JK and LM are perpendicular diameters of a circle. They are each 12 inches long. What is the approximate length of chord LK? With Explanation Please

Answer:

[tex]M(7 ; 4)[/tex]

Step-by-step explanation:

midpoint -M

[tex]M(\frac{x_{1}+x_{2}}{2} ; \frac{y_{1}+y_{2}}{2} )[/tex]

[tex]M(\frac{9+5}{2} ; \frac{6+2}{2} )[/tex]

[tex]M(7 ; 4)[/tex]

Answer:

(7, 4 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

( [tex]\frac{x_{1}+x_{2} }{2}[/tex] , [tex]\frac{y_{1}+y_{2} }{2}[/tex] )

Here (x₁, y₁ ) = (9, 6) and (x₂, y₂ ) = (5, 2) , then

midpoint = ( [tex]\frac{9+5}{2}[/tex] , [tex]\frac{6+2}{2}[/tex] ) = ([tex]\frac{14}{2}[/tex] , [tex]\frac{8}{2}[/tex] ) = (7, 4 )

Answer:

28-8-4x=0

x=5

5 days

Step-by-step explanation:

28-8=20 problems left

then 4 each day

Answer:

2

Step-by-step explanation:

60 divided by 28.3 which gives u 2.120141342756184

so round it which gives u 2

Answer:

1

EXPLAINATION:

y = mx +b

b = y intercept

Answer:

The answer is 440

Step-by-step explanation:

I just took the test and got that question right

Answer:

-18

Step-by-step explanation:

5a2-7-b3

[tex]5*a*2-7-b*3[/tex]

[tex]5\cdot \left(-2\right)\cdot \left(2\right)-7-\left(-3\right)\cdot 3[/tex]

Therefore, -18 is your answer.

Hope this helped! :)

Answer: should be 7819.20

explanation:

7200 * .043 * 2 = 619.20

7200 + 619.20 = 7819.20

Answer:

2) 4/5

3) 3

Step-by-step explanation:

2) 4/8-3=4/5

3)5x-4=3x+2

2x=6

x=3

Answer:

2) 4/5

3)x=3

Step-by-step explanation:

1) 4/8-3= 4/5

2) 5x-4=3x+2

bring 3x to left side and bring -4 to right side (remember signs will be reversed that is 3x becomes -3x and -4 will be 4

5x-3x=4+2

2x=6

x=6/2

x=3

put 3 in place of x

5*3-4=3*3+2

15-4=9+2

11=11

Answer:

it's arm will regenerate and grow back

plz mark me as brainliest

Answer:

B. The arm will regenterate

Answer:

V≈7853.98

Step-by-step explanation:

V=πr2h

r=10

h=25

Solution

V=πr2h=π·102·25≈7853.98163

Answer:

The amount invested in ADB is ¢1363.[tex]\overline 3[/tex] =

The amount invested in Barclays is ¢3,427.[tex]\overline 3[/tex]

The amount invested in GCB is ¢1,713.[tex]\overline 3[/tex]

Step-by-step explanation:

The parameters of the investment Stonewall made are;

The amount in interest he receives from ADB, Barclays and GCB = ¢250

The amount of interest ADP pays = 2% per annum

The amount of interest Barclays pays = 4% per annum

The amount of interest GCB pays = 5% per annum

The amount invested in Barclays = The amount invested in ADB and GCB + ¢350

The amount invested in Barclays = 2 × The amount invested in GCB

a) Let 'x', represent the amount invested in ADB, 'y' represent the amount invested in Barclays, and 'z', represent the amount invested in GCB

We have;

y = x + z + 350

y = 2·z

0.02·x + 0.04·y + 0.05·z = 250

Therefore, we get the three linear equations as follows;

-x + y - z = 350...(1)

y - 2·z = 0...(2)

0.02·x + 0.04·y + 0.05·z = 250...(3)

Using Matrix inversion, we have;

[tex]\left[\begin{array}{ccc}-1&1&-1\\0&1&-2\\0.02&0.04&0.05\end{array}\right] \times \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}350&0&250\end{array}\right][/tex]

The transpose of the 3 by  3 matrix [tex]M^T[/tex] is given as follows;

[tex]M^T = \left[\begin{array}{ccc}-1&0&0.02\\1&1&0.04\\-1&-2&0.05\end{array}\right][/tex]

The Adjugate Matrix is given as follows;

[tex]Adj = \left[\begin{array}{ccc}0.13&-0.09&-1\\-0.04&-0.03&-2\\-0.02&0.06&-1\end{array}\right][/tex]

The inverse of the matrix = Adj/Det where, Det = -0.15, is therefore;

[tex]M^{-1} = \left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right][/tex]

We therefore, get the solution as follows;

[tex]\left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right]\times \left[\begin{array}{c}350&0&250\end{array}\right] = \left[\begin{array}{c}\dfrac{4,090}{3} \\&\dfrac{10,280}{3} \\ & \dfrac{5,140}{3} \end{array}\right][/tex]

[tex]\begin{array}{c}x = \dfrac{4,090}{3} \\&y = \dfrac{10,280}{3} \\ & z = \dfrac{5,140}{3} \end{array}[/tex]

