The voltage V required for a circuit is given by \(v=\sqrt{PR}\) , where P is the power in watts and R is the resistance in ohms.

How many more volts are needed to light a 100-watt light bulb than a 75-watt light bulb if the resistance of both is 110 ohms? Round answer to the nearest tenth.

Answer:

A≈1633.63

Step-by-step explanation:

d Diameter 26

h heigth 7

A=2πd

2h+2π(d

2)2=2·π·26

2·7+2·π·(26

2)2≈1633.62818

To find r when q=1.2, given that q is inversely proportional to r squared and q=30 when r=3:

Calculate the value of k, the constant of proportionality, using the initial values of q and r.

Use the value of k to solve for r when q=1.2.

In an inverse proportion, as one variable increases, the other variable decreases in such a way that their product remains constant. To solve for r when q=1.2, we can follow these steps:

First, establish the relationship between q and r. The given information states that q is inversely proportional to r squared. Mathematically, this can be expressed as q = k/r², where k is the constant of proportionality.

Use the initial values to determine the constant of proportionality, k. Given that q=30 when r=3, substitute these values into the equation q = k/r². Solving for k gives us k = qr² = 30(3²) = 270.

With the value of k, we can solve for r when q=1.2. Substituting q=1.2 and k=270 into the equation q = k/r^2, we have 1.2 = 270/r². Rearranging the equation and solving for r gives us r²= 270/1.2 = 225, and thus r = √225 = 15.

Therefore, when q=1.2 in the inverse proportion q = k/r², the corresponding value of r is 15.

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Answer:

10cm²

Step-by-step explanation:

Area = Length × Width

Area = 4cm × 2.5cm

Area = 10cm²

Answer:

A=10cm²

Step-by-step explanation:

Area of a rectangle=Length × width

A=L×W

where L=4 and W=5/2

we will just substitute for the values

A= 4× 5/2

A=5×2

A=10cm²

To find the 75th percentile of X is A. 1.00, we need to find the value of x such that the probability of X being less than or equal to x is 0.75.

Let f(x) be the density function of X. We know that f(x) is proportional to (1+x^2)^(-1), which means we can write:
f(x) = k(1+x^2)^(-1)
where k is a constant of proportionality. To find k, we use the fact that the total area under the density function is 1:
∫f(x)dx = 1
Integrating both sides, we get:
k∫(1+x^2)^(-1)dx = 1
The integral on the left-hand side can be evaluated using a substitution u = x^2 + 1:
k∫(1+x^2)^(-1)dx = k∫u^(-1/2)du = 2k√(u)
Substituting back for u and setting the integral equal to 1, we get:
2k∫(1+x^2)^(-1/2)dx = 1
Using a trigonometric substitution x = tan(t), we can evaluate the integral on the left-hand side:
2k∫(1+x^2)^(-1/2)dx = 2k∫sec(t)dt = 2kln|sec(t) + tan(t)|
Substituting back for x and simplifying, we get:
2kln|1 + x^2|^(-1/2) = 1
Solving for k, we get:
k = √(2/π)
Now we can write the density function of X as:
f(x) = (√(2/π))(1+x^2)^(-1)
To find the 75th percentile of X, we need to solve the equation:
∫(-∞, x) (√(2/π))(1+t^2)^(-1) dt = 0.75
This integral does not have a closed-form solution, so we need to use numerical methods to approximate the value of x. One way to do this is to use a computer program or a graphing calculator that has a built-in function for finding percentiles of a distribution. Using a graphing calculator, we can enter the function y = (√(2/π))(1+x^2)^(-1) and use the "invNorm" function to find the x-value corresponding to the 75th percentile (which is the same as the z-score for a standard normal distribution).
Doing this, we get:
invNorm(0.75) ≈ 0.6745
Therefore, the 75th percentile of X is approximately:
x = tan(0.6745) ≈ 0.835

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Answer:

B

Step-by-step explanation:

I hope this helps you :)

Answer:

thx for the points

Step-by-step explanation:

Answer:

