Which action is not a step in using paperfolding to find the midpoint of a line segment?

Answer: $0.44

Step-by-step explanation: a) (3 * 0.75) + ((1/4)*2.76) = 2.25 + 0.69 = 2.94

Pays with $5 - $2.94 = $2.06 Change

((1/2) * 3.24) = $1.62

Money left = 2.06 - 1.62 => $0.44

Answer:

B: Ones.

Step-by-step explanation:

Because this number is 4.09, and the decimal is right next to the 4, that means that it is in the ones place. Decimals are always adjacent on the right to the ones place.

Answer:

Ok so on a clock there is 12 numbers where 12 is on top so at 12 am and 12 pm noon and midnight you will be at the top of the clock

Hope This Helps!!!

During the ride on the London Eye, you will be at the same height as the top of the Elizabeth Tower at approximately 21 minutes and 43.16 seconds after the start of the ride.

To determine the time(s) during the ride on the London Eye when you are at the same height as the top of the Elizabeth Tower (commonly known as Big Ben), we need to consider the height of the London Eye and its rotational motion.

Given that the Elizabeth Tower is 320 feet tall, we need to find the position of the London Eye when its height aligns with the top of the tower.

The London Eye has a height of 443 feet, and it completes one full rotation in approximately 30 minutes (or 1800 seconds). This means that it moves at a constant angular velocity of 360 degrees per 1800 seconds.

To find the time(s) when the heights align, we can set up a proportion:

(Height of the Elizabeth Tower) / (Height of the London Eye) = (Angle covered by the London Eye) / 360 degrees

Substituting the given values:

320 / 443 = (Time to align) / 1800

Simplifying the equation:

(Time to align) = (320 / 443) * 1800

Calculating the value:

(Time to align) ≈ 1303.16 seconds

Converting the time to minutes and seconds:

(Time to align) ≈ 21 minutes and 43.16 seconds

Therefore, during the ride on the London Eye, you will be at the same height as the top of the Elizabeth Tower at approximately 21 minutes and 43.16 seconds after the start of the ride.

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Answer:

C $13,220

Step-by-step explanation:

Answer:

The approximate percentage of lightbulb replacement requests numbering between 54 and 63 is of 49.85%.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 54, standard deviation = 3.

What is the approximate percentage of lightbulb replacement requests numbering between 54 and 63?

63 = 54 + 3*3

So between the mean and 3 standard deviations above the mean.

The normal distribution is symmetric, which means that 50% of the values are below the mean and 50% are above.

Of those 50% above, 99.7% are below 63. So

0.5*0.997 = 0.4985

0.4985*100% = 49.85%

The approximate percentage of lightbulb replacement requests numbering between 54 and 63 is of 49.85%.

9514 1404 393

Answer:

Step-by-step explanation:

a) The starting height is h(0) = 7.5 feet, the constant in the quadratic function.

The irrigation system is positioned 7.5 feet above the ground

__

b) The axis of symmetry for quadratic ax^2 +bx +c is x = -b/(2a). For this quadratic, that is x=-10/(2(-1)) = 5. This is the horizontal distance to the point of maximum height. The maximum height is ...

  h(5) = (-5 +10)(5) +7.5 = 32.5 . . . feet

The spray reaches a maximum height of 32.5 feet at a horizontal distance of 5 feet from the sprinkler head.

__

c) The maximum distance will be √32.5 + 5 ≈ 10.7 ft.

The spray reaches the ground at about 10.7 feet away.

Answer:

7.5

32.5

5

maximum

10.7

Step-by-step explanation:

Answer:

[tex]44[/tex]

Step-by-step explanation:

The dimensions of the garden is 12 by 8. If we have a walkway that surrounds the garden, the dimensions of the walkway is 2. Since it surrounds the rectangle all sides add 2 to each of the dimensions so now the dimensions of the garden and walkway is 14×10.

The area of the garden is 96 square ft.

