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4 years ago

PLEASE HELP ME THIS IS FOR A TEST GRADE !!!!! AND I NEED HELP ASAP

Mathematics

2 answers:

kicyunya [14]4 years ago

6 0

$g(t) = t - 2\\h(t) = 3t + 3\\(g \circ h)(t) = g(h(t)) = (3t + 3) - 2 = 3t + 3 - 2 = 3t + 1$

olya-2409 [2.1K]4 years ago

3 0

G (3t+ 3)= 3t + 1

Write the problem as a mathematical expression.

h (t) = 3t + 3 , (g o h) (t),
g (t) = t -2

Set up the composite result function.

g (h (t))

Evaluate g(h(t)) by substituting in the value of h into g

g (3t +3 ) = (3t + 3) -2

Subtract 2 from 3

g (3t +3 ) = 3t + 1

You might be interested in

Ellen says that one and 2/5 equal 5÷7. Is she correct? Explain.

zhuklara [117]

No because 2/5 is the same as 2 divided by 7, so 2/5 = 0.4 where 5 divide by 7 is 0.714….. so they’re not equal.

7 0

2 years ago

Use differentials to estimate the amount of paint needed (in m3) to apply a coat of paint 0.04 cm thick to a hemispherical dome

pogonyaev

The amount of paint is the volume of paint needed.

The amount of paint needed is 0.065312 cubic meters

<h3>How to determine the amount of paint needed</h3>

The volume of a hemisphere is:

$V = \frac 23\pi r^3$

Differentiate the above equation

$V' = 2\pi r^2 r'$

The above equation represents the estimate of the amount of paint needed.

Where:

r represents the radius (r = 52/2 m)
r' represents the thickness (r' = 0.04 cm)

So, we have:

$V' = 2\pi r^2 r'$

$V' = 2 * 3.14 * (52/2\ m)^2 * (0.04cm)$

$V' = 2 * 3.14 * (26\ m)^2 * (0.04cm)$

Express cm as m

$V' = 2 * 3.14 * (26\ m)^2 * (0.0004m)$

$V' = 0.065312 m^3$

Hence, the amount of paint needed is 0.065312 cubic meters

Read more about volumes at:

brainly.com/question/10171109

3 0

2 years ago

The rate of change of the number of mountain lions N(t) in a population is directly proportional to 725 - N(t), where t is the t

wolverine [178]

Answer:

a) The implied differential equation is $\frac{dN}{725 - N} = kdt$

b) The general equation is $N = 725 - C e^{-kt}$

c) The particular equation is $N = 725 - 325 e^{-0.49t}$

d) The population when t = 5, N(5) = 697 = 700( to the nearest 50)

Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, $\frac{dN}{dt} \alpha (725 - N(t))$

$\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt$

Integrating both sides of the equation

$\int {\frac{1 }{725 - N}} \, dN = \int {k} \, dt\\- ln (725 - N) = kt + C\\ ln (725 - N) = -(kt + C)\\725 - N = e^{-(kt + C)} \\725 - N = e^{-kt} * e^{-C} \\725 - N = C e^{-kt}\\N = 725 - C e^{-kt}$