-B}\frac{dB}{dt}=4$

Integrating both sides

$\frac{1}{5-B}\frac{dB}{dt}=4\\\frac{1}{5-B}\cdot dB=4\cdot dt\\\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {4} \, dt$

The integral of left-side is:

$\int\limits {\frac{1}{5-B}} \, dB\\\mathrm{Apply\:u-substitution:}\:u=5-B\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {\frac{1}{u}} \, dB\\\mathrm{du=-dB}\\-\int\limits {\frac{1}{u}} \,du\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-\int\limits {\frac{1}{u}} \,du =-\ln \left|u\right|\\\mathrm{Substitute\:back}\:u=5-B\\-\ln \left|5-B\right|\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\-\ln \left|5-B\right|+C$

The integral of right-side is:

$\int\limits {4} \, dt = 4t + C$

We can join the constants, and this is the implicit general solution

$-\ln \left|5-B\right|+C=4t + C\\-\ln \left|5-B\right|=4t + D$

If we want to find the explicit general solution of the differential equation

We isolate B

$-\ln \left|5-B\right|=4t + D\\\ln \left|5-B\right|=-4t+D\\\left|5-B\right|=e^{-4t+D}$

Recall the definition of |x|

$|x|=\left \{ {{x, \:if \>x\geq \>0 } \atop {-x, \:if \>x0}} \right.$

$\left|5-B\right|=e^{-4t+D}\\5-B= \pm \:e^{-4t+D}\\B=5 \pm \:e^{-4t+D}\\B=5\pm \:e^{-4t}\cdot e^{D}\\B=5+Ae^{-4t}$

where $A=\pm e^{D}$

Now B(1) =30 implies

$B=5+Ae^{-4t}\\30=5+Ae^{-4}\\30-5=Ae^{-4}\\25e^{4}=A$

And the solution is

$B=5+(25e^{4})e^{-4t}\\B=5+25e^{-4t+4}$

8 0

4 years ago

Fill in the blank. Given O below, you can conclude that OH is congruent to _____.

pishuonlain [190]

Answer:

(A) OE

Step-by-step explanation:

From the given figure, it can be seen that TG and LI are the chords of the circle with center O such that TG=LI=10.4 which means that TG is congruent to LI.

Also, OE and OH gives the shortest distance(perpendicular) of the chord LI and TG respectively from center of circle O.

And, We know that "the congruent chords are equidistant from the center of a circle."

Therefore, we can conclude that, OH is congruent to OE.

Hence, option (A) is correct.

5 0

3 years ago

Read 2 more answers

What is the total surface area??? Please help

DedPeter [7]

Answer:

216

Step-by-step explanation:

8(6) + (10+6+8)(7)

48 + 24x7

48 + 168

216

7 0

3 years ago