Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Answers

Answer 1

Answer:

The area bounded by the functions f(x) and g(x) in graph below.

The given function are f(x)=x² and g(x)=√x.

What is the function?

Functions are the fundamental part of the calculus in mathematics. The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input x.

To find the area between two curves defined by functions, integrate the difference of the functions. If the graphs of the functions cross, or if the region is complex, use the absolutevalue of the difference of the functions.

Area bounded = |x²-√x|

Find the domain by finding where the expression is defined.

Interval Notation:

[0,∞)

Set-Builder Notation:{x|x≥0}

Therefore, the area bounded by the functions f(x) and g(x) in graph below.

To learn more about the function visit:

brainly.com/question/28303908.

#SPJ2

Answer 2

Answer:

Answer:

$\large\boxed{1(1)/(3)\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=√(x)\qquad\text{square of both sides}\n\n(x^2)^2=\left(√(x)\right)^2\n\nx^4=x\qquad\text{subtract}\ x\ \text{from both sides}\n\nx^4-x=0\qquad\text{distribute}\n\nx(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\n\nx^3-1=0\qquad\text{add 1 to both sides}\n\nx^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_(0)^1(√(x))dx-\int\limits_(0)^1(x^2)dx=(*)\n\n\int(√(x))dx=\int\left(x^(1)/(2)\right)dx=(2)/(3)x^(3)/(2)=(2x√(x))/(3)\n\n\int(x^2)dx=(1)/(3)x^3\n\n(*)=\left((2x√(x))/(2)\right]^1_0-\left((1)/(3)x^3\right]^1_0=(2(1)√(1))/(2)-(2(0)√(0))/(2)-\left((1)/(3)(1)^3-(1)/(3)(0)^3\right)\n\n=(2(1)(1))/(2)-(2(0)(0))/(2)-(1)/(3)(1)}+(1)/(3)(0)=2-0-(1)/(3)+0=1(1)/(3)$

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Don’t answer just for points i actually need the help

Answers

Answer:

y int is -4. slope is negative. slope is rise/run. in the graph is rises 2 for every 3 it goes to the left. this means your slope is 2/3

put all that together

-2/3x -4 is your equation

Explain the distance formula. Then use it to calculate the distance between A(1,1) and B(7,-7).

Answers

the distance formula:
$d= \sqrt{ (x_(2) - x_(1)) ^(2)+ (y_(2) -y_(1)) ^(2) }$

A(1,1) B(7,-7)

$d= \sqrt{ (7 - 1) ^(2)+ (-7 -1) ^(2) }$

$d= \sqrt{ (6) ^(2)+ (-8) ^(2) }$

$d= √( 36 + 64 )$

$d= √( 100 )$

$d=10$

The answer is 10

Answer:

d=10

Step-by-step explanation:

Solve for x
simplify the answer. this is extremely hard

Answers

D = 1/9 π m² x

Multiply each side by 9 : 9D = π m² x

Divide each side by (π m²) : 9D / (π m²) = x

Truly exhausting !

the solution is in the application

Expanding Brackets when cubed eg(x - 5)3 Answer given is xcubed -15xsquared + 75x -125 How do I work this? I can do two brackets but having trouble with the three brackets. Thanks

Answers

Solve like this - (x-5)^3
= (x-5)(x-5)^2
= (x-5) ( x^2 -10x + 25)
= x^3 -15x^2 + 75 - 125

Hope this helped

What answer describes this sequence? 99, 89.9, 78.8, 68.7,...

a) Neither arithmetic nor geometric
b) Both arithmetic and geometric
c) Arithmetic
d) Geometric

Answers

a) Neither arithmetic nor geometric

a) Neither arithmetic nor Geometric

Explanation:
Since 99-89.9= 9.1, and 89.9 - 78.8 = 11.1, the sequence is not arithmetic.
Like wise 89.9/99=0.908 and 78.8/89.9 = 0.876 makes it not a geomtric sequence

The work of a student to solve a set of equations is shown:Equation A: y = 15 – 2z
Equation B: 2y = 3 – 4z

Step 1: –2(y) = –2(15 – 2z) [Equation A is multiplied by –2.]
2y = 3 – 4z [Equation B]
Step 2: –2y = 15 – 2z [Equation A in Step 1 is simplified.]
2y = 3 – 4z [Equation B]
Step 3: 0 = 18 – 6z [Equations in Step 2 are added.]
Step 4: 6z = 18
Step 5: z = 3
In which step did the student first make an error?

Step1
Step 2
Step 3
Step 4

Answers

Option: step 2.

Because the student divided only the right side by 2. If you divide one side by a number you have to do the same with the other side to maintain the equality.

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