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Anna [14]
3 years ago
8

1. Find the domain of the given function. (1 point)

Mathematics
1 answer:
Vika [28.1K]3 years ago
6 0
1) Find the domain of the given function. f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two.
 using a graphical tool    see the attachment  
the answer is C) x ≥ -3, x ≠ 2  
 2. Identify intervals on which the function is increasing, decreasing, or constant.
g(x) = 2 - (x - 7)2
 using a graphical tool     
see the attachment
 the answer is  C) Increasing: x < 7; decreasing: x > 7
3. Perform the requested operation or operations.
f(x) = 4x + 7, g(x) = 3x2
Find (f + g)(x).
 (f + g)(x) = f(x) + g(x) 
(f + g)(x) = 4x + 7 + 3x^2 
(f + g)(x) = 3x^2 + 4x + 7
 The answer is  C) 4x + 7 + 3x2
 4. Perform the requested operation or operations.
f(x) = x minus five divided by eight. ; g(x) = 8x + 5, find g(f(x)).
 f(x)=(x-5)/8    g(x)=8x+5
g(f(x))=8((x-5)/8)+5=x-5+5=x
the answer is  B) g(f(x)) = x   
5. Find f(x) and g(x) so that the function can be described as y = f(g(x)).y = nine divided by square root of quantity five x plus five. 
 y=f(g(x))=9/((5x+5) ^1/2)
let do
g(x)=5x+5...........so
f(x)= 9/( x^1/2)
the answer is  A) f(x) = nine divided by square root of x. , g(x) = 5x + 5

6. A satellite camera takes a rectangular-shaped picture. The smallest region that can be photographed is a 4-km by 4-km rectangle. As the camera zooms out, the length l and width w of the rectangle increase at a rate of 3 km/sec. How long does it take for the area A to be at least 4 times its original size?
Original size- >4km*4km=16 km2
4 times its original size---------------4*(16km2)-----64 Km2----------- > 8 km by 8 Km
Therefore
3km----------------------------- 1 sec
(8km-4km)---------------------x
X=4/3=1.33 sec
The answer is D) 1.33 sec

7. Find the inverse of the function.
f(x) = the cube root of quantity x divided by seven. - 9
to solve, replace f(x) with y , switch x and y, solve for y and replace y with f⁻¹(x)

f(x)=((x/7)-9) ^(1/3)
replace f(x) with y
y=((x/7)-9) ^(1/3)
switch x and y
x=((y/7)-9) ^(1/3)
solve for y
x^3=((y/7)-9)
x^3+9=y/7
y=7(x^3+9)
the answer is C) f-1(x) = 7(x3 + 9)  

8. Describe how the graph of y= x2 can be transformed to the graph of the given equation.y = (x - 14)2 – 9

using a graphical tool     see the attachment  the answer is C) Shift the graph of y = x2 right 14 units and then down 9 units  

9. Describe how to transform the graph of f into the graph of g. f(x) = alt='square root of quantity x minus nine.' and g(x) = alt='square root of quantity x plus five. '

f(x)=(x-9) ^1/2  g(x)=(x+5) ^1/2 
using a graphical tool     see the attachment

the answer is C) Shift the graph of f left 14 units

10. If the following is a polynomial function, then state its degree and leading coefficient. If it is not, then state this fact.
f(x) = -16x5 - 7x4 – 6
The answer is B) Degree: 5; leading coefficient: -16

11. Write the quadratic function in vertex form.y = x2 + 4x + 7 

Complete the square on the right side of the equation
Use the form ax2+bx+cax2+bx+c, to find the values of a, b, and c.
a=1,b=4,c=7
Consider the vertex form of a parabola.
a(x+d)2+e
Find the value of dd using the formula d=b/2a
d=4/(2*1)=2
Find the value of e using the formula e=c−b2/4a
e=7−4=3
Substitute the values of a, d, and e into the vertex form a(x+d)2+e
(x+2)2+3
The answer is A) y = (x + 2)2+ 3

12. Find the zeros of the function.
f(x) = 3x3 - 12x2 - 15x
using a graphical tool   (see the attachment)x1=-1
x2=0
x3=5
The answer is C) 0, -1, and 5

13. Find a cubic function with the given zeros.7, -3, 2
X1=7
X2=-3
X3=2
f(x)=(x-7)(x+3)(x-2)=(x2-4x-21)(x-2)=x3-6x2-13x+42
the answer is C) f(x) = x3 - 6x2 - 13x + 42  

14. Find the remainder when f(x) is divided by (x - k).f(x) = 7x4 + 12x3 + 6x2 - 5x + 16; k = 3 
 f(x)=7(3)4+12(3)3+6(3)2-5(3)+16=946
  The answer is the B) 946

15.  Use the Rational Zeros Theorem to write a list of all potential rational zerosf(x) = x3 - 10x2 + 4x - 24

The constant term of () is -24

 The leading coefficient is 1.

We have to only consider the factors of the constant (leading coefficient = 1)

The factors are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 24, -24

The answer is A) ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

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X^2-2=x
x^2-2+x=0
x^2+x-2=0
(x+2)(x-1)=0
Therefore, x+2=0 or x-1=0
Hence, x=-2 or x=1
Solutions= -2 or 1
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Hope this helps

Have a great day/night

Feel free to ask any questions
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The problem statement gives a relation between the amount removed from one bag and the amount removed from the other. It asks for the amount remaining in each bag. Thus, there are several choices for variables in this problem, some choices resulting in more complicated equations than others.

Let's do it this way: let x represent the amount remaining in bag 1. Then the amount removed from bag 1 is (100-x). The amount remaining in bag 2 is 2x, so the amount removed from that bag is (100-2x). The problem statement tells us the relationship between amounts removed:

... (100 -x) = 3(100 -2x)

... 100 -x -3(100 -2x) = 0 . . . . . . subtract the right side

... 5x -200 = 0 . . . . . . . . . . . . . . eliminate parentheses and collect terms

... x -40 = 0 . . . . . . . . . . . . . . . . .divide by 5

... x = 40 . . . . . . . . . . . . . . . . . . . add 40

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<u>Check</u>

The amount removed from the first bag is 60 kg. The amount removed from the second is 20 kg. The amount removed from the first bag is 3 times the amount removed from the second bag, as described.


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Step-by-step explanation:

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