Find the specified element in the inverse of the given matrix

Answer:

19 members voted.

Step-by-step explanation:

Percentage problems can be solved by a rule of three.

25% of the 76 members of the club voted in the election. How many members voted?

So 76 is 100% = 1. How much is 0.25?

76 - 1

x - 0.25

[tex]x = 76*0.25[/tex]

[tex]x = 19[/tex]

19 members voted.

A G I = Adjustable Gross income,

1 exception = $ 8200,

A G I= Total Income (Gross Income + Amount earned through other means) - (Adjustments +Deduction)

= ($ 23670+$8200)-($ 0 + $0)

 = $ 31,870

1 exception = $ 8200, there may be many reasons for exception.

So, Taxable Income = Gross Income - 1 Exception

                                  = $ 23670 - $ 8200

                                  =  $ 15,470

Answer:

Hence, AB=12.

Step-by-step explanation:

We are given that the perpendicular bisector of side AB of ∆ABC intersects side BC at point D.

this means that side AE=BE.

Also we could clear;ly observe that

ΔBED≅ΔAED

( since AE=BE, side ED common, ∠BED=∠AED

so by SAS congruency the two triangles are congruent)

Now we are given that:

the perimeter of ∆ABC is 12 cm larger than the perimeter of ∆ACD.

i.e. AB+AC+BC=AC+AD+CD+12

AB+BC=AD+CD+12

as AD=BD

this means that AD+CD=BD+CD=BC

AB+BC=BC+12

AB=12

Hence AB=12

Answer:

The required length of [tex]AB[/tex] is [tex]12\rm\;{cm}[/tex].

Step-by-step explanation:

Given: The perpendicular bisector of side [tex]AB[/tex] of [tex]\bigtriangleup{ABC}[/tex] intersects side [tex]BC[/tex] at point [tex]D[/tex] and the perimeter of  [tex]\bigtriangleup{ACD}[/tex].

From the figure,

[tex]AE=BE[/tex]         .......(1)              (as [tex]DE[/tex] is perpendicular bisector of side [tex]AB[/tex])

Now, In [tex]\bigtriangleup{BED}[/tex] and [tex]\bigtriangleup{AED}[/tex]

     [tex]AE=BE[/tex]                                     ( from equation 1 )

[tex]\angle {BED} =\angle {AED}[/tex]                               ( Both [tex]90^\circ[/tex] )

    [tex]ED=ED[/tex]                                     ( Common side)

[tex]\bigtriangleup{BED}\cong\bigtriangleup{AED}[/tex]                              ( by SAS congruence rule)

      [tex]BD=AD[/tex]    .........(2)                   (by CPCT)

As per question,

The perimeter of ∆ABC is with 12 cm larger than the perimeter of ∆ACD.

[tex]AB+BC+AC=AC+CD+AD+12[/tex]

         [tex]AB+BC=AD+CD+12\\AD+CD=BD+CD\\AB+BC=BC+12\\[/tex]

                   [tex]AB=12\rm\;{cm}[/tex]

Hence, the length of [tex]AB[/tex] is [tex]12\rm\;{cm}[/tex].

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Answer:

$48

Step-by-step explanation:

List price = $60.00

Discount = 20% .

Hence the new price is (100-20)% = 80% of original price .

=> price = $ 60* 80/100

=> price = $ 48

Answer: There are 95 students who scored between 35 and 61.

Step-by-step explanation:

Since we have given that

The following data : 11,35,61,70,79.

So, the median of this data would be = 61

First two data belongs to "First Quartile " i.e. Q₁

and the second quartile is the median i.e. 61.

The last two quartile belongs to "Third Quartile" i.e. Q₃

And we know that each quartile is the 25th percentile.

And we need "Number of students who scored between 35 and 61."

So, between 35 and 61 is 25% of total number of students.

So, Number of students who scored between 35 and 61 is given by

[tex]\dfrac{25}{100}\times 380\\\\=\dfrac{1}{4}\times 380\\\\=95[/tex]

Hence, There are 95 students who scored between 35 and 61.

The number of students who scored between 35 and 61 is 95

The 5 number summary is the value of the ;

The total Number of students = 380

The lower quartile (Lower 25%) = 35

The median (50%) = 61

This means that 25% of the total students scored between 35 and 61.

Hence, 95 students scored between 35 and 61.