-9 + ( -16) = -25

( -4) – (-6) + (-5) – (8) =-11

( 3) – (-9) + (-3) – (-2) =11

- ( 5) – (-7) + (-8) =-6

- ( -8) – (-9) + (-4) – (-1) =14

( -5) – (-11) + (10) – (8) =8

( -1) – (-2) + (-3) – (-6) =4

( -4) – ( 4) – (6) + (-5) – (-8) =-11

(-6) + (-5) – (-8) =-3

Step-by-step explanation:

Hope it helps

Answers:

-9 + ( -16) = -25

( -4) – (-6) + (-5) – (8) = -11

( 3) – (-9) + (-3) – (-2) = 11

- ( 5) – (-7) + (-8) = -6

- ( -8) – (-9) + (-4) – (-1) = 14

( -5) – (-11) + (10) – (8) = 8

( -1) – (-2) + (-3) – (-6) = 4

( -4) – ( 4) – (6) + (-5) – (-8) = -11

(-6) + (-5) – (-8) = -3

The number of papers xavier have remaining after working for 12 hours is 18

Xavier can grade 6 papers per hour, so in 12 hours he can grade:

6 papers/hour x 12 hours = 72 papers

Therefore, after working for 12 hours, Xavier would have

90 - 72 = 18 papers remaining to grade.

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Answer:

Step-by-step explanation:

a. To sketch the plane curve defined by the vector function r(t) = ti + ln(t)j when t = 2, we can first evaluate the position vector at t = 2:

r(2) = 2i + ln(2)j

Next, we plot the point (2, ln(2)) on the x-y plane. Since the curve is two-dimensional, we don't need to determine the vectors T(t) and N(t) for this sketch.

b. To find the scalar tangential component of acceleration aT for the particle with the position vector r(t) = e^t⟨cos(4t), sin(4t), 0⟩, we need to differentiate the position vector twice with respect to time to obtain the acceleration vector a(t).

First, let's differentiate r(t) with respect to t:

r'(t) = (d/dt) (e^t⟨cos(4t), sin(4t), 0⟩)

      = e^t⟨-sin(4t), cos(4t), 0⟩ + e^t⟨-4cos(4t), 4sin(4t), 0⟩

      = e^t⟨-sin(4t) - 4cos(4t), cos(4t) + 4sin(4t), 0⟩

Now, we differentiate r'(t) with respect to t to find the acceleration vector:

a(t) = (d/dt) (e^t⟨-sin(4t) - 4cos(4t), cos(4t) + 4sin(4t), 0⟩)

     = e^t⟨-4cos(4t) + 16sin(4t), -4sin(4t) - 16cos(4t), 0⟩

To find the scalar tangential component of acceleration aT, we project the acceleration vector a(t) onto the tangent vector T(t). Since the tangent vector is the derivative of the position vector with respect to t, we have T(t) = r'(t):

T(t) = e^t⟨-sin(4t) - 4cos(4t), cos(4t) + 4sin(4t), 0⟩

Finally, we can find the scalar tangential component of acceleration aT by taking the dot product of a(t) and T(t):

aT = a(t) · T(t)

   = e^t⟨-4cos(4t) + 16sin(4t), -4sin(4t) - 16cos(4t), 0⟩ · e^t⟨-sin(4t) - 4cos(4t), cos(4t) + 4sin(4t), 0⟩

   = (-4cos(4t) + 16sin(4t))(-sin(4t) - 4cos(4t)) + (-4sin(4t) - 16cos(4t))(cos(4t) + 4sin(4t))

The resulting expression for aT is quite complex and difficult to simplify without a specific value of t.

The new coordinates of point qqq after a 180-degree rotation about the origin are (-x, -y).

The point qqq was rotated about the origin (0,0) by 180 degrees.
To rotate a point about the origin by 180 degrees, we can use the following steps:

1. Identify the coordinates of the point qqq. Let's say the coordinates are (x, y).

2. Apply the rotation formula to find the new coordinates. The formula for a 180-degree rotation about the origin is: (x', y') = (-x, -y).

3. Substitute the values of x and y into the formula. In this case, the new coordinates will be: (x', y') = (-x, -y).

So, the new coordinates of point qqq after a 180-degree rotation about the origin are (-x, -y).

