What is the volume of a sphere with a radius of 6.3 cm, rounded to the nearest tenth of a cubic centimeter?

Answer:

30 x 4 + 4 x 4 and 4 x 4 + 4 x 30

Step-by-step explanation:

Answer:

35.6 feets

Step-by-step explanation:

To obtain tree height :

(Height of stick / shadow of stick = height of tree / shadow of tree)

Height of stick = 2 feets

Shadow of stick = 0.8 feets

Shadow of tree = 14.24 feets

Height of tree = h

(2 / 0.8 = h /14.24)

Cross multiply

0.8h = 14.24 * 2

0.8h = 28.48

h = 28.48 / 0.8

h = 35.6 feets

Answer:

B

Step-by-step explanation:

D) - y = (x + 5)(x - 3)(x + 1).

Table in this case would look like this:

Write coefficients of x³, x² ,x and the constant in a row and divisor would be the x value obtained by equation x + 5 = 0.

The sequence of multiplications would be as shown in picture.

x² - 2x - 3

x² - 3x + x - 3

x(x - 3) + 1(x - 3)

(x + 1)(x - 3)(x + 5).

Answer:

Is that your answer

Answer:

D. 42

2X3X7=42

Step-by-step explanation:

It's just volume baseXwidthXhight. Hope this helps.

Answer:

1/6 or 16.67% chance

Step-by-step explanation:

Assuming its a normal six sided dice. You would have a 16.67% chance of getting a 3.

The equation of the median PM in triangle PQR with vertices P(2, 3), Q(-3, 7), and R(-1, -3) is y = (1/3)x + 7/3.

To find the midpoint of QR, we calculate the average of the x-coordinates and the average of the y-coordinates. The x-coordinate of point M is (-3 + (-1))/2 = -2/2 = -1, and the y-coordinate of point M is (7 + (-3))/2 = 4/2 = 2.

Therefore, the coordinates of point M are (-1, 2). Now, we have two points, P (2, 3) and M (-1, 2), and we can find the equation of the line passing through these points using the point-slope form.

The slope of the line passing through P and M is (2 - 3)/(-1 - 2) = -1/-3 = 1/3. Using the point-slope form, we have:

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

Expanding and rearranging the equation, we get:

y = (1/3)x + 7/3

Therefore, the equation of the median PM in triangle PQR is y = (1/3)x + 7/3.

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The proof shows that if we assume the premises are true and the conclusion is false, it leads to a contradiction. Therefore, the argument is valid. The modus ponens and conjunction are used.

To construct a proof for the given argument, we'll use a proof by contradiction. We'll assume the premises are true and the conclusion is false, then we'll derive a contradiction. If a contradiction is reached, it means the original assumption was false, and thus the argument is valid.

Argument:

H ⊃ I

J ⊃ K

~K

H ∨ J

Conclusion: I

Proof by contradiction:

H ⊃ I (Premise)

J ⊃ K (Premise)

~K (Premise)

H ∨ J (Premise)

~I (Assumption for proof by contradiction)

H (Disjunction elimination from 4)

I (Modus ponens using 1 and 6)

~J (Assumption for proof by contradiction)

K (Modus ponens using 2 and 8)

~K ∧ K (Conjunction introduction of 3 and 9)

Contradiction: ~I ∧ I (Conjunction introduction of 5 and 7)

Conclusion: I (Proof by contradiction)

The proof shows that if we assume the premises are true and the conclusion is false, it leads to a contradiction. Therefore, the argument is valid.

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Answer:

0

Step-by-step explanation:

-9 x (0/-3)

-9 x 0

0

Answer:

1 1/4 cups

Step-by-step explanation:

3 3/4 cups = 3.75

1/3 = .33333

3.75 x .33333 = 1.25

1.25 = 1 1/4 cups

The MGF of Y using the fact that Y = μ + oZ where Z ~ N(0, 1) is e^(tμ + t²/2).

The MGF of Y is given by,

E[exp(tY)] = E[exp(t(μ+Z))]

We know that if X is a normal random variable, X~N(μ, σ²) with μ as the mean and σ² as the variance.

