For the function f(x)=|4x−2|+10, evaluate f(−3).

Answer:

The volume of the solid = 1444

Step-by-step explanation:

Given that:

The region of the solid is bounded by the curves [tex]y = 38 \sqrt{cos \ x}[/tex] and the axis on [tex][-\dfrac{\pi}{2}, \dfrac{\pi}{2}][/tex]

using the slicing method

Let say the solid object extends from a to b and the cross-section of the solid perpendicular to the x-axis has an area expressed by function A.

Then, the volume of the solid is ;

[tex]V = \int ^b_a \ A(x) \ dx[/tex]

However, each perpendicular slice is an isosceles leg on the xy-plane and vertical leg above the x-axis

Then, the area of the perpendicular slice at a point [tex]x \ \epsilon \ [-\dfrac{\pi}{2},\dfrac{\pi}{2}][/tex] is:

[tex]A(x) =\dfrac{1}{2} \times b \times h[/tex]

[tex]A(x) =\dfrac{1}{2} \times(38 \sqrt{cos \ x})^2[/tex]

[tex]A(x) =\dfrac{1444}{2} \ cos \ x[/tex]

[tex]A(x) =722 \ cos \ x[/tex]

Applying the general slicing method ;

[tex]V = \int ^b_a \ A(x) \ dx \\ \\ V = \int ^{\dfrac{\pi}{2} }_{-\dfrac{\pi}{2}} (722 \ cos x) \ dx \\ \\ V = 722 \int ^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} cosx \dx[/tex]

[tex]V = 722 [ sin \ x ] ^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}[/tex]

[tex]V = 722 [sin \dfrac{\pi}{2} - sin (-\dfrac{\pi}{2})][/tex]

[tex]V = 722 [sin \dfrac{\pi}{2} + sin \dfrac{\pi}{2})][/tex]

[tex]V = 722 [1+1][/tex]

[tex]V = 722 [2][/tex]

V = 1444

∴ The volume of the solid = 1444

Answer:

35.75

Step-by-step explanation:

6.5 times 5.50 = 35.75

or

5.50+5.50+5.50+5.50+5.50+5.50=35.75

Steps to solve:

2(n + 3) = 2n + 3

~Distribute left side

2n + 6 = 2n + 3

~Subtract 6 to both sides

2n = 2n - 3

~Subtract 2n to both sides

n = -3

Best of Luck!

Answer:

10 feet

Step-by-step explanation:

if it is 10 feet above and needs to touch the ground it needs to drop that same amount

Answer:

10ft.

Step-by-step explanation:

if the balloon is floating 10ft from the ground the it can only drop 10ft to reach the ground.

*hope this helps. please give me brainliest, i only need four more *

Answer:

The parabola [tex]y=x^{2}[/tex] at the points [tex](-2+\sqrt{17},21-4\sqrt{17})[/tex] and [tex](-2-\sqrt{17},21+4\sqrt{17})[/tex] when [tex]t=t_{1}=-4+\sqrt{17}[/tex] and [tex]t=t_{2}=-4-\sqrt{17}[/tex]

Step-by-step explanation:

We have the following line written in parametric form :

[tex]L(t)=(2+t,5-4t)[/tex] with [tex]t[/tex] ∈ IR.

In order to find the intersection between [tex]L(t)[/tex] and the parabola [tex]y=x^{2}[/tex] we know that ''[tex]2+t[/tex]'' is the x-coordinate of the line and ''[tex]5-4t[/tex]'' is the y-coordinate of the line. Now, to solve this problem we need to find the values of ''[tex]t[/tex]'' in which the intersection occurs. We can do this by replacing the components    ''[tex]x[/tex]'' and ''[tex]y[/tex]'' of [tex]L(t)[/tex] in the equation of the parabola ⇒

[tex]L(t)=(2+t,5-4t)[/tex] = ( x component , y component ) = ( x , y ) ⇒

In the parabola : [tex]y=x^{2}[/tex] ⇒ [tex]5-4t=(2+t)^{2}[/tex]