The area of the garden and walkway is 140 so let subtract the area of the garden from the total area of both the garden and walkway.

[tex]140 - 96 = 44[/tex]

The area is 44.

Answer:

120 square feet

Step-by-step explanation:

(8+2*2)

(18+2*2) - 8*18 = 120 square feet.

Answer:

22

Step-by-step explanation:

3x-14= 4(x-9)

3×-14= 4x-36

4x-36-3x+14=0

×-22÷0

x=22

(1) Recall that

sin(x - y) = sin(x) cos(y) - cos(x) sin(y)

sin²(x) + cos²(x) = 1

Given that α lies in the third quadrant, and β lies in the fourth quadrant, we expect to have

• sin(α) < 0 and cos(α) < 0

• sin(β) < 0 and cos(β) > 0

Solve for cos(α) and sin(β) :

cos(α) = -√(1 - sin²(α)) = -3/5

sin(β) = -√(1 - cos²(β)) = -5/13

Then

sin(α - β) = sin(α) cos(β) - cos(α) sin(β) =  (-4/5) (12/13) - (-3/5) (-5/13)

==>   sin(α - β) = -63/65

(2) In the second identity listed above, multiplying through both sides by 1/cos²(x) gives another identity,

sin²(x)/cos²(x) + cos²(x)/cos²(x) = 1/cos²(x)

==>   tan²(x) + 1 = sec²(x)

Rewrite the equation as

3 sec²(x) tan(x) = 4 tan(x)

3 (tan²(x) + 1) tan(x) = 4 tan(x)

3 tan³(x) + 3 tan(x) = 4 tan(x)

3 tan³(x) - tan(x) = 0

tan(x) (3 tan²(x) - 1) = 0

Solve for x :

tan(x) = 0   or   3 tan²(x) - 1 = 0

tan(x) = 0   or   tan²(x) = 1/3

tan(x) = 0   or   tan(x) = ±√(1/3)

x = arctan(0) + nπ   or   x = arctan(1/√3) + nπ   or   x = arctan(-1/√3) + nπ

x = nπ   or   x = π/6 + nπ   or   x = -π/6 + nπ

where n is any integer. In the interval [0, 2π), we get the solutions

x = 0, π/6, 5π/6, π, 7π/6, 11π/6

(3) You only need to rewrite the first term:

[tex]\dfrac{\sin(x)}{1-\cos(x)} \times \dfrac{1+\cos(x)}{1+\cos(x)} = \dfrac{\sin(x)(1+\cos(x))}{1-\cos^2(x)} = \dfrac{\sin(x)(1+\cos(x)}{\sin^2(x)} = \dfrac{1+\cos(x)}{\sin(x)}[/tex]

Then

[tex]\dfrac{\sin(x)}{1-\cos(x)}+\dfrac{1-\cos(x)}{\sin(x)} = \dfrac{1+\cos(x)+1-\cos(x)}{\sin(x)}=\dfrac2{\sin(x)}[/tex]

Answer:

sin A = 4/5

cos A = 3/5

Step-by-step explanation:

SOHCAHTOA

cot A = 1/tan A

tan A = opp/adj

cot A = 1/tan A = adj/opp

cot A = 3/4

adj = 3; opp = 4

adj^2 + opp^2 = hyp^2

3^2 + 4^2 = hyp^2

9 + 16 = hyp^2

hyp = 5

sin A = opp/hyp

sin A = 4/5

cos A = adj/hyp

cos A = 3/5

Answer:

sin A = 4/5

cos A = 3/5

Answer:

6.

Step-by-step explanation:

.

Answer:

[tex]C)\:8[/tex]

8 units tiles must be added

--------------------------------------

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Answer:

[tex]82\sqrt{21}\text{ or approximately 375.77 miles per hour}[/tex]

Step-by-step explanation:

Please refer to the diagram below. R is the radar station and x is the distance from the station to the plane.

