A swimmer swims 2 kilometers every 15 minutes. If he comes to swim at the same rate which graph best represents his swimming speed in kilometers per minute

Answer:

y=2x+5

Step-by-step explanation:

u find the slope and since the y intercept is (0,5), that's the b value

hope this helps

Answer: 2x+5

Step-by-step explanation: calculate the slope of the points with the formula m= (y2-y1)/(x2-x1) then use the y-int to complete the equation

The last one as we can see that as the wight increases, the miles per gallon decreases.

Answer: 80

Step-by-step explanation:

Answer:

True

Step-by-step explanation:

It pases vertical line test but does not have an inverse

Answer:

the second one is the answer

Step-by-step explanation:

hope that helps

Answer:

B; {x|–5 < x < 6}

Step-by-step explanation:

Answer:

true

Step-by-step explanation:

Answer:

55

Step-by-step explanation:

Answer:

$7969.24

Step-by-step explanation:

5000*1.06 to power of 8 = 7969.24037265

round to 2dp = $7969.24

Answer:

7400

Step-by-step explanation:

Answer:

Step-by-step explanation:

17x + 14 + 4x -2 = 21x + 12

21x + 12 = 180

-12         -12

21x = 168

x = 8 degrees

[tex]\boxed{\large{\bold{\green{ANSWER~:) }}}}[/tex]

Here,

The two lines l and m are parallel,

cut by a transversat line t

So The two angles are supplementary each other.

The sum of two supplementary angles are 180°

(17x+14)°+(4x-2)°=180°

17x+14°+4x-2°=180°

17x+4x+14°-2°=180°

21x+12°=180°

21x=180°-12°

21x=168°

x=168°/21

x=8°

The value of x is 8°

75% of the population has an IQ greater than 89.95.

(a)What percentage of people have an IQ lower than 91?The given distribution is the normal distribution, with the mean 100 and standard deviation 15. It is required to calculate the percentage of people having an IQ score lower than 91.

To calculate the percentage of people having an IQ score lower than 91, standardize the given IQ score of 91 using the formula of z-score.z=(x−μ)/σwherez is the standardized score,x is the raw score,μ is the mean, andσ is the standard deviation.

The values can be substituted as follows.z=(91−100)/15=−0.6Now, find the probability of having a z-score less than or equal to -0.6 using the standard normal distribution table.

The value in the table is 0.2743, which means the probability of having a z-score less than or equal to -0.6 is 0.2743.Thus, 27.43% of people have an IQ score lower than 91.

(a) 27.43% of people have an IQ lower than 91.(b)Fill in the blank. 75% of the population have an IQ that is greater than X.

In order to find X, the z-score can be calculated using the formula of z-score.z=(x−μ)/σwherez is the standardized score,x is the raw score,μ is the mean, andσ is the standard deviation.

The z-score for the given problem can be calculated as follows:z = (x - μ)/σ (standardized score formula)z = (x - 100)/15 (values substituted)To find the value of x for which 75% of the population have an IQ greater than x, we need to determine the z-score that corresponds to the 25th percentile.

This is because 75% of the population is above the 25th percentile and below the 100th percentile.Using a standard normal distribution table, we can find the z-score that corresponds to the 25th percentile. The z-score is approximately -0.67.

Now that we have the z-score, we can solve for x as follows.-0.67 = (x - 100)/15 (substitute z-score)-10.05 = x - 100 (multiply both sides by 15)-89.95 = x

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The value of e varies based on A. Oe=b - Pb e = 0 Oe=AtAb would be  (AT A)-1 AT b.

The given projection matrix is P = A(AT A)-1A".

We have been asked to find the value of e if A is invertible. Let's proceed further and solve this problem. First, we need to find the product of A and its transpose, i.e., AT A.A.T.A = [a11 a12 ... a1n] [a21 a22 ... a2n] ... [an1 an2 ... ann] = [Σ(ai1)(aj1) Σ(ai1)(aj2) ... Σ(ai1)(ajn)] [Σ(ai2)(aj1) Σ(ai2)(aj2) ... Σ(ai2)(ajn)] ... [Σ(ain)(aj1) Σ(ain)(aj2) ... Σ(ain)(ajn)]

The inverse of AT A is (AT A)-1. Thus, (AT A)-1 AT A = I.Where I is the identity matrix. So we get P = A(AT A)-1 A".

