Need help please.. :(

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:

[tex]{\boxed{\boxed { ⎆ Answer :- }}} \ [/tex]

[tex]2x + 4 = x + 40 \\ 2x - x = 40 - 4 \\ x = 36[/tex]

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꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐

Answer:

6.

Step-by-step explanation:

.

Answer:

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

8 units tiles must be added

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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:

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]

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:

The length is 14 and the area is 196 cm².

The side length of the cube is 14 meters.

We have,

Volume of Cube = 2744 m³

To find the side length of a cube when given its volume, you can use the formula:

Side length = ∛(Volume)

So, substitute this value into the formula to calculate the side length:

Side length = ∛(2,744)

= ∛ 14  x 14 x 14

= 14 m

Therefore, the side length of the cube is 14 meters.

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Answer:

d

Step-by-step explanation:

Answer:

12.259-12.25 890654321

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:

Top left: not a function

Top right: not a function

Bottom left: function

Bottom right: not a function

Step-by-step explanation:

A function is a relationship where each x value has it's own y value ( note that domain = x values and range = y values)

For the one on the top left.

S and n have more than one y value.

Because s and n have more than one y value the relation is not a function

For the one of the top right.

There x value "c" has multiple y values therefore the relation is not a function

For the one on the bottom left

Each x value has it's own y value therefore it is a function ( note that the y values can repeat. It's only the x values that can't repeat. )

For the one on the bottom right

The x value "-5" has multiple y values therefore the relation is not a function

Answer:

D

Step-by-step explanation:

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:

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]

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:

10 pounds of potatoes would cost $15.

Step-by-step explanation:

Set up proportion.

4/6=10/x

simplify 4/6 into 2/3,

2/3=10/x

cross product,

2*x=3*10

2x=30

x=30/2

x=15

lemme just add some to the great reply above,

[tex]\begin{array}{ccll} lbs&\$\\ \cline{1-2} 4&6\\ 10&x \end{array}\implies \cfrac{4}{10}=\cfrac{6}{x}\implies 4x = 60\implies x = \cfrac{60}{4}\implies x = 15[/tex]

her profit is 204 dollars and 68 cents= 204.68

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

Step-by-step explanation:

this is the correct answer for the question

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%.

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

Answer:

130 km/hr

Step-by-step explanation:

Average Speed = Total Distance / Total Time.

Let us just make up 2 distances from a time

A: 55 km/hour for 2 hours = 110 km

B: 75 km/hour for 2 hours = 150 km

Total Distance = 260 km

Total Time = 2 hours.

Average Speed = 260 / 2 = 130 km/hr

Now let's try it again. If we get a different answer, then the problem is unanswerable.

A: 55 km/hr  for 3 hours = 165 km

B: 75 km/hr for 3 hours = 225 km

Total distance  = 390 km

Total time = 3 hours

Average speed = 390 / 3 = 130 which is the same answer we got before.

(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:

[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:

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!