We are given that the plane is flying horizontally at an altitude of two miles and at a speed of 410 mph. And we want to find the rate at which the distance from the plane to the station is increasing when it is five miles away from the station.

In other words, given da/dt = 410 and x = 5, find dx/dt.

From the Pythagorean Theorem:

[tex]a^2+4=x^2[/tex]

Implicitly differentiate both sides with respect to time t. Both a and x are functions of t. Hence:

[tex]\displaystyle 2a\frac{da}{dt}=2x\frac{dx}{dt}[/tex]

Simplify:

[tex]\displaystyle a\frac{da}{dt}=x\frac{dx}{dt}[/tex]

Find a when x = 5:

[tex]a=\sqrt{5^2-2^2}=\sqrt{21}[/tex]

Therefore, dx/dt when da/dt = 410, x = 5, and a = √(21) is:

[tex]\displaystyle \frac{dx}{dt}=\frac{(\sqrt{21})(410)}{5}=82\sqrt{21}\approx 375.77\text{ mph}[/tex]

The rate at which is distance from the plane to the radar station is increasing at a rate of approximately 375.77 miles per hour.

Answer:

Step-by-step explanation:

(0, -1)  & (3 ,1)

[tex]Slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{1-[-1]}{3-0}\\\\=\frac{1+1}{3}\\\\=\frac{2}{3}[/tex]

Let's see

[tex]\\ \tt\leadsto A=P(1+r/n)^{nt}[/tex]

[tex]\\ \tt\leadsto A=1200(1+0.055)^t[/tex]

[tex]\\ \tt\leadsto A=1200(1.055)^t[/tex]

Answer:

[tex]\sf A=1200(1.055)^t[/tex]

Step-by-step explanation:

Annual Compound Interest Formula

[tex]\large \text{$ \sf A=P\left(1+r\right)^{t} $}[/tex]

where:

Given:

Substitute the given values into the equation:

[tex]\implies \sf A=1200(1+0.055)^t[/tex]

[tex]\implies \sf A=1200(1.055)^t[/tex]

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Answer:

Following are the response to the given points:

Step-by-step explanation:

For question 5.11:

For point a:

For all the particular circumstances, it was not an appropriate sampling strategy as each normal distribution acquired is at a minimum of 30(5) = 150 or 2.5 hours for a time. Its point is not absolutely fair if it exhibits any spike change for roughly 10 minutes.

For point b:

The problem would be that the process can transition to an in the state in less than half an hour and return to in the state. Thus, each subgroup is a biased selection of the whole element created over the last [tex]2 \frac{1}{2}[/tex] hours. Another sampling approach is a group.

For question 5.12:

This production method creates 500 pieces each day. A sampling section is selected every half an hour, and the average of five dimensions can be seen in a [tex]\bar{x}[/tex]line graph when 5 parts were achieved.

This is not an appropriate sampling method if the assigned reason leads to a sluggish, prolonged uplift. The difficulty would be that gradual or longer upward drift in the procedure takes or less half an hour then returns to a controlled state. Suppose that a shift of both the detectable size will last hours [tex]2 \frac{1}{2}[/tex] . An alternative type of analysis should be a random sample of five consecutive pieces created every [tex]2 \frac{1}{2}[/tex] hour.

Answer:

The cost of 16 onions is $ 1.20.

Step-by-step explanation:

To determine what is the cost, in dollars, of 16 onions if 3 onions weigh 1.5 lb and the price of onions is 33 cents per kilogram, the following calculation must be performed:

1.5 pounds = 0.68 kilos

0.68 / 3 = 0.22666 kilos each onion

16 x 0.22666 = 3.626 kilos

0.33 x 3.626 = 1.20

Therefore, the cost of 16 onions is $ 1.20.

her profit is 204 dollars and 68 cents= 204.68

Using a linear approximation, the estimated cube root of 217 is  6.00925.

Given that the number is,

The cube root of 217

Now, for the cube root of 217 using a linear approximation, use differentials.