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Answer:

2 : 3

Step-by-step explanation:

You need to show the ratio using the smallest whole numbers possible.

Multiply both numbers by 3.

2/3 : 1

2/3 × 3 : 1 × 3

2 : 3

Answer:

(x+12, y-24)

Step-by-step explanation:

The rule is (x, y) -> (x+12, y-24)

Answer:

y=-3x+5

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-4-(-1))/(3-2)

m=(-4+1)/1

m=-3/1

m=-3

y-y1=m(x-x1)

y-(-1)=-3(x-2)

y+1=-3x+6

y=-3x+6-1

y=-3x+5

In order to divide these polynomials, let's use the following property:

So we have:

As per the given standard deviation, the probability that the first-year student shows a higher emotional intelligence test score in a randomly selected senior/first-year pair is 0.72

Here we have given that the score, S, of a randomly selected senior student at a large university has a mean of 123 and a standard deviation of 29.

And the score, F, of a randomly selected first-year student at a large university has a mean of 104 and a standard deviation of 34.

Then the probability is calculated as,

=> 123/104

=> 1.182

Where as the based on the standard deviation, the value is calculated as

=> 29/34

=> 0.852

Then the total probability of the given situation is calculated as,

=> 0.852/1.182

=> 0.72

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If a function f is differentiable at a point a and f(a) is not equal to zero, then the absolute value function |f| is also differentiable at that point.

The proof involves considering two cases based on the sign of f(a) and showing that the limit of the difference quotient exists for |f| at point a in both cases. However, it is important to note that |f| is not differentiable at the point where f(a) equals zero.

To prove that if f is differentiable at a, then |f| is also differentiable at a, provided that f(a) ≠ 0, we need to show that the limit of the difference quotient exists for |f| at point a.

Let's consider the function g(x) = |x|. The absolute value function is defined as follows:

g(x) = {

x if x ≥ 0,

-x if x < 0.

Since f(a) ≠ 0, we can conclude that f(a) is either positive or negative. Let's consider two cases:

Case 1: f(a) > 0

In this case, we have g(f(a)) = f(a). Since f is differentiable at a, the limit of the difference quotient exists for f at point a:

lim (x→a) [(f(x) - f(a)) / (x - a)] = f'(a).

Taking the absolute value of both sides, we have:

lim (x→a) |(f(x) - f(a)) / (x - a)| = |f'(a)|.

Since |g(f(x)) - g(f(a))| / |x - a| = |(f(x) - f(a)) / (x - a)| for f(a) > 0, the limit on the left-hand side is equal to the limit on the right-hand side, which means |f| is differentiable at a when f(a) > 0.

Case 2: f(a) < 0

In this case, we have g(f(a)) = -f(a). Similarly, we can use the same reasoning as in Case 1 and conclude that |f| is differentiable at a when f(a) < 0.

Since we have covered both cases, we can conclude that if f is differentiable at a and f(a) ≠ 0, then |f| is also differentiable at a.

Note: It's worth mentioning that at the point where f(a) = 0, |f| is not differentiable. The proof above is valid when f(a) ≠ 0.

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Step 1

Given;

Step 2

A) Find the value of b²-4ac

Step 3

B) Determine how many real solutions quadratic has.

Answer:

In order to describe the pattern of the number of dots as a function of the picture number you have to take the average values of each number and find the line of best fit between each dot. This gives you a function, such as, f(x) = mx+b.

Answer:

First find x and y at given parameters:

Maximum height is at vertex. Time is:

Maximum height is:

Horizontal distance:

Time to hit the ground:

Answer:

a

Step-by-step explanation:

Answer:the anS is(A):The mode for the students is{{ 2 }and the mode for the teachers is {4}

Step-by-step explanation:HOPE MY ANS HELPS CUZ I ATTEMPTED THIS QUESTION IN A

QUIZ BEFORE

The correct answer is (c) approximately normal for large samples.

The central limit theorem states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution. In the case of the difference of two sample means (x1 - x2), the central limit theorem still applies, and the distribution becomes approximately normal as the sample size (n) increases. Therefore, for large sample sizes, the sampling distribution of (x1 - x2) can be approximated by a normal distribution, and the properties of the normal distribution can be used to make statistical inferences about the population mean difference.