The MGF of X is given by,

MGF_X(t) = E[e^(tx)] = e^(μt + (σ²t²)/2)

Here, Y ~ N(μ, 2) we have Y = μ + oZ where Z ~ N(0, 1)

MGF_Y(t) = E[exp(tY)] = E[exp(t(μ+Z))]MGF_

Y(t) = E[e^(tμ+tZ)]MGF_

Y(t) = e^(tμ) E[e^(tZ)]

We know that the MGF of Z is already derived in the lecture 1,

It is MGF_Z(t) = e^(t²/2)MGF_

Y(t) = e^(tμ) e^(t²/2)MGF_

Y(t) = e^(tμ + t²/2)

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Given information is that Y~ N(μ, 2), let's find the MGF of Y using the fact that Y = μ + oZ where Z~ N(0, 1).

The MGF of Y becomes:

MGF of [tex]Y = e^{t} \mu+ MGF\ of\ o \times e^{((t^2)/2)}[/tex]

Hence, the MGF of Y is [tex]e^{t}\mu + MGF\ of\ o \times e^{((t^2)/2)}[/tex].

The MGF of Y is as follows:

MGF of Y = MGF of μ + MGF of oZ

The MGF of Y = MGF of μ + MGF of oMGF of Z

Since the mean of Y is μ, we can substitute the above equation with the following:

[tex]MGF\ of\ Y = e^{t}\mu + MGF\ of\ oMGF\ of\ Z[/tex]

Now let's find the MGF of Z: We know that the MGF of Z is given by;

MGF of [tex]Z = e^{((t^2)/2)}[/tex]

Therefore, the MGF of Y becomes: MGF of [tex]Y = e^{t}\mu + MGF\ of\ o \times e^{((t^2)/2)}[/tex]

Hence, the MGF of Y is [tex]e^{t}\mu + MGF\ of\ o \times e^{((t^2)/2)}[/tex].

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Answer:

110.45 inches cubed

Step-by-step explanation:

5in x 4.7in x 4.7in

Answer:

7/15

Step-by-step explanation:

7+8=15

there's 7 girls so you have 7/15 chance of getting a girl

Answer:

8/15

Step-by-step explanation:

possibilities/sample size= 8 girls/ 15 students

The first three nonzero terms of two linearly independent solutions about x = 0 can be obtained by Taylor expanding the solutions in terms of the exponent r and truncating the series to the desired order.

To determine if x = 0 is a regular singular point of the differential equation xy'' + y = 0, we substitute y = x^r into the equation and solve for the exponent r. Differentiating y twice with respect to x, we have y'' = r(r - 1)x^(r - 2). Substituting these expressions into the differential equation, we get [tex]x(x^r)(r(r - 1)x^(r - 2)) + x^r = 0[/tex]. Simplifying, we obtain r(r - 1) + 1 = 0, which yields r^2 - r + 1 = 0. Solving this quadratic equation, we find that the exponents at the singular point x = 0 are complex and given by r = (1 ± i√3)/2.

To find the first three nonzero terms of two linearly independent solutions about x = 0, we can use the Taylor series expansion. Let's consider the solution y1(x) corresponding to the exponent r = (1 + i√3)/2. Expanding y1(x) as a series around x = 0, we have y1(x) =[tex]x^r = x^((1 +[/tex]i√3)/2) = x^(1/2) *[tex]x^(i√3/2[/tex]). Using the binomial series expansion and Euler's formula, we can write [tex]x^(1/2) and x^(i√3/2)[/tex] as infinite series.

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The estimate of the number of gallons of oil in the tank at t = 14 hours is 100.8 gallons. The correct answer is option C.

To estimate the number of gallons of oil in the tank at t = 14 hours using a trapezoidal approximation,

we need to calculate the total change in oil volume over the given time period.

The trapezoidal approximation involves dividing the time interval into subintervals and approximating the change in volume as the sum of trapezoidal areas.