So, the derivative of the function [tex]f(x) = x^{(1/3)[/tex] at a known point.

Taking the derivative of [tex]f(x) = x^{(1/3)[/tex], we get:

[tex]f'(x) = (\dfrac{1}{3} )x^{-2/3[/tex]

Now, we can choose a point near 217 to evaluate the linear approximation.

Let's use x = 216, which is a perfect cube.

Substituting x = 216 into the derivative, we get:

[tex]f'(216) = (\dfrac{1}{3} )(216)^{-2/3[/tex]

            [tex]= 0.00925[/tex]

Next, use the linear approximation formula:

Δy ≈ f'(a)Δx

Since our known point is a = 216 and we want to estimate the cube root of 217,

since 217 - 216 = 1

Hence, Δx = 1

Δy ≈ f'(216)

Δx ≈ 0.00925 × 1

      ≈ 0.00925

Finally, add this linear approximation to the known value at the known point to get our estimate:

Estimated cube root of 217 ;

f(216) + Δy = 6 + 0.00925

                 = 6.00925

Therefore, the estimated cube root of 217 is 6.00925.

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Answer:

d

Step-by-step explanation:

Step-by-step explanation:

this is the correct answer for the question

Answer:

D

Step-by-step explanation:

Answer:

12.259-12.25 890654321

Answer:

Step-by-step explanation:

There are 2 different ways to do this: calculus and by completing the square. In this particular instance, calculus is WAY easier, and since I don't know for what class you are doing this, I'll do both ways. First the calculus way. We know the position equation, and the first derivative of the position is velocity. We also know that when the velocity is equal to 0 is when the object is at its max height. So we'll find the derivative first, then solve it for t:

If [tex]s(t)=-3t^2+20t+2[/tex] then the first derivative is

v(t) = -6t + 20 Solving for t requires that we set the velocity equal to 0 (again, this is where the object is at its max height), so

0 = -6t + 20 and

-20 = -6t so

t = 3.3 seconds. Now that we know that at 3.3 seconds the object is at its highest point, we sub that time into the position function to see where it is at that time:

s(3.3) = [tex]-3(3.3)^2+20(3.3)+2[/tex] and

s(3.3) = 35.3 meters.

Now onto the more difficult way...completing the square. Begin by setting the position function equal to 0 and then move over the constant to get:

[tex]-3t^2+20t=-2[/tex] Since the leading coefficient is not a 1 (it's a 3), we have to factor out the 3, leaving us with:

[tex]-3(t^2-\frac{20}{3}t)=-2[/tex] Now the rule is to take half the linear term, square it, and add it to both sides. Our linear term is [tex]\frac{20}{3}[/tex] and half of that is [tex]\frac{20}{6}[/tex]. Squaring that:

[tex](\frac{20}{6})^2=\frac{400}{36}=\frac{100}{9}[/tex]. We will add that in to both sides. On the left it's easy, but on the right we have to take into account that we still have that -3 sitting out front, refusing to be ignored. So we have to multiply it in when we add it to the right. Doing that gives us:

[tex]-3(t^2-\frac{20}{3}t+\frac{100}{9})=-2-\frac{100}{3}[/tex] We will clean this up a bit now. The reason we do this is because on the left we have created a perfect square binomial which will give us the time we are looking for to answer this question. Simplifying the right and at the same time writing the perfect square binomial  gives us:

[tex]-3(t-\frac{20}{6})^2=-\frac{106}{3}[/tex] Now the last step is to move the constant back over and set the quadratic back equal to y:

[tex]y=-3(t-\frac{20}{6})^2+\frac{106}{3}[/tex].  The vertex of this quadratic is

[tex](\frac{20}{6},\frac{106}{3})[/tex] where

[tex]\frac{20}{6}=3.3[/tex] as the time it takes for the ball to reach its max height of

[tex]\frac{106}{3}=35.3[/tex] meters.