Solving the equation we find that :

[tex]t^{2}+8t-1=0[/tex]

Using the quadratic formula with

[tex]a=1[/tex] , [tex]b=8[/tex] and [tex]c=-1[/tex]

We find that the two possible values for t :

[tex]t_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}[/tex]     and  [tex]t_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}[/tex]

are [tex]t_{1}=-4+\sqrt{17}[/tex] and [tex]t_{2}=-4-\sqrt{17}[/tex]

This values [tex]t_{1}[/tex] and [tex]t_{2}[/tex] are the values of the parameter t where the line intersects the parabola so we can find the points by replacing the values of the parameter in the equation [tex]L(t)[/tex] :

[tex]L(t_{1})=(-2+\sqrt{17},21-4\sqrt{17})[/tex] and

[tex]L(t_{2})=(-2-\sqrt{17},21+4\sqrt{17})[/tex]

The final answer is

The parabola [tex]y=x^{2}[/tex] at the points [tex](-2+\sqrt{17},21-4\sqrt{17})[/tex] and [tex](-2-\sqrt{17},21+4\sqrt{17})[/tex] when [tex]t=t_{1}=-4+\sqrt{17}[/tex] and [tex]t=t_{2}=-4-\sqrt{17}[/tex]

Answer:

I think it's 352. 02 g/mol

Step-by-step explanation:

googled it :)

Answer:

352.02 g/mol

In chemistry, the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles.

Answer:

1.17.9

2. 120.61

Step-by-step explanation:

Answer:

17.9

120.61

Step-by-step explanation:

Answer:

25m

Step-by-step explanation:

10/6 = length of shadow per metre of pole

so 15 x 10/6 = shadow of 15m high pole

150/6 = 25m

Answer:

[tex]f^{-1}(x)=\frac{1}{6}\ln(x+9)[/tex]

Step-by-step explanation:

So we have the function:

[tex]f(x)=e^{6x}-9[/tex]

To solve for the inverse of a function, change f(x) and x, change the f(x) to f⁻¹(x), and solve for it. Therefore:

[tex]x=e^{6f^{-1}(x)}-9[/tex]

Add 9 to both sides:

[tex]x+9=e^{6f^{-1}(x)}[/tex]

Take the natural log of both sides:

[tex]\ln(x+9)=\ln(e^{6f^{-1}(x)})[/tex]

The right side cancels:

[tex]\ln(x+9)=6f^{-1}(x)[/tex]

Divide both sides by 6:

[tex]f^{-1}(x)=\frac{1}{6}\ln(x+9)[/tex]

And we're done!

Answer:

Option (B)

Step-by-step explanation:

Sue's weight at the start of the year = 82 kg

Sue weighed herself at the end of every year, so that the straight line on the graph shows the gradual reduction in the weight every month.

Slope of the line will describe the reduction in the weight per month.

Let the two points on the given line are (1, 80) and (11, 60).

Slope of the line will be,

m = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

   = [tex]\frac{60-80}{11-1}[/tex]

   = -2 kg per month

(Here minus sign denotes the reduction of weight per month)

Option (B) will be the answer.

Answer:

2 kg

Step-by-step explanation:

Answer:

$1,079.04

Step-by-step explanation:

multiply: 89.92x12=1078.04

Answer:-0.5

Step-by-step explanation:

Answer: x = -1/2y + 2

Step-by-step explanation:

12x + 6y = 24  To solve for x  first start by subtracting 6y from both sides

      - 6y    -6y

12x =  24 - 6y    Now divide each side by 12

x= 2 - 1/2y

Turn the percentage into a decimal.

8.25% / 100 = 0.0825

Multiply.

39.89 * 0.0825 ≈ $3.29

Subtract.

39.89 + 3.29 = $43.18

Best of Luck!

