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Harlamova29_29 [7]
3 years ago
14

One solution to a quadratic equation is 4 - 5i. Find the equation.

Mathematics
1 answer:
marishachu [46]3 years ago
4 0

Answer:

The answer would be Equation B:

x2 - 8x + 41

Step-by-step explanation:

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Which could be the entire interval over which the
Alex73 [517]

Answer:

(1,4)

Step-by-step explanation:

7 0
3 years ago
Find an equation for the line tangent to the circle x^2 +y^2=25 at the point (3, -4)
Lostsunrise [7]
Hello : 
the center of this cercle is : O( 0, 0)
calculate the slope  m of the line : OA    .. A (3,-4)
m= (yA - yO) / (xA- xO) 
m = (-4-0)/(3-0)
m= -4/3
if : k the slope of the tangent : k×m = -1   (the tangent perpendiclar of : OA)
k× (-4/3) = -1
k = 3/4
an equation of this tangent is : y - (-4) = (3/4)(x-3)

7 0
3 years ago
Find the slope of each of the lines below:
kolezko [41]
When a line in a slope graph has no rise/run like this one, there is no slope.
The slope is "Undefined".
3 0
3 years ago
C=2(y-k) solve for y​
marshall27 [118]

Answer:

\blue{y = \dfrac{C + 2k}{2}}

Step-by-step explanation:

C = 2(y - k)

C = 2y - 2k

C + 2k = 2y

\dfrac{C + 2k}{2} = y

y = \dfrac{C + 2k}{2}

5 0
3 years ago
2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar
tia_tia [17]

9514 1404 393

Answer:

  a)  E = 6500 -50d

  b)  5000 kWh

  c)  the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

<u>Given</u>:

  starting excess (E): 6500 kWh

  usage: 50 kWh/day (d)

<u>Find</u>:

  a) E(d)

  b) E(30)

  c) E(150)

<u>Solution</u>:

a) The exces is linearly decreasing with the number of days, so we have ...

  E(d) = 6500 -50d

__

b) After 30 days, the excess remaining is ...

  E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

__

c) After 150 days, the excess remaining would be ...

  E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

3 0
3 years ago
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