Now, the value of e can be calculated as: Oe = b - Pe = b - A(AT A)-1 A" b = A x (AT A)-1 x AT b

This is the expression for the solution of the least square problem and if A is invertible, we can find the solution by directly calculating A-1 x b which is nothing but e. Thus, the value of e is e = A-1b.

Substituting the given expression of e, we get e = (AT A)-1 AT b.

Thus, the correct answer is e = (AT A)-1 AT b.

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Answer:

9!

Step-by-step explanation:

To determine the local minimum and local maximum of the function A(x) = ∫₀ˣ f(t) dt on the interval (0, 6), we need to analyze the behavior of A(x) and its derivative.

Let's denote F(x) as the antiderivative of f(x), which means that F'(x) = f(x).

To find the local minimum and maximum, we need to look for points where the derivative of A(x) changes sign. In other words, we need to find the values of x where A'(x) = 0 or A'(x) is undefined.

Using the Fundamental Theorem of Calculus, we have:

A(x) = ∫₀ˣ f(t) dt = F(x) - F(0)

Taking the derivative of A(x) with respect to x, we get:

A'(x) = (F(x) - F(0))'

Since F(0) is a constant, its derivative is zero, and we are left with:

A'(x) = F'(x) = f(x)

Now, let's analyze the behavior of f(x) based on the given figure to determine the local minimum and maximum of A(x) on the interval (0, 6). Without the specific information about the shape of the graph, it is not possible to determine the exact values of x that correspond to local minimum or maximum points.

To find the local minimum, we need to locate a point where f(x) changes from decreasing to increasing. This point would correspond to x = x_min.

To find the local maximum, we need to locate a point where f(x) changes from increasing to decreasing. This point would correspond to x = x_max.

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The particular solution that does not involve any terms from the homogeneous solution is given by:[tex]Y(t) = C3 + C4te^(-t/2).[/tex]

To find a particular solution of the given differential equation using the method of variation of parameters, we follow these steps:

Solve the associated homogeneous equation: 4y" - 4y + y = 0.

The characteristic equation is:

[tex]4r^2 - 4r + 1 = 0.[/tex]

Solving the quadratic equation, we find two repeated roots: r = 1/2.

Therefore, the homogeneous solution is given by: y_h(t) = C1[tex]e^(t/2)[/tex] + C2t[tex]e^(t/2),[/tex] where C1 and C2 are constants.

Find the particular solution using the variation of parameters.

Let's assume the particular solution has the form:

[tex]y_p(t) = u1(t)e^(t/2) + u2(t)te^(t/2).[/tex]

To find u1(t) and u2(t), we differentiate this expression:

[tex]y_p'(t) = u1'(t)e^(t/2) + u1(t)(1/2)e^(t/2) + u2'(t)te^(t/2) + u2(t)e^(t/2) + u2(t)(1/2)te^(t/2).[/tex]

We equate the coefficients of e^(t/2) and te^(t/2) on both sides of the original equation:

[tex](1/2)(u1(t) + u2(t)t)e^(t/2) = 16e^(t/2).[/tex]

From this, we can deduce that u1(t) + u2(t)t = 32.

Differentiating again:

[tex]y_p''(t) = u1''(t)e^(t/2) + u1'(t)(1/2)e^(t/2) + u1'(t)(1/2)e^(t/2) + u1(t)(1/4)e^(t/2) + u2''(t)te^(t/2) + u2'(t)e^(t/2) + u2'(t)(1/2)te^(t/2) + u2(t)e^(t/2) + u2(t)(1/2)te^(t/2).[/tex]

Setting the coefficient of [tex]e^(t/2)[/tex]equal to zero:

[tex](u1''(t) + u1'(t) + (1/4)u1(t))e^(t/2) = 0.[/tex]

Similarly, setting the coefficient of [tex]te^(t/2)[/tex]equal to zero:

[tex](u2''(t) + u2'(t) + (1/2)u2(t))te^(t/2) = 0.[/tex]

These two equations give us a system of differential equations for u1(t) and u2(t):

u1''(t) + u1'(t) + (1/4)u1(t) = 0,

u2''(t) + u2'(t) + (1/2)u2(t) = 0.