Answer:

u=-1

Step-by-step explanation:

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

u= -1

Answer:

375 miles

Step-by-step explanation:

m=miles

I set it up as an equation.                                                    $75+.60m=.80m                      

And solved from there                                                                                                                                                                                                                                                                                

Answer:

375 miles :D

Step-by-step explanation:

Answer:

if a student did not achieve 90 percent on the geometry final, then the student did not pass geometry class

Step-by-step explanation:

Answer:(-2,2)

Step-by-step explanation:

21.406 in standard form

Answer:

21406 X 10^3

Answer:

Yes

Step-by-step explanation:

(5*2)x = 10x

5*2 = 10 so

10x = 10x

They are equal.

Answer:

Yes

Step-by-step explanation:

5 x 2 is 10, and so 5(2)x is 10x

No.

Here is why:

If you mean equally divided then no because of the little 5 in the ones's place. If that 5 was a 0, it would definitely be divided equally.

Hope this makes sense!

Answer:

no

Step-by-step explanation:

Answer:

"abscissa"

Step-by-step explanation:

The first number in an ordered pair of numbers in a coordinate system is generally called abscissa.

Answer:

x value is the correct answer

Step-by-step explanation:

I just took the test 2021 thank me later

Answer:

−6y+24

Step-by-step explanation:

Answer:

-4

Step-by-step explanation:

parenthesis come first (PEMDAS)

2×8=16

2×3y=6y

9+16-6y-1

Add like terms

9+16=25

25-6y-1

Subtract like terms

25-1=24

-6y=24

Multiple each sides by using fractions

1/(-)6×24/1=4

Answer:

a).To break even

They need to enroll 87 student

B) Profit=$100000

C) yeah they can think of that.

Profit= $70000

Step-by-step explanation:

Initial start up cost = $350000

Cost from adminstrative bodies

= 12000 for first 100 student

= $1200000

Total cost =$1200000+$350

Totral cost =$ 1550000

Tuition cost

= $18000 per student

To break even

The college needs to enroll

1550000/18000= 86.1

They need to enroll 87 student

B) if they enroll 75 students

Adminstrative cost = 12000*75

Adminstrative cost= 900000

Total cost= 350000+900000

Total cost =$ 1250000

Tuition fee= $18000*75

Tuition fee=$ 1350000

Their profit= 1350000-1250000

Profit=$100000

C).if tuition is $24000 and enrollment number= 35

Adminstrative cost = 35*12000

Adminstrative cost = $420000

Total cost= $770000

Tuition fee= $24000*35

Tuition fee= $840000

Profit= $840000-$770000

Profit= $70000

Answer:

-16

Step-by-step explanation:

4 + 5(8-10)2

4 + 5(-2)2

4 + 10(-2)

4 - 20

-16

The answer is: 24

I assume this is a question from k12

  4 + 5(8 - 10)^2

  4 + 5(-2)^2

  4 + 5(4)

  4 + 20

= 24

Hope this helps ^^

Answer:

0.966

Step-by-step explanation:

Given that:

Probability of DVD player breaking down before the warranty expires = 0.034

To find:

The probability that the player will not break down before the warranty expires = ?

Solution:

Here, The two events are:

1. The DVD player breaks down before the warranty gets expired.

2. The DVD player breaks down after the warranty gets expired

In other words, the 2nd event can be stated as:

The DVD does not break down before the warranty gets expired.

The two events here, have nothing in common i.e. they are mutually exclusive events.

So, Sum of their probabilities will be equal to 1.

[tex]\bold{P(E_1)+P(E_2)=1}\\\Rightarrow 0.034+P(E_2)=1\\\Rightarrow P(E_2)=1-0.034\\\Rightarrow P(E_2)=\bold{0.966}[/tex]

Don’t know the answer

Answer:

A)

[tex]f'(x) = \left\{ \begin{array}{lIl} \frac{1}{2} & \quad x >2 \\ 0& \quad x =2\\1&\quad x<2 \end{array} \right.[/tex]

B) Continuous but not differentiable.

