If EFGH is a parallelogram, what is the value of x?

To obtain the maximum profit, the company should set the price of each pair of socks to $5. This answer is derived from the vertex of the given quadratic profit function.

The student question asks at which price the sock selling company should sell its product to earn a maximum profit, given the profit function: P(x)=-3(x-5)^2+12. This function is a standard form of a quadratic equation, and in such equation, the maximum value is achieved at the vertex.

In this case, the vertex (h,k) of the quadratic function is (5,12), where 'h' is the value of 'x' that gives the maximum profit 'k'. So the company should set the price at $5 to earn a maximum profit.

This economic model analysis' underlying principle relies on the assumption of maximizing total revenues and minimizing total costs to yield the maximum profit. Hence, recognizing the nature of the profit function is crucial.

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the answer to your question is

 Unit walking rate in miles per hour = 5 2/3 / 2 2/3 = 17/3 / 8/3 = 17/3 x 3/8 = 17/8 = 2 1/8 miles per hour.

Unit walking rate in hours per mile = 1/ 17/8 = 8/17 hours per mile

Answer:

Rachel enjoys exercising outdoors. Today she walked 5 2/3 miles or 5.67 miles in 2 2/3 hours or 2.67 hours.

Rachel’s unit walking rate in miles per hour is = [tex]\frac{5.67}{2.67}[/tex] = 2.125 miles per hour or 2 1/8 miles per hour.

Rachel’s unit walking rate in hours per mile = [tex]\frac{2.67}{5.67}[/tex] = 0.47 hours per mile.

Group terms that contain the same variable, and move the constant to the opposite side of the equation

Complete the square. Remember to balance the equation by adding the same constants to each side.

(A²-3A+2.25)=-1+2.25

Rewrite as perfect squares

(+/-)(A-1.5)=1.12

first solution

(A-1.5)=1.12--------> A=1.12+1.5-----> A=2.62-----> sin x=2.62

second solution

-(A-1.5)=1.12----> -A+1.5=1.12----> A=1.5-1.12----> A=0.38---> sin x=0.38

the answer is

the equation was factored incorrectly

Its is b. hope this halps

Answer:

I think it's b, please correct me if I'm wrong.

Step-by-step explanation:

I'm taking the test rn so I hope its right.

Answer:

its b i just did it

Step-by-step explanation:

Length is [tex]\boldsymbol{4.4}[/tex] feets and width is equal to [tex]\boldsymbol{8.8}[/tex] feets.

The area of a two-dimensional region, form, or planar lamina in the plane is the quantity that expresses its extent.

A quadrilateral having four right angles is known as a rectangle.

Let [tex]\boldsymbol{w}[/tex] feets denotes width of a rectangle.

Length of a rectangle [tex]=\boldsymbol{w-4}[/tex] feets

Area of a rectangle [tex]=\boldsymbol{42}[/tex] square feet

Length [tex]\times[/tex] Width [tex]=42[/tex]

      [tex]w(w-4)=42[/tex]

[tex]w^2-4w-42=0[/tex]

[tex]w=\frac{4\pm \sqrt{16+168}}{2}[/tex]

   [tex]=2\pm \sqrt{46}[/tex]

As dimension can not be negative, [tex]2-\sqrt{46}[/tex] is rejected.

So,

[tex]w=2+ \sqrt{46}[/tex]

   [tex]=\boldsymbol{8.8}[/tex] feets

Length [tex]=8.8-4[/tex]

            [tex]=\boldsymbol{4.4}[/tex] feets

So, length is [tex]4.4[/tex] feets and width is equal to [tex]8.8[/tex] feets.

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The perimeters of two rhombi with equal areas are sometimes equal, but not always.

Let's consider the formula for the area of a rhombus, which is given by [tex]\( A = \frac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \)[/tex] are the lengths of the diagonals. For the perimeter P, we have [tex]\( P = 4 \times s \)[/tex], where s is the length of one side of the rhombus.

For two rhombi to have equal areas, the product of their diagonals must be the same. That is, if we have two rhombi with diagonals [tex]\( (d_1, d_2) \) and \( (d_1', d_2') \), then \( d_1 \times d_2 = d_1' \times d_2' \)[/tex].

However, the perimeter depends only on the length of one side, s, and not on the diagonals. Therefore, two rhombi with the same area can have different side lengths and thus different perimeters.