Let's calculate the approximate volume of oil at t = 14 hours using the given data and the trapezoidal approximation: Interval 1 (2 to 5 hours):

Average rate = (R(2) + R(5)) / 2 = (8.2 + 7.8) / 2 = 16 / 2 = 8 gallons per hour.

Volume change =

[tex]Average rate \times time = 8 \times (5 - 2)[/tex]

= 24 gallons.

Interval 2 (5 to 10 hours):

Average rate = (R(5) + R(10)) / 2 = (7.8 + 8.6) / 2 = 16.4 / 2 = 8.2 gallons per hour

Volume change =

[tex]Average rate \times time = 8.2 \times (10 - 5) [/tex]

= 41 gallons

Interval 3 (10 to 14 hours):

Average rate = (R(10) + R(14)) / 2 = (8.6 + 9.3) / 2 = 17.9 / 2 = 8.95 gallons per hour

Volume change =

[tex]Average rate \times time = 8.95 \times (14 - 10)[/tex]

= 35.8 gallons.

Total volume change = Interval 1 + Interval 2 + Interval 3 = 24 + 41 + 35.8 = 100.8 gallons.

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If A is equal to [0, 4] and I is equal to [1, 0], [0, 1], then [0 -  4][1 0] equals 0 and [0 -  4] equals 0 and [2 - 4] equals 0. Accordingly, the eigenvalues of the matrix

[dt/dy] + [0, 4] [x] = [0] can be written as the differential equation above in a matrix. Here, [0, 4] is the coefficient network and [x] is the variable grid. Given, arrangement of differential conditions, dt/dy + 4x = 0. Let [0, 4] be the framework's eigenvalue, and then [0, 4] [x] = [x] => (A-I) [x] = 0, where An represents the coefficient grid, I represents the character lattice, and x represents the variable network.

The determinant of [A-I] is 0 if for a non-trivial solution, [A-I] [x] = 0. On the off chance that An is equivalent to [0, 4] and I is equivalent to [1, 0], [0, 1], then [0 - 4][1 0] equivalents 0 and [0 - 4] equivalents 0 and [2 - 4] equivalents 0. As a result, the matrix's eigenvalues

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Answer:

y=0.12/1(x-5)^2 -3

y=1/10(x-10)^2 -4

Step-by-step explanation:

Given the directrix and focus of the parabolas, the equation of the parabolas are [tex]y=\frac{1}{6}(x^{2} +10x - 2)[/tex] and [tex]y=\frac{1}{20}(-x^{2} +20x - 80)[/tex].

Equation of a parabola is given by-

Distance of a point (x, y) on parabola from directrix =  Distance of a point (x, y) on parabola from focus

focus = (-5, -3)

directrix = y = -6

[tex]\sqrt{(x+5)^{2}+(y+3)^{2} } = (y+6)\\\\ (x+5)^{2}+(y+3)^{2} = (y+6)^{2}\\\\x^{2} +25+5x = 6y+27\\\\y=\frac{1}{6}(x^{2} +10x - 2)[/tex]

focus = (10,-4)

directrix = y = 6

[tex]\sqrt{(x-10)^{2}+(y+4)^{2} } = (y-6)\\\\ (x-10)^{2}+(y+4)^{2} = (y-6)^{2}\\\\x^{2} +100-20x = -20y+20\\\\y=\frac{1}{20}(-x^{2} +20x - 80)[/tex]

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Answer:

D.1767.1 in³

Step-by-step explanation:

volume of a sphere=4/3πr³

where r=radius of the sphere

Given diemeter= 15in

so radius=15/2=7.5in

volume=4/3×π×(7.5)³

=4/3×22/7×(7.5)³

=1767.1in³

hope it helps...

have a great day!!

The solution to the triple integral in spherical coordinates is 22000π. This can be obtained by evaluating the integral in three steps: integrating with respect to r, then with respect to θ, and finally with respect to φ.

To change the triple integral to spherical coordinates, we need to express the integrand and the limits of integration in terms of spherical coordinates.