Solving these equations, we obtain:

u1(t) = C3[tex]e^(-t/2)[/tex] + C4t[tex]e^(-t/2),[/tex]

u2(t) = -4C3[tex]e^(-t/2)[/tex] - 4C4t[tex]e^(-t/2).[/tex]

Substitute the values of u1(t) and u2(t) into the assumed particular solution:

[tex]y_p(t) = (C3e^(-t/2) + C4te^(-t/2))e^(t/2) - 4C3e^(-t/2) - 4C4te^(-t/2).[/tex]

Simplifying further:

[tex]y_p(t) = C3 + C4te^(-t/2) - 4C3e^(-t/2) - 4C4te^(-t/2).[/tex]

So, the particular solution that does not involve any terms from the homogeneous solution is given by:

[tex]Y(t) = C3 + C4te^(-t/2).[/tex]

Here, C3 and C4 are arbitrary constants that can be determined using initial conditions or boundary conditions if provided.

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Answer:

The general solution to the given linear system is x = 3z - 1, y = -z + 1, where z is a free variable. This means that the solution consists of infinitely many points that lie on a straight line in three-dimensional space.

To solve the linear system, we can use the method of elimination or Gaussian elimination. Here, we'll use Gaussian elimination to find the general solution.

We start by writing the augmented matrix of the system:

[1 1 -2 | 0]

[2 2 3 | 1]

[3 3 1 | 7]

To simplify the matrix, we perform row operations to create zeros in the first column below the first entry. We subtract twice the first row from the second row and subtract three times the first row from the third row:

[1 1 -2 | 0]

[0 0 7 | 1]

[0 0 7 | 7]

Next, we divide the second and third rows by 7 to create leading ones:

[1 1 -2 | 0]

[0 0 1 | 1/7]

[0 0 1 | 1]

Now, we perform row operations to create zeros in the second column below the second entry. We subtract the third row from the second row:

[1 1 -2 | 0]

[0 0 1 | 1/7]

[0 0 0 | 0]

From the last row, we can see that 0z = 0, which means that z is a free variable. We can assign a parameter to z, say t, and solve for x and y in terms of t. From the first row, we have x + y - 2z = 0. Plugging in the values for x and y, we get x = 3z - 1 and y = -z + 1. Therefore, the general solution to the linear system is x = 3z - 1, y = -z + 1, where z is a free variable.

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Answer:

There will be 13 gifts given in the party.

Step-by-step explanation:

Given that Ming invited 12 people to her chrismas party, with a total of 13 people at her party, and each person have each other person a present, to determine how many presents were given the following logical reasoning must be performed:

Given that there are 13 people at the party, and each of them will bring a gift, by mathematical logic there will be 13 gifts distributed in the party.

An isometry on R^n must be regular and the set of isometries forms a group under matrix multiplication.

An isometry is a linear transformation that preserves distances, meaning the norm of the transformed vector is equal to the norm of the original vector. To show that an isometry must be regular (i.e., invertible), we can assume there exists a non-invertible isometry matrix A. In this case, there exists a nonzero vector x such that Ax = 0. However, this contradicts the property of an isometry since ||Ax|| = ||0|| = 0, but ||x|| ≠ 0. Thus, an isometry must be regular.

The set of isometries forms a group under matrix multiplication because it satisfies the group axioms: closure (the product of two isometries is an isometry), associativity (matrix multiplication is associative), identity (identity matrix is an isometry), and inverses (the inverse of an isometry is also an isometry).

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Given the differential equation is y''+5y = e^(at) with initial conditions y(0) = 0 and y'(0) = 0,  the correct answer is: y(t) = (-1/5√(5)) cos (√(5)t)+(1/5) sin (√(5)t).

To solve the given initial value problem using Laplace transform, we need to apply Laplace transform on both sides of the differential equation.

y''+5y=e^(at) L{y''+5y} = L{e^(at)) s^2Y(s)-sy(0)-y'(0)+5Y(s)=1/(s-a) [by Laplace transform formula] s^2Y(s)+5Y(s)=1/(s-a) ... [i]

Applying Laplace transform on both sides, we get:

L{y''+5y}=L{e^(at))

Using the initial conditions, we get Y(s)=1/[(s-a)(s^2+5)] Y(s) = [A/(s-a)] + [(Bs+C)sin(t)+ (Ds+E)cos(t)]/√(5) (i)

To find the values of A, B, C, D, and E, we take the inverse Laplace transform of both sides of equation (i) using partial fraction expansion. Let's solve for A:

Y(s)=A/(s-a)+(Bs+C)sin(t)/√(5)+(Ds+E)cos(t)/√(5)

Multiplying by s-a on both sides: (s-a)Y(s)=A+Bssin(t)/√(5)+Csincos(t)/√(5)+Dscos(t)/√(5)+Esin(t)/√(5)

Taking the inverse Laplace transform: y(t)=Ae^(at)+(B/√(5))sin(√(5)t)+(C/√(5))cos(√(5)t)+(D/√(5))cos(√(5)t)+(E/√(5))sin(√(5)t)

Differentiating y(t) with respect to t, we get:

y'(t)=Aae^(at)+Bcos(√(5)t)-Csin(√(5)t)-Dsin(√(5)t)+Ecos(√(5)t)

Using the initial conditions, y(0)=0 and y'(0)=0 in equation (iii), we get:

0=A+E ...(iv)0=A+B/√(5)+D/√(5) ...(v)

Solving equations (iv) and (v) simultaneously, we get A=0, B=√(5)/5, D=-√(5)/5, and E=0

Substituting these values in equation (iii), we get: y(t)=(√(5)/5)sin(√(5)t)-(√(5)/5)cos(√(5)t)

Therefore, the correct answer is: y(t)= (-1/5√(5))cos(√(5)t)+(1/5)sin(√(5)t).

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Answer:

B. There is only one value for s

Step-by-step explanation:

I hope this works for u.. :3

Can I have a brainliest plz :))

Answer:c?

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:

Step-by-step explanation:

1) a + a

2) 2n+4 = 6n

3)   1

4n = 4

n = 1

The approximation of the integral I is   -5.1941 using the composite Trapezoidal rule with n = 4.

We need to divide the interval [0, 2] into subintervals and apply the Trapezoidal rule to each subinterval.

The formula for the composite Trapezoidal rule is given by:

I = (h/2) × [f(x₀) + 2f(x₁) + 2f(x₂) + ... + 2f(xₙ₋₁) + f(xₙ)]

Where:

h = (b - a) / n is the subinterval width

f(xi) is the value of the function at each subinterval point

In this case, n = 4, a = 0, and b = 2. So, h = (2 - 0) / 4 = 0.5.

Now, let's calculate the approximation:

[tex]f\left(x_0\right)\:=\:f\left(0\right)\:=\:\left(0\:-\:3\right)e^{\left(0^2\right)}\:=\:-3[/tex]

[tex]f\left(x_1\right)\:=\:f\left(0.5\right)\:=\:\left(0.5\:-\:3\right)e^{\left(0.5^2\right)}\:=-2.535[/tex]

[tex]f\left(x_2\right)\:=\:f\left(1\right)\:=\:\left(1\:-\:3\right)e^{\left(1^2\right)}\:=\:-1.716[/tex]

[tex]f\left(x_3\right)\:=\:f\left(1.5\right)\:=\:\left(1.5\:-\:3\right)e^{\left(1.5^2\right)}\:=\:-1.051[/tex]

[tex]f\left(x_4\right)\:=\:f\left(2\right)\:=\:\left(2\:-\:3\right)e^{\left(2^2\right)}\:=\:-0.065[/tex]

Now we can plug these values into the composite Trapezoidal rule formula:

I = (0.5/2) × [-3 + 2(-2.535) + 2(-1.716) + 2(-1.051) + (-0.065)]

= (0.25)× [-3 - 5.07 - 3.432 - 2.102 - 0.065]

= -5.1941

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Answer:

3. 102

4. c

Step-by-step explanation:

For number 3:

(c^2 - a) + b becomes (10^2 - 3) + 5

10^2 = 100

100 - 3 + 5 = 102

For number 4:

Usually when you see distributive property, you're multiplying into parentheses. This time you're factoring out.

