In the process oif finding the difference quoitent for the function f(x)=π,Joan writes f(π+h)-f(π)/h

Answers

Answer 1

Answer:

Joan's expression for the difference quotient is incorrect. The correct form is [f(π+h) - f(π)] / h, where f(x) = π, indicating the difference between two function values divided by h.

Joan's expression, f(π+h) - f(π) / h, is written incorrectly. The correct expression for the difference quotient for the function f(x) = π should be:

[f(π+h) - f(π)] / h

So, the correct expression is to subtract f(π) from f(π+h) and then divide the result by h. Joan made an error in the placement of the brackets, and the corrected expression accurately represents the difference quotient for this function.

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What are the formulas for the volume of a cylinder and come ?

Answers

Volume of a cylinder: π * r^2 * h

Volume of a cone: 1/3 of the volume of a cylinder. V = π * r^2 * h/3

Explanation: r is the radius, h is the height, and if needed, you can use 3.14 for π

Answer:

The volume of a cylinder is: π × r2 × h

The volume of a cone is: 1 3 π × r2 × h

Determine the general solution of:

sin 3x - sin x = cos2x

Answers

sin³ x-sin x=cos ² x

we know that:
sin²x + cos²x=1 ⇒cos²x=1-sin²x
Therefore:

sin³x-sin x=1-sin²x
sin³x+sin²x-sin x-1=0

sin³x=z

z³+z²-z-1=0

we divide by Ruffini method:
1   1 -1 -1
1 1 2 1 z=1
-------------------------------------
1   2 1 0
-1   -1 -1 z=-1
--------------------------------------
1 1 0 z=-1

Therefore; the solutions are z=-1 and z=1

The solutions are:
if z=-1, then
sin x=-1   ⇒x= arcsin -1=π+2kπ (180º+360ºK) K∈Z

if z=1, then
sin x=1   ⇒ x=arcsin 1=π/2 + 2kπ (90º+360ºK) k∈Z

π/2 + 2kπ U   π+2Kπ=π/2+kπ k∈Z ≈(90º+180ºK)

Answer: π/2 + Kπ or 90º+180ºK K∈Z
Z=...-3,-2,-1,0,1,2,3,4....

Angles help find values

Answers

Here's the rule. I'm SURE you learned it in Middle School. Or,
I guess I should say: I'm SURE it was taught in Middle School.

Vertical angles are equal.

"Vertical angles" are the pair of angles that don't share a side,
formed by two intersecting lines.

AND ... even if you forgot it since hearing it in Middle School,
it was clearly explained in the answer to the question that
you posted 9 minutes before this one.

In #8, 'x' and 'z' are vertical angles.
'y' and 116° are vertical angles.

In #9, 'B' and 131° are vertical angles.

In #10, 'B' and 135° are vertical angles.

For all of these, it'll also help you to remember that all the angles
on one side of any straight line add up to 180°.

Angles around a point = 360°
Opposite angles are the same

8. y=116°
   x=64°
   z=64°

9.B=131°
   x=49°

10.B=135°
     x=45°

Refer to the figure and find the volume V generated by rotating the given region about the specified line.R3 about AB

Answers

Answer:

Hence, volume is: $(34\pi)/(45)$ cubic units.

Step-by-step explanation:

We will first express our our equation of the curve and the line bounded by the region in terms of the variable y.

i.e. the curve is re $x=(1)/(16)y^4$

and the line is given as: $x=(1)/(2)y$

Since after rotating the given region $R_(3)$ about the line AB.

we see that for the following graph

the axis is located at x=1.

and the outer radius(R) is: $(1)/(16)y^4$

and the inner radius(r) is: $(1)/(2)y$

Now, the area of the graph= area of the disc.

Area of graph= $\pi(R^2-r^2)$

Now the volume is given as:

$Volume=\int\limits^2_0 {Area} \, dy$

On calculating we get:

Volume= $(34\pi)/(45)$ cubic units.

The volume V generated by rotating the given region about the specified line R3 about AB is $\boxed{\frac{{34\pi }}{{45}}{\text{ uni}}{{\text{t}}^3}}.$

Further explanation:

Given:

The coordinates of point A is $\left( {1,0} \right).$

The coordinates of point B is $\left( {1,2} \right).$

The coordinate of point C is $\left( {0,2} \right).$

The value of y is $y = 2\sqrt[4]{x}.$

Explanation:

The equation of the curve is $y = 2\sqrt[4]{x}.$

Solve the above equation to obtain the value of x in terms of y.

$\begin{aligned}{\left( y \right)^4}&={\left( {2\sqrt[4]{x}} \right)^4} \n{y^4}&=16x\n\frac{1}{{16}}{y^4}&= x\n\end{aligned}$

The equation of the line is $x = (1)/(2)y.$

After rotating the region ${R_3}$ is about the line AB.

From the graph the inner radius is ${{r_2} = (1)/(2)y$ and the outer radius is ${{r_1}=\frac{1}{{16}}{y^4}.$

${\text{Area of graph}}=\pi\left( {{r_1}^2 - {r_2}^2} \right)$

$Area = \pi\left( {{{\left({\frac{1}{{16}}{y^4}} \right)}^2} - {{\left({(1)/(2)y} \right)}^2}}\right)$

The volume can be obtained as follows,

$\begin{aligned}{\text{Volume}}&=\int\limits_0^2 {Area{\text{ }}dy}\n&=\int\limits_0^2{\pi \left( {{{\left({\frac{1}{{16}}{y^4}} \right)}^2} - {{\left( {(1)/(2)y} \right)}^2}} \right){\text{ }}dy}\n&= \pi \int\limits_0^2 {\left( {\frac{1}{{256}}{y^8} - (1)/(4){y^2}} \right){\text{ }}dy}\n\end{aligned}$

Further solve the above equation.

$\begin{aligned}{\text{Volume}}&=\pi \left[ {\int\limits_0^2 {\frac{1}{{256}}{y^8}dy - } \int\limits_0^2{(1)/(4){y^2}{\text{ }}dy} } \right]\n&= \frac{{34\pi }}{{45}}\n\end{aligned}$

The volume V generated by rotating the given region about the specified line R3 about AB is $\boxed{\frac{{34\pi }}{{45}}{\text{ uni}}{{\text{t}}^3}}.$

Learn more:

1. Learn more about inverse of the functionbrainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Volume of the curves

Keywords: area, volume of the region, rotating, generated, specified line, R3, AB, rotating region.

To find the difference of 7- 3 5/12 how do yu rename the 7

Answers

Since we need them to be fractions, 7 could be 7/1. but now we need a common denominator, so lets find one!

7*12 = 84
1*12 = 12

Answer: 84/12

Solve for x: log6x-log2=1

Answers

$log6x-log2=1;\ D:6x > 0\to x > 0\to x\in\mathbb{R^+}\n\nlog6x-log2=1\n\nlog\left(6x:2\right)=log10^1\n\n3x=10\ /:3\n\nx=(10)/(3)\n\nx=3(1)/(3)\in D$

$a > 0\ \wedge\ b > 0\ \wedge\ c > 0\ \wedge\ d\in\mathbb{R}\n\nlog_ab-log_ac=log_a(b:c)\n\nlog_ab=d\iff\ a^d=b\ \ (log_ab=log_aa^d)$

Log(6x/2)=1
Log(3x)=1
This is the same as saying 10^1=3x
10=3x
x=10/3 or 3 and 1/3

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