The given integrand is f(x, y, z) = 6x² + y² + z².

In spherical coordinates, the integrand becomes f(r, θ, φ) = 6(rsinθcosφ)² + (rsinθsinφ)² + (rcosθ)².

The limits of integration are as follows:

- The bounds for r are from 0 to 10, as the region Q is bounded by the upper hemisphere x² + y² + z² = 100.

- The bounds for θ are from 0 to π/2, as we are considering the upper hemisphere.

- The bounds for φ are from 0 to 2π, as φ covers a complete revolution around the z-axis.

The triple integral in spherical coordinates is then given by:

∭Q f(r, θ, φ) r² sinθ dr dθ dφ,

which becomes:

∫(φ=0 to 2π) ∫(θ=0 to π/2) ∫(r=0 to 10) [6(rsinθcosφ)² + (rsinθsinφ)² + (rcosθ)²] r sinθ dr dθ dφ.

To solve the given triple integral, we'll start by evaluating the innermost integral with respect to r, then the middle integral with respect to θ, and finally the outer integral with respect to φ.

The integrand is:

[6(rsinθcosφ)² + (rsinθsinφ)² + (rcosθ)²] r² sinθ

First, let's evaluate the innermost integral with respect to r, while treating θ and φ as constants:

∫(r=0 to 10) [6(rsinθcosφ)² + (rsinθsinφ)² + (rcosθ)²] r² sinθ dr

= ∫(r=0 to 10) [6(sin²θcos²φ)r⁴ + (sin²θsinφ)r⁴ + (cos²θ)r⁴] sinθ dr

= ∫(r=0 to 10) [(6sin²θcos²φ + sin²θsin²φ + cos²θ) r⁴] sinθ dr

= [(6sin²θcos²φ + sin²θsinφ + cos²θ) ∫(r=0 to 10) r⁴] sinθ dr

= [(6sin²θcos²φ + sin²θsin²φ + cos²θ) * (10^5/5)] sinθ

= [(6sin²θcos²φ + sin²θsin²φ + cosθ) * 2 × 10⁵] sinθ

Next, let's evaluate the middle integral with respect to θ, while treating φ as a constant:

∫(θ=0 to π/2) [(6sin²θcos²φ + sin²θsin²φ + cos²θ) * 2 × 10⁵] sinθ dθ

= 2 × 10⁵ ∫(θ=0 to π/2) [6sin²θcos²φ + sin²θsin²φ + cos²θ] sinθ dθ

= 2 × 10⁵ [2/3cos²φ + 1/4 + 1/3]

= 2 × 10⁵ [2/3cos²φ + 7/12]

Finally, let's evaluate the outer integral with respect to φ:

[tex][\int_{0}^{2\pi} 2\times10^5 \left( \frac{2}{3}\cos^2\phi + \frac{7}{12} \right) d\phi \\\\= 2\times10^5 \left( \frac{2}{3}\pi + \frac{7}{12}(2\pi) \right)][/tex]

= 22π × 10000

= 22000π

Therefore, the solution to the given triple integral is 22000π.

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Complete question :

Change the triple integral to spherical coordinates: SIS 6x2 + y2 + z2) av (༴ AV Where Q is bounded by the upper hemisphere: x2 + y2 +22=100 : 21 10 ("S", p's p. sino dpdooo 2T pº sino dododo 21 10 2 3 sino doopde 0 0 0 10 ["S" p2 sino apdoce

The value of [v]B1 is [[1][0]][[0][0]]

Suppose [v]B2 is as follows:

[v]B2 = [[11][14]]

[13][14]]

[7][6]]

[10]]

If the ordered bases are B1 = {a, b} and B2 = {c, d}, we want to find [v]B1.

To find [v]B1, we need to express the columns of [v]B2 in terms of the basis vectors of B1.

The first column of [v]B2 is [11, 13, 7, 10]. We want to express this column in terms of the basis vectors of B1: [a, b].

