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3 years ago

Please help x-3=x^(2)-10x+25

Mathematics

1 answer:

schepotkina [342]3 years ago

8 0

Answer:

x=7,4

Step-by-step explanation:

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The A&M Hobby Shop carries a line of radio-controlled model racing cars. Demand for the cars is assumed to be constant at a

Bond [772]

Answer:

Step-by-step explanation:

A) Demand per month= 40 cars

Annual Demand (D)= 12*40 = 480

Fixed Cost per order (K)= 15

Holding Cost= 20% of cost= 60 *0.2 = 12

a. Economic Order Quantity=

Q^{*}={\sqrt {{\frac {2DK}{h}}}}

= √(2*480*15)/12

=34.64 ~ 35

Total Cost =P*D+K(D/EOQ)+h(EOQ/2) P= Cost per unit

= 60*480+ 15(480/35) + 12(35/2)

= 28800+ 205.71+ 210

=$29215.71

B). Backorder Cost (b)= $45

Qbo= Q* × √( b+h/ h)

= 35*√(12+45/ 45)

= 35* 1.12

=39.28 ~ 39

Shortage (S)= Qbo * (K/K+b)

= 39* (15/15+45)

= 39* 0.25

= 9.75

Total Cost Minimum=( bS2/ 2Qbo) + P (Qbo- S)2/2Qbo + K(D/Qbo)

=45* 9.752 / 2* 392 + 60 (39-9.75)2/ 2* 392 + 15 ( 480/39)

= 1.40+ 21.9.+ 184.61

=$207.91

C)Length of backorder days (d) = Demand ÷ amount of working days

d = 480 ÷ 300

d = 1.6

Calculate the backorders as the maximum number of backorders divided by the demand per day

s/d = 9.75/1.6 = 6.09 days (answer)

D) Calculate the difference in total between not using backorder:

$207.85 + $207.85 - 207.91 = $207.79

The saving in using backorder is $207.79.

Therefore I would recommend using a backorder

6 0

3 years ago

I need the answer in it’s simplest form

Lina20 [59]

3/4 changed to 6/8 makes it a 7/8 when adding making this most simplified into 7/8

3 0

3 years ago

The table shows the battery lives in hours of ten Brand A batteries and ten Brand B batteries.

defon

Plotting the data (attached photo) roughly shows that the data is skewed to the left. In other words, data is skewed negatively and that the long tail will be on the negative side of the peak.

In such a scenario, interquartile range is normally the best measure to compare variations of data.

Therefore, the last option is the best for the data provided.

5 0

3 years ago

Read 2 more answers

Hans invests $550 at a rate of x% per year compound interest. At the end of 10 years, the value of the investment is $638.30, co

cricket20 [7]

Answer:

1.5%.

Step-by-step explanation:

x per cent is written as 0.01x as a decimal fraction.

638.30 = 550(1 + 0.01x)^10

(1 + 0.01x)^10 = 638.30/550

Taking logarithms:

10 ln (1 + 0.01x) = ln ( 638.30/550)

ln (1 + 0.01x) = ln ( 638.30/550)/ 10

ln(1 + 0.01x) = 0.014889

Converting the logs:-

1 + 0.01x = e^0.014889 = 1.015

0.01x = 1 - 1.015

0.01x = 0.015

x = 0.015/ 0.01

= 1.5.

8 0

3 years ago

What would be the answer for 16.058-1.2, and how would you solve it. Please and thankyou.

Annette [7]

Answer:

14.858

Step-by-step explanation:

16.058−1.2

=16.058−1.2

=16.058+−1.2

=14.858

7 0

3 years ago

Read 2 more answers