The amount invested in ADB, x = ¢4,090/3 = ¢1363.[tex]\overline 3[/tex]

The amount invested in Barclays, y = ¢10,282/3 = ¢3,427.[tex]\overline 3[/tex]

The amount invested in GCB, z = ¢5,140/3 = ¢1,713.[tex]\overline 3[/tex]

Answer: 6

Step-by-step explanation:

The total amount needed by Kallen for the gift and the wrapping paper will be:

= $43.87 + $2.94

= $46.81

Since he earns $8.25 for each lawn he mows, the number of lawns they he will need to complete to buy the gift and wrapping will be:

= $46.81 / $8.25

= 5.67

= 6 lawns approximately

He'll need to mow 6 lawns

Answer:

The exact same

Step-by-step explanation:

^ just means put it up like 3^2=3²

The inverse Laplace transform of 1 / (3s² + 5s + 1) is f(t) = 1/2 × [tex]e^{(-t)[/tex]- 1/2 × [tex]e^{(-t/3)[/tex].

To find the inverse Laplace transform of the function F(s) = 1 / (3s² + 5s + 1), we can use partial fraction decomposition and reference tables for Laplace transforms.

Step 1: Factorize the denominator

Factorize the denominator of the function 3s² + 5s + 1 to find its roots:

3s² + 5s + 1 = (s + 1)(3s + 1)

Step 2: Write the partial fraction decomposition

Write the function F(s) as a sum of partial fractions:

F(s) = A / (s + 1) + B / (3s + 1)

Step 3: Determine the values of A and B

To find the values of A and B, we can multiply both sides of the equation by the common denominator and equate the numerators:

1 = A(3s + 1) + B(s + 1)

Expand the right side and collect like terms:

1 = (3A + B)s + (A + B)

By equating the coefficients of s and the constant terms on both sides, we get a system of equations:

3A + B = 0

A + B = 1

Solving this system of equations, we find A = 1/2 and B = -1/2.

Step 4: Write the inverse Laplace transform

Using the partial fraction decomposition, we can now write the inverse Laplace transform:

F(s) = 1/2 × (1 / (s + 1)) - 1/2 × (1 / (3s + 1))

Referring to Laplace transform tables, we find that the inverse Laplace transform of 1 / (s + a) is [tex]e^{(at)[/tex], and the inverse Laplace transform of 1 / (s - a) is [tex]e^{(at)[/tex]. Therefore, applying these results, we have:

f(t) = 1/2 × [tex]e^{(-t)[/tex] - 1/2 × [tex]e^{(-t/3)[/tex]

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Answer:

the slope is 3

the y intercept is (0,-7)

Step-by-step explanation:

3x-y=7 solve for y

add y to both sides

3x=7+y

subtract 7 from both sides

y=3x-7

y=mx+b

m=slope

b=y-intercept

3=slope

-7=y-intercept

Hope that helps :)

(1) To find the value of B that makes f_(XY)(x,y) a valid joint density function, we need to ensure that the total probability over the entire domain is equal to 1. In this case, the domain is |x|<1 and |y|<1.

The integral of f_(XY)(x,y) over the given domain should be equal to 1:

∫∫ f_(XY)(x,y) dx dy = 1

∫∫ B(1+xy) dx dy = 1

To solve this integral, we integrate with respect to x first and then with respect to y:

∫(∫ B(1+xy) dx) dy

∫[Bx + B(xy^2)/2] dy, integrating with respect to x

Bxy + B(xy^2)/2 + C, integrating with respect to y

Now, evaluate the integral over the given domain:

∫[-1,1] [Bxy + B(xy^2)/2 + C] dy

[Bxy^2/2 + B(xy^3)/6 + Cy] evaluated from -1 to 1

[B/2 + B/6 + C] - [-B/2 - B/6 - C]

(B/2 + B/6 + C) - (-B/2 - B/6 - C)

2B/3 = 1

Solving for B:

B = 3/2

Therefore, the value of B that makes f_(XY)(x,y) a valid joint density function is B = 3/2.

(2) To determine if X and Y are uncorrelated, we need to calculate the covariance between X and Y. If the covariance is zero, then X and Y are uncorrelated.

Cov(X, Y) = E[XY] - E[X]E[Y]

To calculate E[XY], we need to find the joint expectation:

E[XY] = ∫∫ xy f_(XY)(x,y) dx dy

E[XY] = ∫∫ xy (3/2)(1+xy) dx dy

Integrating over the domain |x|<1 and |y|<1, we can calculate E[XY].

Similarly, we need to calculate E[X] and E[Y] to determine Cov(X, Y).

If Cov(X, Y) is found to be zero, then X and Y are uncorrelated.

(3) To prove or disprove independence between X and Y, we need to check if the joint probability density function (pdf) can be factorized into the product of the marginal pdfs of X and Y.

If f_(XY)(x,y) = f_X(x)f_Y(y), then X and Y are independent.

To determine if this factorization holds, we need to compare the joint pdf f_(XY)(x,y) with the product of the marginal pdfs f_X(x) and f_Y(y). If they are equal, then X and Y are independent. Otherwise, they are dependent.