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Answer:

5.1

Step-by-step explanation:

Answer:

x=2,y=5

Step-by-step explanation:

y=8x-11 ----(1)

y=-7x+19----(2)

Replace (1) in (2)

8x-11 =-7x+19

8x+7x=19+11

15x =30

x =30÷2

x =2

If x=2,then replace x=2 in (1)

y=8(2)-11

y=16-11

y=5

The rate at which the length of the cylinder is changing can be determined using the formula for the volume of a cylinder and applying the chain rule of differentiation. The rate of change of the length h is found to be -0.8192 in/sec.

The volume V of a cylinder is given by the formula V = πr²h, where r is the radius and h is the length of the cylinder. We are given that V = 128π cubic inches.

Differentiating both sides of the equation with respect to time, we get dV/dt = d(πr²h)/dt. Using the chain rule, this becomes dV/dt = π(2r)(dr/dt)h + πr²(dh/dt).

Since the radius r is decreasing at a constant rate of 0.05 inches per second (dr/dt = -0.05), and the volume V is constant (dV/dt = 0), we can substitute these values into the equation. Additionally, we know that r = 2.5 inches.

0 = π(2(2.5)(-0.05))h + π(2.5)²(dh/dt).

Simplifying the equation, we have -0.25πh + 6.25π(dh/dt) = 0.

Solving for dh/dt, we find that dh/dt = -0.25h/6.25 = -0.04h.

Substituting h = 8 (since V = πr²h = 128π, and r = 2.5), we get dh/dt = -0.04(8) = -0.32 in/sec.

Therefore, the rate at which the length h is changing when the radius r is 2.5 inches is -0.32 in/sec, which is equivalent to -0.8192 in/sec (rounded to four decimal places). The correct answer is (b) -0.8192 in/sec.

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Answer:

588 women 252 men

Step-by-step explanation:

70% of 840 is 588 so to get the men you subtract 840 - 588 + 252

Hope this helps!

Answer: No you can not make any triangles with 2 inches, 2 inches and 4 inches. hoped this helped!

Step-by-step explanation:

The ratiο οf the time Sam spends wοrking fοr his mοther tο the time he spends wοrking fοr his father per year is 1/6 οr 1:6. Thus, οptiοn D is cοrrect.

The relative size οf twο quantities expressed as the quοtient οf οne divided by the οther; the ratiο οf a tο b is written as a:b οr a/b.

If Sam wοrks fοr his father fοr 2/5 οf the year, then he wοrks fοr his mοther fοr the remaining time, which is 1 - 2/5 = 3/5 οf the year.

Tο find the ratiο οf the time Sam spends wοrking fοr his mοther tο the time he spends wοrking fοr his father per year, we can write:

Sam wοrked fοr his father = 2/3 οf year

Remaining time = 1/3

Sam wοrked fοr his mοther = 1/3(remainder year)=(1/3)*(1/3)

     =1/9

Ratiο οf time spent fοr mοther and father is

= (1/9):(2/3)

=(1/3):2

= 1:6

Sο, οptiοn d) 1/6 is cοrrect answer

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Complete question:

Sam works for his father for 2/3; of the year and works for his mother = of the remainder year. What is the

ratio of the time Sam spends working for his mother to the time he spends working for his father per year?

A function of the type A cos(Lx) is not an appropriate solution for the particle in a one-dimensional box since it does not satisfy the required boundary conditions for the system.

In one-dimensional box case, the particle is confined within a definite region, represented by the interval [0, L]. This boundary conditions requires that the wave function of a particle must be zero, i.e., ψ(0) = ψ(L) = 0. This allows that the probability density of finding the particle outside the box is also zero.

The function of the type A cos (Lx) does not satisfy these boundary conditions. If x is 0, the function evaluates to A cos(0) = A, which is not zero. Similarly, when x is L, the function evaluates to A cos(L), which is usually not zero unless L is an integer multiple of π.

It has physical interpretations related to the quantization of energy levels in the finite system, which is a fundamental aspect of quantum mechanics.

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Answer:

\(thank \: you\)