I'd say if you plan on taking calculus cuz you're not there yet, you'll see that many of these types of problems become much simpler when you know it!

Answer:

[tex]x = 77[/tex]

Step-by-step explanation:

Given

[tex]First \to Second[/tex]

[tex]35 \to x[/tex]

[tex]20 \to 44[/tex]

[tex]20 \to 44[/tex]

Required

Find x by SSS

Represent the triangle sides as a ratio

[tex]35 : x = 20 : 44[/tex]

Express as fraction

[tex]\frac{x}{35} = \frac{44}{20}[/tex]

Multiply by 35

[tex]x = \frac{44}{20} * 35[/tex]

[tex]x = \frac{44 * 35}{20}[/tex]

[tex]x = \frac{1540}{20}[/tex]

[tex]x = 77[/tex]

Answer:

ccccccccccccccccccccccccccc

Step-by-step explanation:

Answer:

4

Step-by-step explanation:

The mean is the average of a data set. It can be found by adding up all of the values in a data set and then dividing it by the number of values in the data set.

The values in this data set;

[tex]7,5,5,3,2,2[/tex]

The number of values in this data set,

[tex]6[/tex]

Find the mean;

[tex]\frac{sum\ of\ vlaues}{number\ of\ values}[/tex]

[tex]=\frac{7+5+5+3+2+2}{6}\\\\=\frac{24}{6}\\\\=4[/tex]

Answer:4778.3 cm^3

Step-by-step explanation: The formula for volume of a cone is V=1/3h pi r^2. By plugging in the height and the radius we get our answer.

Answer:

4778.4 :)

Step-by-step explanation:

The power consumed in the lamp marked 100W - 110V is 15.68W

The power consumed in the lamp marked 100W - 220V is 62.73W

Given:

First lamp rating

Power (P) = 100W

Voltage (V) = 110V

Second lamp rating

Power (P) = 100W

Voltage (V) = 220V

Source

Voltage = 220V

i. Get the resistance of each lamp.

Remember that power (P) of each of the lamps is given by the quotient of the square of their voltage ratings (V) and their resistances (R). i.e

P = [tex]\frac{V^2}{R}[/tex]

Make R subject of the formula

⇒ R = [tex]\frac{V^2}{P}[/tex]             ------------------(i)

For first lamp, let the resistance be R₁. Now substitute R = R₁, V = 110V and P = 100W into equation (i)

R₁ = [tex]\frac{110^2}{100}[/tex]

R₁ = 121Ω

For second lamp, let the resistance be R₂. Now substitute R = R₂, V = 220V and P = 100W into equation (i)

R₂ = [tex]\frac{220^2}{100}[/tex]

R₂ = 484Ω

ii. Get the equivalent resistance of the resistances of the lamps.

Since the lamps are connected in series, their equivalent resistance (R) is the sum of their individual resistances. i.e

R = R₁ + R₂

R  = 121 + 484

R = 605Ω

iii. Get the current flowing through each of the lamps.

Since the lamps are connected in series, then the same current flows through them. This current (I) is produced by the source voltage (V = 220V) of the line and their equivalent resistance (R = 605Ω). i.e

V = IR [From Ohm's law]

I = [tex]\frac{V}{R}[/tex]

I = [tex]\frac{220}{605}[/tex]

I = 0.36A

iv. Get the power consumed by each lamp.

From Ohm's law, the power consumed is given by;

P = I²R

Where;

I = current flowing through the lamp

R = resistance of the lamp.

For the first lamp, power consumed is given by;

P = I²R           [Where I = 0.36 and R = 121Ω]

P = (0.36)² x 121

P = 15.68W

For the second lamp, power consumed is given by;

P = I²R           [Where I = 0.36 and R = 484Ω]

P = (0.36)² x 484

P = 62.73W

Therefore;

The power consumed in the lamp marked 100W - 110V is 15.68W

The power consumed in the lamp marked 100W - 220V is 62.73W