Let's consider an example:

Rhombus 1:

[tex]- Diagonals \( d_1 = 8 \) units and \( d_2 = 4 \) units[/tex]

[tex]- Area \( A = \frac{1}{2} \times 8 \times 4 = 16 \) square units[/tex]

[tex]- Side length \( s \), using the Pythagorean theorem (since the diagonals bisect each other at right angles), is \( s = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \) units[/tex]

[tex]- Perimeter \( P = 4 \times s = 4 \times \sqrt{20} \) units[/tex]

Rhombus 2:

[tex]- Diagonals \( d_1' = 10 \) units and \( d_2' = 3.2 \) units[/tex]

[tex]- Area \( A' = \frac{1}{2} \times 10 \times 3.2 = 16 \) square units (equal to the area of Rhombus 1)[/tex]

[tex]- Side length \( s' \), using the Pythagorean theorem, is \( s' = \sqrt{1.6^2 + 2.56^2} = \sqrt{2.56 + 6.5536} = \sqrt{9.1136} \) units[/tex]

[tex]- Perimeter \( P' = 4 \times s' = 4 \times \sqrt{9.1136} \) units[/tex]

Comparing the perimeters:

[tex]- \( P = 4 \times \sqrt{20} \approx 4 \times 4.472 = 17.888 \) units[/tex]

[tex]- \( P' = 4 \times \sqrt{9.1136} \approx 4 \times 3.0188 = 12.0752 \) units[/tex]

Clearly, the perimeters are not equal, even though the areas are the same.

However, it is possible for two rhombi with equal areas to have equal perimeters if their side lengths are the same. For instance, if Rhombus 1 and Rhombus 2 both had side lengths of s = s' = 4 units, then their perimeters would both be P = P' = [tex]4 \times 4 = 16 \)[/tex] units, regardless of the lengths of their diagonals, as long as the diagonals satisfy the area condition [tex]\( d_1 \times d_2 = d_1' \times d_2' \)[/tex].

In conclusion, the perimeters of two rhombi with equal areas can sometimes be equal, specifically when the side lengths are the same, but they are not always equal, as demonstrated by the example above.

Answer:

Option 4.

Step-by-step explanation:

Answer:

I just took the test the answer is

Step-by-step explanation:

B. 87,000, and the graph would have a y-intercept at 67,000.

The word BOOKKEEPER has 10 letters, but with repetitions. There are 30240 distinct permutations. On the circular table, with students of the same major sitting together, there are 1440 different ways to seat the students.

The word BOOKKEEPER has 10 letters, but the letter O appears twice, the letter K appears twice, and the letter E appears three times. To find the number of distinct permutations, we need to divide the total number of permutations by the repetitions. So, we have 10!/(2!2!3!) = 30240 distinct permutations.

In the second question, we need to consider the students with the same major sitting together as a single entity. We can treat the circular table as a linear arrangement by fixing one student in a position and arranging the rest. So, the total number of different ways the students can be seated around the table is (3!2!5!) × 2! = 1440.

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y= (−1/6)x^2+7x−80      multiply both sides by -6

-6y  =    x^2   - 42x  +  480       subtract 480 from both sides

 -6y  - 480   =  x^2  -  42x    take (1/2)  of 42  = 21.....square this  = 441  and add to both sides

 -6y - 480 + 441  =   x^2  - 42x  +  441       simplify the left, factor the right

 -6y - 39   =  (x  - 21)^2      factor the left side as

 -6 (y  +  39/6)   =  ( x - 21)^2       (1)

 Using the form

 4p (y - k)  =  ( x  - h)^2        we  can   write (1)  as

 4 (-3/2)(y - (-39/6) )   =  ( x  - 21)^2

 The vertex  = ( h, k)  =  ( 21, -39/6)   and    p  = -3/2

 And the directrix   is given by 

 y  = k - p  →   y  = -39/6 - (-3/2)  =  -39/6 + 3/2  = -39/6 + 9/6  =

-30/6  = - 5-------> y=-5

the answer Part 1) is

y=-5

Part 2) (y+2)^2= 129X-5 

Using the form

 4p (x - h)  =  ( y  - k)^2        we  can   write (1)  as

 4*(129/4)*(x-5/129)=(y+2)²

 The vertex  = ( h, k)  =  ( 5/129, -2)   and    p  = 129/4

 And the directrix   is given by 

x  = h - p  →   x = 5/129 - (129/4)  =  (20-16641)/516-----> -16621/516

x=-32.21

the answer part 2) is