To do this, we set up the following equation:

[11][13][7][10] = [a][b]

Solving this equation, we find that:

11a + 13b = 11

13a + 14b = 13

7a + 6b = 7

10a = 10

From the last equation, we can see that a = 1.

Substituting this value of a into the first three equations, we can solve for b:

11 + 13b = 11

13 + 14b = 13

7 + 6b = 7

Simplifying these equations, we find that b = 0.

Therefore, [v]B1 is as follows:

[v]B1 = [[1][0]]

[0][0]]

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Answer:

Fraction = 120/100 | Decimal = 1.2

Step-by-step explanation:

Fraction:

100% is 100/100 but 120% is 20% over so the fraction is now 120/100 (20% = 20/100)

Decimal:

100% is 1 but as I said earlier, 120% is 20% over so the decimal is now 1.2 (20% = 0.2)

Answer:

[tex]y=2x-3[/tex]

Step-by-step explanation:

This is just algebraic manipulation. In order to solve for y, you need to isolate it. Start this by moving the 2x from the left side of the equation. You can do this by subtracting 2x from both sides and you should end up with:

[tex]-y=-2x+3[/tex]

After this, you still have a negative y, which means you just need to divide both sides of the equation by -1 to get rid of the negative. That should reverse the signs of all the variables in the equation, making it look like:

[tex]y=2x-3[/tex]

The complex number Z = (20 + 120i) can be expressed in polar form as Z = 2√370(cos(1.405) + isin(1.405)).

To express the complex number Z = (20 + 120i) in polar form, we need to find its magnitude (r) and argument (θ).

The magnitude of a complex number Z = a + bi is given by the formula:

|r| = √(a^2 + b^2)

In this case, a = 20 and b = 120.

Therefore, the magnitude of Z is:

|r| = √(20^2 + 120^2) = √(400 + 14400) = √14800 = 2√370.

The argument (θ) of a complex number Z = a + bi is given by the formula:

θ = arctan(b/a)

In this case, a = 20 and b = 120. Therefore, the argument of Z is:

θ = arctan(120/20) = arctan(6) ≈ 1.405 radians.

Now we can express Z in polar form as Z = r(cosθ + isinθ), where r is the magnitude and θ is the argument:

Z = 2√370(cos(1.405) + isin(1.405)).

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Answer:

[tex]23\frac{1}{16}[/tex] pulgada

Step-by-step explanation:

[tex]\frac{16}{9} =\frac{41}{y}[/tex]

16 × y = 9 × 41

16y = 369

16y ÷ 16 = 369 ÷ 16

[tex]y=23\frac{1}{16}[/tex]

By reading off the graph as shown in the question, we can see that the time that is required is 20 minutes.

The half life is the time that it taken for only half or 50% of the isotopes that were originally present in the sample to remain. We know that the half life does differ by the kind of sample that is used.

In this case, we want to determine  how long does it take a sample of francium-223 to decay by 50%. This could easily be done from the graph of the decay as shown in the question.

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To determine whether the function f(z) = 1/(z^2 + 1) has an antiderivative on a given domain, we need to check if the function is analytic on that domain.

(a) For the domain G = C\{i, -i}, the function f(z) = 1/(z^2 + 1) is analytic on G. This is because it is a rational function and does not have any singularities (poles) within the domain. Hence, it has an antiderivative on G.

(b) For the domain G = {z Re(z) > 0}, the function f(z) = 1/(z^2 + 1) does not have an antiderivative on G. This is because the function has singularities at z = i and z = -i, which lie on the imaginary axis. Since the domain excludes these points, f(z) is not analytic on G and does not have an antiderivative on G.In summary, the function f(z) = 1/(z^2 + 1) has an antiderivative on the domain G = C\{i, -i} but does not have an antiderivative on the domain G = {z Re(z) > 0}.

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Answer:

The desired product is (x + 2)(x + 4).

Step-by-step explanation:

If x is an even integer, x + 2 is the next consecutive even integer and x + 4 the next.  

The desired product is (x + 2)(x + 4).

