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lapo4ka [179]
3 years ago
11

I need help pleaseeeee

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer:

-7

82

Step-by-step explanation:

Just add the two wings(edges) then divide by 2 , simple

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Triangle ABC has vertices A(0,0), B(0,4), and C(3,0). Write the equation for the line containing altitude AR in standard...

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4 0
3 years ago
Factor completely 2x^2 − 2x − 40. 2(x − 5)(x 4) (2x − 10)(x 4) (x − 5)(2x 8) 2(x − 4)(x 5)
Zina [86]
2x^2 - 2x - 40 = 2x^2 -10x + 8x - 40 = 2x(x - 5) + 8(x - 5)

= (x-5)(2x+8) = 2(x-5)(x+4)
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3 years ago
Find the monthly interest payment in the situation described below. Assume that the monthly interest rate is 1 divided by 12 of
asambeis [7]

The monthly interest payment is 162

<h3>What is Interest?</h3>

Interest, in its most simple form, is calculated as a percent of the principal.

First,

monthly interest rate,

= 0.24/12

= 0.02.

Now,

Average monthly interest,

=0.02×$675

= $13.50.

Thus, the annual interest payments,

= 12×$13.50

= $162

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3 0
1 year ago
1. If B is between A and C, AB = 12 in., and AC = 19 in., then what is BC?
Oksanka [162]

Answer:

BC = 7.

Step-by-step explanation:

subtract 12 inches from 19 and then you get the result for BC.

8 0
3 years ago
A manufacturer has been selling 1000 flat-screen TVs a week at $500 each. A market survey indicates that for A manufacturer has
luda_lava [24]

Answer:

a) Demand function: q=6000-10\cdot p

b) The rebate should be of $200, so the sale price becomes $300 per unit.

c) The rebate should be of $150, so the sale price becomes $350 per unit.

Step-by-step explanation:

a) In this case, we have a known point of the demand function (1000 units sold at $500), and the slope of a linear function (increase by 100 units fora decrease in $10).

We can express the demand function (linear) as:

q=b+m\cdot p

To calculate the slope m, we use:

m=\Delta q/\Delta p=(+100)/(-10)=-10

To calculate b, we use the known point and the calculated slope:

q=b+m\cdot p\\\\b=q-m\cdot p=1000-(-10)\cdot (500)=1000+5000=6000

Then the demand function is:

q=6000-10\cdot p

b) The revenue can be expressed as:

R=q\cdot p = (6000-10p)\cdot p=6000p-10p^2

To maximize, we can derive and equal to zero

dR/dp=6000-2*10p=0\\\\20p=6000\\\\p=300

The rebate should be of $200, so the sale price becomes $300 per unit.

c) If we take into account the cost, we have that

R=q\cdot p-C=(6000p-10p^2)-(68000+100q)\\\\R=(6000p-10p^2)-(68000+100(6000-10p))\\\\R=6000p-10p^2-(68000+600000-1000p)\\\\R=-10p^2+7000p-668000

To maximize, we can derive and equal to zero

dR/dp=-20p+7000=0\\\\p=7000/20=350

The rebate should be of $150, so the sale price becomes $350 per unit.

5 0
3 years ago
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