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2 years ago

If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?

Mathematics

1 answer:

Vladimir79 [104]2 years ago

5 0

Answer:

x ≥ 1

Step-by-step explanation:

There are no domain restrictions on a(x), so we can find the domain of the composite function by looking at that composition:

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

The argument of the square root function must be non-negative, so we must have ...

3(x -1) ≥ 0

x -1 ≥ 0 . . . . . divide by 3

x ≥ 1 . . . . . . . add 1

The domain of b(a(x)) is x ≥ 1.

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54.85 rounded to the nearest 1/10th percent

Iteru [2.4K]

Answer:

54.9

Step-by-step explanation:

You always round up if its 0.5 or anything .5

54.58 broken down:

The 5 is the Tens place

The 4 is the Ones place

The 8 is the Tenth place (what the question is asking for)

The 5 is the Hundredths place.

:) Hope this helps! Please mark brainliest! :)

8 0

2 years ago

Please help with this math question thanks!!!

melisa1 [442]

The surface area is
A1=7.5*1.5*2=22.5 ft2 top and bottom view
A2=3*1.5*2=9 ft2 laterals view
A3=(7.5*1.5+2.5*1.5)*2=30 ft2 front and rear view

At=22.5+9+30=61.5 ft2
the answer is 61.5 ft2

5 0

3 years ago

Read 2 more answers

ASAP <br>What is the domain?

WARRIOR [948]

real numbers hope it helps

7 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago

Do these ratios make a proportion? 8/11 and 88/111

mixas84 [53]

Answer:

Step-by-step explanation:

111 divided by 11 is 10.0909090909

and 88 divided by 8 is 11 so no it`s not proportional.

3 0

2 years ago