Answer:

Marcy paid $4.15

Step-by-step explanation:

Given

Represent hotdogs with x and chips with y.

So, we have:

Jamal

[tex]3x + 2y = 4.85[/tex]

Kenny

[tex]5x + 6y = 9.55[/tex]

See attachment for complete question

Required

Determine the amount for 2x and 3y

From Jamal's and Kenny's orders we have:

[tex]3x + 2y = 4.85[/tex] --- (1)

[tex]5x + 6y = 9.55[/tex] --- (2)

Multiply (1) by 3

[tex]3 * [3x + 2y = 4.85][/tex]

[tex]9x + 6y = 14.55[/tex] --- (3)

Subtract (2) and (3)

[tex]9x - 5x + 6y - 6y = 14.55 - 9.55[/tex]

[tex]9x - 5x = 5[/tex]

[tex]4x = 5[/tex]

Solve for x

[tex]x = \frac{5}{4}[/tex]

[tex]x = 1.25[/tex]

Substitute [tex]x = 1.25[/tex] in [tex]3x + 2y = 4.85[/tex]

[tex]3 * 1.25 + 2y = 4.85[/tex]

[tex]3.75 + 2y = 4.85[/tex]

Solve for y

[tex]y = \frac{4.85 - 3.75}{2}[/tex]

[tex]y = \frac{1.10}{2}[/tex]

[tex]y = 0.55[/tex]

So, the cost of 2x and 3y is:

[tex]Cost = 2x + 3y[/tex]

[tex]Cost = 2*1.25 + 3*0.55[/tex]

[tex]Cost = \$4.15[/tex]

1. This is a valid approach to prove the argument.

2. This is the first step of the contrapositive proof.

3. This statement is true since if one of them is rational, the other one could also be rational or irrational.

4. This statement is true because rational numbers are those numbers that can be expressed as a ratio of two integers.

5. This is true because any rational number can be expressed as a fraction of two integers.

6. This is true since it's the sum of two fractions.

7. This is also true since the sum and product of two integers are always integers.

8. This is true since the product of any non-zero number with another non-zero number is also non-zero.

9. This statement is true since x+y is the quotient of two integers, and since both integers are non-zero, then the quotient is also non-zero and therefore rational.

Therefore, the given argument is correct.

Each step in the argument is logically valid, and the argument follows a correct proof by contrapositive to show that if x is rational and y is rational, then x + y is rational.

The given argument is correct. Let us evaluate each line of the proof and make sure that it is accurate and logical.

Proposition: For every pair of real numbers x and y, if x + y is irrational, then x is irrational or y is irrational

1. We proceed by contrapositive proof.

This is a valid approach to prove the argument.

2. We assume for real numbers x and y that it is not true that x is irrational or y is irrational and we prove that x + y is rational.

This is the first step of the contrapositive proof.

3. If it is not true that x is irrational or y is irrational then neither x nor y is irrational.

This statement is true since if one of them is rational, the other one could also be rational or irrational.

4. Any real number that is not irrational must be rational. Since x and y are both real numbers then x and y are both rational.

This statement is true because rational numbers are those numbers that can be expressed as a ratio of two integers.

5. We can therefore express x as a/b and y as c/d as a, where a, b, c, and d are integers and b and d are both not equal to 0.

This is true because any rational number can be expressed as a fraction of two integers.

6. The sum of x and y is: x + y = a/b + c/d = (ad+bc)/bd

This is true since it's the sum of two fractions.

7. Since a, b, c, and d are integers, the numerator ad + bc and the denominator bd are integers.

This is also true since the sum and product of two integers are always integers.

8. Furthermore since b and d are both non-zero, bd is also non-zero.

This is true since the product of any non-zero number with another non-zero number is also non-zero.

9. Therefore, x + y is a rational number.

This statement is true since x+y is the quotient of two integers, and since both integers are non-zero, then the quotient is also non-zero and therefore rational.

Therefore, the given argument is correct.

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Answer:

3

Step-by-step explanation:

the scale is 2 so you need to figure out 3y + 5 = 14