20c - 8d

It's not a, because that would be 20c + 8d

It's not b, because that would be 20c -32d

It is c, because 4 x 5c = 20 c and 4 x -2d = -8d

It's not d, because that would be 20c- 8cd

Answer:

Quotient: p + 7

Remainder: 10

Step-by-step explanation:

To find the quotient of the expression (2p² + 7p - 39) ÷ (2p - 7), we can use long division or synthetic division. Let's use long division:

           ____________________

2p - 7  |   2p² + 7p - 39

We start by dividing the first term of the dividend by the first term of the divisor, which gives us 2p² ÷ 2p = p. We then multiply p by the divisor (2p - 7) and subtract it from the dividend:

         p

    ____________________

2p - 7  |   2p² + 7p - 39

        - (2p² - 7p)

       14p - 39

    ____________________

2p - 7  |   2p² + 7p - 39

        - (2p² - 7p)

        ___________

               14p - 39

We repeat the process by dividing the first term of the new dividend (14p - 39) by the first term of the divisor (2p - 7). This gives us (14p - 39) ÷ (2p - 7) = 7. We then multiply 7 by the divisor (2p - 7) and subtract it from the new dividend:

         p + 7

    ____________________

2p - 7  |   2p² + 7p - 39

        - (2p² - 7p)

        ___________

               14p - 39

        - (14p - 49)

        ___________

                10

We are left with a remainder of 10. Therefore, the quotient is p + 7 with a remainder of 10.

Quotient: p + 7

Remainder: 10

Hope this helps!

p + 7 should be it.

I am not 100% sure?

Answer:

yo is your math class teacher named wicker? I think I'm ur classmate lol

By using the method of variation of parameters to solve a nonhomogeneous DE with W = -3 W2 = e 112 and W = er, we have

Given: W1=-3, W2=e^t and W3=er.The general solution of the non-homogeneous differential equation, y" + p(t) y' + q(t) y = g(t) , where p(t) and q(t) are functions of t and g(t) is non-zero function is given by;{eq}y = y_c + y_p {/eq}Where {eq}y_c {/eq} is complementary function and {eq}y_p {/eq} is particular function obtained by using variation of parameters.The solution is as follows:The given differential equation is{eq}y''+3y'+2y=-3e^{-t}+e^{t}+re^t{/eq}Characteristic equation is{eq}m^2+3m+2=0{/eq}Solving above equation gives us, {eq}m=-1,-2{/eq}Therefore, complementary function {eq}y_c=c_1e^{-t}+c_2e^{-2t} {/eq}Now, we find the particular solution by using the method of variation of parameters.Let {eq}y_p=u_1e^{-t}+u_2e^{-2t}{/eq}be a particular solution where {eq}u_1{/eq} and {eq}u_2{/eq} are functions of {eq}t.{/eq}Here W is a Wronskian and is given as:{eq}W=\begin{vmatrix}W_1&W_2\\W_1'&W_2'\\\end{vmatrix}=\begin{vmatrix}-3&e^t\\-1&e^t\\\end{vmatrix}=2e^{2t}+3e^{t}{/eq}Now, we find {eq}u_1{/eq} and {eq}u_2{/eq} as follows:{eq}u_1=\frac{-\int W_2 g(t) dt}{W}=\frac{-\int e^t(-3e^{-t}+e^{t}+re^t)dt}{2e^{2t}+3e^{t}}=-\frac{r}{5}-\frac{7}{10}+\frac{3}{10}e^{t}{/eq}Similarly,{eq}u_2=\frac{\int W_1 g(t) dt}{W}=\frac{\int -3e^{-t}(-3e^{-t}+e^{t}+re^t)dt}{2e^{2t}+3e^{t}}=-\frac{r}{5}-\frac{1}{10}+\frac{3}{10}e^{-2t}{/eq}Hence, the general solution of the differential equation is {eq}y=y_c+y_p=c_1e^{-t}+c_2e^{-2t}-\frac{r}{5}-\frac{7}{10}+\frac{3}{10}e^{t}-\frac{r}{5}-\frac{1}{10}+\frac{3}{10}e^{-2t}{/eq}So, Option D, {eq}-4u_2=0,~u_1=-\frac{r}{5}-\frac{7}{10}+\frac{3}{10}e^{t},~u_2=-\frac{r}{5}-\frac{1}{10}+\frac{3}{10}e^{-2t}{/eq} is correct.

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Answer:

20.00

Step-by-step explanation:

Answer:

A. Shading above a solid line.

Step-by-step explanation:

Let [tex]f(x) = 2\sqrt{3}\cdot x + 2[/tex], whose domain is all real numbers, since it is a first order polynomial (linear function) and meaning that for all element of [tex]x[/tex] exists one and only one value for [tex]f(x)[/tex], meaning a solid line. If [tex]g(x) > f(x)[/tex], then the range of all possible results is a shade area above the solid line.

Hence, the correct answer is A.