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To solve for the problem, the formula is:

n = [z*s/E]^2

where:

z is the z value for the confidence interval

s is the standard deviation

E is the number of units

So plugging that in our equation, will give us:

= [1.96*9.38/3]^2

= (18.3848/3)^2

= 6.1284^2

= 37.5 or 38

The zeros of the function y = -3(x - 2)² + 6 are x = 2 + √2 and x = 2 - √2, which represent the hot dog prices at which the hot dog stand makes $0.00 profit.

The zeros of a function are the values of x that make the y-coordinate equal to zero. In this case, the function y = -3(x - 2)² + 6 represents the daily profit, and we need to find the x-values that result in a profit of $0.00. Setting the profit, y, to zero and solving for x, we get:

0 = -3(x - 2)² + 6

Adding 3(x - 2)² to both sides and simplifying the equation gives:

3(x - 2)² = 6

Dividing both sides by 3, we have:

(x - 2)² = 2

Take the square root of both sides, remembering to consider both the positive and negative square roots:

x - 2 = ±√2

Adding 2 to both sides gives us the final solutions:

x = 2 ± √2

So, the zeros of this function are x = 2 + √2 and x = 2 - √2. These are the hot dog prices at which the hot dog stand makes $0.00 profit.

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The derivative[tex]\( \frac{dw}{dt} \) is \( \frac{7t^6 e^{4-t}}{2+9t} - \frac{t^7 e^{4-t}}{2+9t} - \frac{9t^7 e^{4-t}}{(2+9t)^2} \).[/tex]

To find [tex]\( \frac{dw}{dt} \)[/tex] using the chain rule for the given function[tex]\( w = \frac{x e^y}{z} \), where \( x = t^7 \), \( y = 4 - t \), and \( z = 2 + 9t \)[/tex], follow these steps:

1. **Express ( w ) in terms of ( t ):**

  Substitute ( x ), ( y ), and ( z ) into ( w ):

[tex]\[ w = \frac{x e^y}{z} = \frac{(t^7) e^{(4 - t)}}{2 + 9t} \][/tex]

2. **Apply the chain rule:**

  The chain rule states that for a function ( w(t) ) defined implicitly by ( w = f(x(t), y(t), z(t)) ), the derivative [tex]\( \frac{dw}{dt} \)[/tex] is given by:

[tex]\[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt} \][/tex]

3. **Compute partial derivatives of ( w ) with respect to ( x ), ( y ), and ( z ):**

[tex]\( \frac{\partial w}{\partial x} = \frac{e^y}{z} \)[/tex]  

  [tex]\( \frac{\partial w}{\partial y} = \frac{x e^y}{z} \)[/tex]  

[tex]\( \frac{\partial w}{\partial z} = -\frac{x e^y}{z^2} \)[/tex]

4. **Compute [tex]\( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \):**[/tex]

[tex]\( \frac{dx}{dt} = 7t^6 \)[/tex]  

[tex]\( \frac{dy}{dt} = -1 \)[/tex]

[tex]\( \frac{dz}{dt} = 9 \)[/tex]

5. **Substitute these into the chain rule formula:**

[tex]\[ \frac{dw}{dt} = \frac{e^y}{z} \cdot 7t^6 + \frac{x e^y}{z} \cdot (-1) + \left(-\frac{x e^y}{z^2}\right) \cdot 9 \][/tex]

6. **Substitute[tex]\( x = t^7 \), \( y = 4 - t \), \( z = 2 + 9t \)[/tex] into the expression:**

[tex]\( e^y = e^{4 - t} \)[/tex]

  Substitute these values into the formula for [tex]\( \frac{dw}{dt} \):[/tex]

[tex]\[ \frac{dw}{dt} = \frac{e^{4 - t}}{2 + 9t} \cdot 7t^6 - \frac{t^7 \cdot e^{4 - t}}{2 + 9t} - \frac{9t^7 \cdot e^{4 - t}}{(2 + 9t)^2} \][/tex]

Therefore, [tex]\( \frac{dw}{dt} \)[/tex] is:

[tex]{\frac{dw}{dt} = \frac{7t^6 e^{4 - t}}{2 + 9t} - \frac{t^7 e^{4 - t}}{2 + 9t} - \frac{9t^7 e^{4 - t}}{(2 + 9t)^2} } \][/tex]

The percentile scores for a distribution with a mean of 100 and a standard deviation of 15, the z-scores for 112 and 82 are 0.8 and -1.2, respectively.

Calculating Percentile Scores

To determine percentile scores for the values given in the original student question, we need to calculate the z-scores for 112 and 82 from a distribution with a mean of 100 and a standard deviation of 15.

