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3 years ago

If a(x) = 3x + 1 and (x)= x-4, what is the domain of (boa)(x)?

Mathematics

1 answer:

lesantik [10]3 years ago

8 0

Answer:

$(b\circ a)(x)=3x-3$ ; the domain is all real numbers

Step-by-step explanation:

$a(x) = 3x+1$

$b(x)=x-4$

$(b\circ a)(x)=b(a(x))=b(3x+1)=(3x+1)-4=3x+1-4=3x-3$

Therefore, the domain of the composite function is all real numbers.

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Which is a factor of each term of the polynomial? (72−12) CLEAR CHECK 7 2 7

Inessa [10]

Answer:

Is the greatest factor

Step-by-step explanation:

All factors include

2,3,4,6,12

Thats it

3 0

3 years ago

What is the solution of the equation? -6 = 2/3x a. = -9 b. = -6 c. = 4 d. = -4

erastovalidia [21]

-6 = 2/3 x
x = -6*3 /2
x = -18/2
x = -9

Answer is Option A

Hope this helps!

8 0

4 years ago

At 3 p.m. an oil tanker traveling west in the ocean at 14 kilometers per hour passes the same spot as a luxury liner that arrive

schepotkina [342]

N(t) = 16t ; Distance north of spot at time t for the liner. W(t) = 14(t-1); Distance west of spot at time t for the tanker. d(t) = sqrt(N(t)^2 + W(t)^2) ; Distance between both ships at time t. Let's create a function to express the distance north of the spot that the luxury liner is at time t. We will use the value t as representing "the number of hours since 2 p.m." Since the liner was there at exactly 2 p.m. and is traveling 16 kph, the function is N(t) = 16t Now let's create the same function for how far west the tanker is from the spot. Since the tanker was there at 3 p.m. (t = 1 by the definition above), the function is slightly more complicated, and is W(t) = 14(t-1) The distance between the 2 ships is easy. Just use the pythagorean theorem. So d(t) = sqrt(N(t)^2 + W(t)^2) If you want the function for d() to be expanded, just substitute the other functions, so d(t) = sqrt((16t)^2 + (14(t-1))^2) d(t) = sqrt(256t^2 + (14t-14)^2) d(t) = sqrt(256t^2 + (196t^2 - 392t + 196) ) d(t) = sqrt(452t^2 - 392t + 196)

3 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

Solve for x.38(2x+16)−2=13 Enter your answer in the box.x =

lys-0071 [83]

Given:

$\frac{3}{8}(2x+16)-2=13$ $\frac{3}{8}(2x+16)=13+2$ $\frac{3}{8}(2x+16)=15$ $(2x+16)=15\times\frac{8}{3}$ $(2x+16)=40$ $2x=40-16$ $2x=24$ $x=\frac{24}{2}$ $x=12$

5 0

1 year ago