(4) To prove or disprove the independence between X^2 and Y^2, we follow a similar approach as in (3). We compare the joint pdf of X^2 and Y^2 with the product of their marginal pdfs. If they are equal, X^2 and Y^2 are independent. Otherwise, they are dependent.

By examining the factorization of the joint pdfs and comparing them with the product of the marginal pdfs, we can determine the independence relationships between the variables X, Y, X^2, and Y^2.

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The last entry in the synthetic division table is not zero, 3i is not a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36.

To determine if 3i is a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36, we can use synthetic division.

First, we set up the synthetic division table:

   3i | 1   -4   9   -36

Performing the synthetic division:

      | (1)  (-4)   (9)   (-36)

   3i |        3i    9i²   27i

      | (1)  (-4 + 3i)   (9 + 9i²)   (-36 + 27i)

Simplifying the last row, we have:

      | (1)  (-4 + 3i)   (9 - 9)   (-36 + 27i)

      | (1)  (-4 + 3i)   (0)   (-36 + 27i)

Therefore, the last entry in the synthetic division table is not zero, 3i is not a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36.

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The given recurrence relation is an = 400 - 5an-2 + 200-3, with initial values a0 = 1 and a1 = 2. We need to find an explicit formula for a(n).

To find an explicit formula for a(n), we will first solve the recurrence relation and then generalize the pattern. Let's expand the relation for a few terms to observe the pattern:

a2 = 400 - 5a0 + 200-3 = 400 - 5(1) + 200-3 = 397

a3 = 400 - 5a1 + 200-3 = 400 - 5(2) + 200-3 = 388

a4 = 400 - 5a2 + 200-3 = 400 - 5(397) + 200-3 = -9603

a5 = 400 - 5a3 + 200-3 = 400 - 5(388) + 200-3 = -1260

From the given examples, we can observe that the recurrence relation alternates between positive and negative values. This suggests that the relation might have a periodic pattern. Since the given recurrence is a second-order relation (in terms of n), it is reasonable to assume that the pattern repeats every two terms.

Based on this observation, we can establish the following explicit formula for a(n):

a(n) = (-1)^(n mod 2) * (200n - 5^(n/2) + 200-3)

In this formula, the term (-1)^(n mod 2) alternates between -1 and 1 depending on the parity of n, ensuring the alternating pattern. The other terms account for the linear and exponential components of the recurrence relation.

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Equation system that corresponds to the graph:

1. y ≤ 3x 2. y > –2x – 1

To find the system of equations that corresponds to the provided graph, we must first analyze it and locate the regions that fulfil the specified requirements.

1. Begin by locating the darkened region underneath the line y 3x. This line has a slope of 3 and intersects the origin (0,0). Shade the area beneath the line.

2. After that, locate the darkened region above the line y > -2x - 1. The slope of this line is -2, while the y-intercept is -1. The area above the line should be shaded.

3. The solution space that meets both requirements is represented by the overlapping shaded region between the two lines. The common area is located below y 3x and above y > -2x - 1.

4. The equation system that corresponds to this common area is: - y 3x - y > -2x - 1

The space for the addition in the open area next to the present building is defined by these two formulae.

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Answer:

(5,6)

Step-by-step explanation:

I'm guessing you mean the coordinates for the point where the lines intersect

Answer:

I believe it is b: (5,6)

Explanation:

The point (5,6) is the only point on the graph that both lines go through.

Answer:

We meet again

Step-by-step explanation:

When a rectangle is dilated by a scale factor k, the length and width of the rectangle are both multiplied by k. In this case, the length of the rectangle is being dilated by a scale factor of 5. So the length of the rectangle after the dilation will be 20 * 5 = **100 meters**.

Answer:

It has been answered for you

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Answer:

Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.

Step-by-step explanation:

Please give brainliest

please help me

Step-by-step explanation:

mark me brilliant

Answer:

6 ft

Step-by-step explanation:

The triangle CDE is half of the size of triangle ABC, so if you match up the corresponding sides, DE is half of AC. AC is 12 so DE is 6.

hope is helped!

Answer:

It would decrease, but not necessarily by 8%

If the teacher had used a 90% confidence interval rather than a 98% confidence interval, the width of the interval would decrease, but not necessarily by 8%.

Option D is the correct answer.

A confidence interval is a range of values that is likely to contain the true value of an unknown parameter, such as a population means or proportion, based on a sample of data from that population.

It is a statistical measure of the degree of uncertainty or precision associated with a statistical estimate.

We have,

The width of a confidence interval is determined by the level of confidence and the sample size.

A higher confidence level requires a wider interval, and a larger sample size typically results in a narrower interval.

Since the sample size is fixed in this case, changing the confidence level will result in a change in the width of the interval.

As the confidence level increases, the width of the interval also increases. This is because a higher confidence level requires a larger margin of error, which is added to the point estimate to create the interval.

Therefore,

If the teacher had used a 90% confidence interval rather than a 98% confidence interval, the width of the interval would decrease, but not necessarily by 8%.

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