To find the z-score, use the formula:

Z = (X - mu) / sigma

Where X is the raw score, \\mu is the mean, and \\sigma is the standard deviation.
For 112:

Z = (112 - 100) / 15

Z = 12 / 15

Z = 0.8
For 82:

Z = (82 - 100) / 15

Z = -18 / 15

Z = -1.2

Once the z-scores are calculated, we can look up the corresponding percentiles in a standard normal distribution table.

A z-score of 0.8 corresponds to approximately the 78.81st percentile, meaning about 78.81% of the scores are below 112.

A z-score of -1.2 corresponds to approximately the 11.54th percentile, meaning about 11.54% of the scores are below 82.

sin(a+b) = -1/7 +√455/42 = 0.8721804464845457

cos(a-b) = √13/42 - √35/7 =  -0.7761476987942811

tan(a+b )= (6√13/13 + √35) / (1 - 6√455/13) = -0.525

Given sin(a) = 6/7 and cos(b) = -1/6, with a in quadrant II and b in quadrant III, we need to utilize trigonometric identities to find sin(a+b), cos(a-b), and tan(a+b).

Firstly, since a is in quadrant II, cos(a) is negative. We use the identity sin²(a) + cos²(a)=1 to find cos(a):

Similarly, since b is in quadrant III, sin(b) is also negative. We use the identity sin²(b) + cos²(b)=1 to find sin(b):

Now we can use the angle addition and subtraction formulas:

1. sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

2. cos(a - b) = cos(a)cos(b) + sin(a)sin(b)

3. tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))

Answer:

5836200.

Step-by-step explanation:

Given  :  5836197 .

To find : Round 5836197 to the nearest hundred.

Solution : We have given 5836197

Step 1 : First, we look for the rounding place which is the hundreds place.

Step 2 : Rounding place is 97.

Step 3 : 97  is greater than  50 then it would be rounded up mean next number to 97 would be increase to 1 and 97 become 00.

Step 4 : 5836200.

Therefore, 5836200.

The correct answer is C.

Jonathan’s answer is reasonable because 2 times 8 is 16, and 16 is close to 15.77.

Hope this helps.

Answer:= 5/6

Step-by-step explanation:hope this helps

The answer provides an expression for the course average and a compound inequality to earn a B in a wellness course.

Expression representing the course average: 1/5(70 + 86 + 81 + 83 + x)

Compound inequality for earning a B: 80 ≤ 1/5(70 + 86 + 81 + 83 + x) < 90

(a) The function describes a parabola that opens downward. The value of A(x) is zero when x=0 and when x=100. The vertex (maximum) is halfway between those zeros, at x=50. The dimensions are 50 and 100-50 = 50. The maximum area pen will be 50 ft square.

(b) A(50) = (50 ft)² = 2500 ft² . . . the maximum area

(c) A(x) will be negative if x < 0 or x > 100, so the domain is 0 ≤ x ≤ 100.

The value of A(x) ranges from 0 to its maximum, 2500. Hence the range is 0 ≤ A(x) ≤ 2500.

For a rectangular fenced area, the dimensions that maximize the area are 50 feet by 50 feet, given a total of 200 feet of fence. The maximum possible area is 2500 square feet. The domain for this function is [0, 100] and the range is [0, 2500].

This problem deals with maximizing an area given a certain length of fencing. It is a part of optimization problems often encountered in calculus. Firstly, we understand that a rectangle will give maximum area for a fixed perimeter. Considering a rectangle, the dimensions would be x and 100-x, where x is one of the sides of the rectangle.

(a) To find the dimension that maximizes the area, we first need to find the derivative A'(x) of A(x), which equals to -2x + 100. Setting A'(x) to zero gives us x = 50. That means the dimensions that will give us maximum area are 50 feet by 50 feet.

(b) Substituting x = 50 back into the original function, we find that the maximum area is A(50)=50(100-50)=50*50=2500 square feet.

(c) For the domain of A in this application, it would be all possible values of x. So, x could be any number from 0 to 100 (both inclusive). The range of A would be from 0 to the maximum value, 2500 square feet.

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The reduction factor of 0.8 means the lengths in the reduced triangle are 0.8 times those of the original.

Then the original 6 inch length is reduced to 0.8×6 inches = 4.8 inches in the reduced triangle.

Answer:

4.8 inches

Step-by-step explanation:

The scale factors are used to convert a figure into another one with similar characteristics but different lengths, in this example is a triangle, and in order to calculate the measure of the height you just have to multiply the original height by the scale factor:

6 inches * scale factor

6 inches* 0.8= 4.8 inches

So the resultant triangle will have a height of 4.8 inches.