Step-by-step explanation:

So we have the piecewise function:

[tex]f(x) = \left\{ \begin{array}{lIl} \frac{1}{2}x+5 & \quad x >2 \\ 6& \quad x =2\\x +4&\quad x<2 \end{array} \right.[/tex]

A)

To write the differentiated piecewise function, let's differentiate each equation separately. Thus:

1)

[tex]\frac{d}{dx}[\frac{1}{2}x+5}][/tex]

Expand:

[tex]\frac{d}{dx}[\frac{1}{2}x]+\frac{d}{dx}[5][/tex]

The derivative of a linear equation is just the slope. The derivative of a constant is 0. Thus:

[tex]\frac{d}{dx}[\frac{1}{2}x+5}]=\frac{1}{2}[/tex]

2)

[tex]\frac{d}{dx}[6][/tex]

Again, the derivative of a constant is 0. Thus:

[tex]\frac{d}{dx}[6]=0[/tex]

3)

We have:

[tex]\frac{d}{dx}[x+4][/tex]

Expand:

[tex]\frac{d}{dx}[x]+\frac{d}{dx}[4][/tex]

Simplify:

[tex]=1[/tex]

Now, let's substitute our original equations for the differentiated equations. The inequalities will stay the same. Therefore:

[tex]f'(x) = \left\{ \begin{array}{lIl} \frac{1}{2} & \quad x >2 \\ 0& \quad x =2\\1&\quad x<2 \end{array} \right.[/tex]

B)

For a function to be differentiable at a point, the function must be a) continuous at that point, and b) the left and right hand derivatives must be equivalent.

Let's first determine if the function is continuous at the point. Remember that a function is continuous at a point if and only if:

[tex]\lim_{x \to n^-} f(n)= \lim_{x \to n^+}f(n)=f(n)[/tex]

Let's find the left hand limit of f(x) at it approaches 2.

[tex]\lim_{x \to 2^-}f(x)[/tex]

Since it's coming from the left, let's use the third equation:

[tex]\lim_{x \to 2^-}f(x)\\=\lim_{x \to 2^-}(x+4)[/tex]

Direct substitution:

[tex]=(2+4)=6[/tex]

So:

[tex]\lim_{x \to 2^-}f(x)=6[/tex]

Now, let's find the right-hand limit:

[tex]\lim_{x \to 2^+}f(x)[/tex]

Since we're coming from the right, let's use the first equation:

[tex]\lim_{x \to 2^+}(\frac{1}{2}x+5)[/tex]

Direct substitution:

[tex](\frac{1}{2}(2)+5)[/tex]

Multiply and add:

[tex]=6[/tex]

So, both the left and right hand limits are equivalent. Now, find the limit at x=2.  

From the piecewise function, we can see that the value of f(2) is 6.

Therefore, the function is continuous at x=2.

Now, let's determine differentiability at x=2.

For a function to be differentiable at a point, both the right hand and left hand derivatives must be equivalent.

So, let's find the derivative of the function as x approahces 2 from the left and from the right.

From the differentiated piecewise function, we can see that as x approaches 2 from the left, the derivative is 1.

As x approaches 2 from the right, the derivative is 1/2.

Therefore, the right and left hand derivatives are not the same.

Thus, the function is continuous but not differentiable.

Answer:

The two numbers are 32 and 60.

Step-by-step explanation:

We can create two simultaneous equations to solve this problem using the information given in the question.

Let number 1 = x

Let number 1 = y

x = y - 28 -> ( 1 )

x + y = 92 -> ( 2 )

We can solve simultaneous equations using substitution or elimination. For this question, we will use substitution as it is the easier option.

Substitute ( 1 ) into ( 2 ):

x + y = 92 -> ( 2 )

( y - 28 ) + y = 92

2y - 28 = 92

2y = 120

y = 120 / 2

y = 60 -> ( 3 )

Substitute ( 3 ) into ( 1 ):

x = y - 28 -> ( 1 )

x = ( 60 ) - 28

x = 32

Therefore:

x = 32 , y = 60