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skad [1K]
3 years ago
5

Hello!

Mathematics
1 answer:
Lena [83]3 years ago
5 0

9514 1404 393

Answer:

   square with side length √17 units, so area 17 square units.

Step-by-step explanation:

Each side has a slope with a rise of 1 and a run of 4, or a rise of -4 and a run of 1. The ratios of these numbers are opposite reciprocals, so we know the sides are perpendicular. The sum of squares of these numbers is 17, so we know the sides are √17 and all are the same length. (The Pythagorean theorem tells us this.)

A quadrilateral with equal-length perpendicular sides is a square. This square has side length √17 units and area (√17)² = 17 square units.

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What is the mean absolute deviation of 1, 2, 2, 3, 5, 6, 6, 8, 8, 9
nikdorinn [45]

Answer:

5

Step-by-step explanation:

1+2+2+3+5+6+6+8+8+9/10=5

6 0
3 years ago
The Fine Line Pen Company makes two types of ballpoint pens: a silver model and a gold model. The silver model requires 1 minute
tresset_1 [31]

Answer:

Optimal production = 600 gold pens

Revenue  = 600*7 = $4200 gold pens

Step-by-step explanation:

The Fine Line Pen Company makes two types of ballpoint pens: a silver model and a gold model.

A. The silver model requires 1 minute in a grinder and 3 minutes in a bonder.

B. The gold model requires 3 minutes in a grinder and 4 minutes in a bonder.

Because of maintenance procedures,

C. the grinder can be operated no more than 30 hours per week and

D. the bonder no more than 50 hours per week.

The company makes

E. $5 on each silver pen and

F. $7 on each gold pen.

How many of each type of pen should be produced and sold each week to maximize profits?

Solution:

We will solve the problem graphically, with number of silver pens, x, on the x axis, and number of gold pens, y, on the y axis, i.e.

1. From A and C, the maximum number of silver pens

x <= 30*60 / 1 = 1800 and

x <= 50*60 /3 = 1000  ....................(1)   bonder governs

2. from A & D, the maximum number of gold pens

y <= 30*60 / 3 = 600 .....................(2) grinder governs

y <= 50*60 / 4 = 750

3. From D,

x + 3y <= 30*60 = 1800  ...................(limit of grinder) ..... (3)

3x + 4y <= 50*60 = 3000 .................(limit of bonder)  .......(4)

Need to maximize profit,

Z(x,y) = 5x+7y, represented by parallel lines y = -5x/7 + k such that all constraints of (3) and (4) are satisfied.

The maximum is obtained when Z passes through (360,480), i.e. at intersection of constraints (3) and (4).  Using slope intercept form,

(y-480) = -(5/7)(x-360)

or y=-(5/7)x + (737+1/7)    [the purple line] which violates the red line, so not a solution.

Next try the point (0,600)

(y-600) = -(5/7)(x-0), or

y = 600 - (5/7)x   [the black line]

As we can see all point on the black (in the first quadrant) satisfy the constraints, so it is a feasible solution, and is the optimal solution, with a revenue of

Revenue  = 600*7 = 4200 gold pens

8 0
4 years ago
Pls help me........
Natasha_Volkova [10]
I think it’s the first one
5 0
3 years ago
Another KA question pls answer lol
FinnZ [79.3K]

Answer:

-40m + 24

Step-by-step explanation:

to answer this question you should distribute the -8 to all the digits

-8m + 24 -32m

then you can add like terms

-8m and -32m become -40m

-40m + 24 is the answer

4 0
2 years ago
Read 2 more answers
Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. x
miskamm [114]

Answer:

\frac{784}{15} \pi

Step-by-step explanation:

According to the given situation, the calculation of volume of the solid is shown below:-

Here we will consider the curves that is

x = 7y^2, x = 7

Now, rotating the line for the line x which is equals to 7

7y^2 = 7\\\\y^2 = 1\\\\ y = \pm1

So, the inner radio is

7 - 7 = 0

and the outer radius is

7y^2 - 7\\\\ = 7(y^2 - 1)

Now, the area of cross section is

A(y) = \pi(7(y^2 - 1))^2\\\\ = 49\pi(y^4 - 2y^2 + 1)

The volume is

V = \int\limits^1_{-1} A(y)dy

now we will put the values into the above formula

= \int\limits^1_{-1} 49\pi(y^4 - 2y^2 + 1)dy\\\\ = 49\pi(\frac{y^5}{5}  - \frac{2y^3}{3} + y)^{-1}\\\\ = 49\pi(\frac{1}{5} - \frac{2}{3}  + 1 + \frac{1}{5} - \frac{2}{3} + 1)\\\\ = 49\pi(2 + \frac{2}{5} - \frac{4}{3} )\\\\ = 49\pi(\frac{30+6-20}{15} )\\\\ = \frac{49\pi}{15} (16)

After solving the above equation we will get

= \frac{784}{15} \pi

6 0
4 years ago
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