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3 years ago

Y=1000x+b how do i solve for b?

Mathematics

1 answer:

Irina18 [472]3 years ago

6 0

Answer:

put it into a graphing calculator, the point where the line intercepts the y-axis would be B

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If x√(1+y) + y√(1+x) = 0 , then dy/dx = ?

frutty [35]

$DIFFERENTIAL \: \: \: CALCULUS \\ \\ \\ Given \: expression \: - \\ \: \\ x \sqrt{1 + y} \: = \: y \sqrt{1 + x} \\ \\ x \sqrt{1 + y} \: = \: - y \sqrt{1 + x} \\ \\ \\ Squaring \: both \: sides \: , \: We \: get \: - \: \\ \\ {x}^{2} (1 + y) = {y}^{2} (1 + x) \\ \\ {x}^{2} + {x}^{2} y - {y}^{2} - x {y}^{2} = 0 \\ \\ ( {x}^{2} - {y}^{2} ) + xy(x - y) = 0 \\ \\ (x - y)(x + y + xy) = 0 \\ \\ Therefore \: , \: either \: \: y=x \: \: or \: \\ \: \: \: \: \: \: \: \: \: \: \: x+y+xy=0 \\ \\ \\ Since \: , \: \: x=y \: \: doesn't \: satisfy \: the \\ given \: function \: , \: we \: reject \: it \\ \\ x + y + xy = 0 \\ \\ y \: \: = \: \: \frac{ - x}{1 + x} \\ \\ \frac{dy}{dx} \: \: = \: - \: \frac{(1 + x) - x.1}{ {(1 + x)}^{2} } \\ \\ \\ \frac{dy}{dx} = \frac{ - 1}{( {1 + x)}^{2} } \: \: \: \: \: \: \: \: \: Ans.$

6 0

2 years ago

Solve for x. Enter your answer in the box. x = °

Reika [66]

x = 109°

3*180° - 100° - 102° - 108° - 121° = 109°

6 0

3 years ago

Read 2 more answers

Each person plans to take 2 slices of pizza.mark concludes that he would get more pizza by taking 1 slice from each pizza.instea

aalyn [17]

Answer:

No if you only get two slices, it doesn't matter which pizza you take it from you still only get 2 slices.

Step-by-step explanation:

8 0

2 years ago

Please help with both! :)

Vaselesa [24]

Answer:

Step-by-step explanation:

for table the equation is

A) y = 2x + 5

for graph equation is

y = 1/2x -1

4 0

2 years ago

If C(x) is the total cost incurred in producing x units of a certain commodity, then the average cost of producing x units of th

nataly862011 [7]

Answer:

x = 2000 cameras

Step-by-step explanation:

C(x) Total cost in producing x units

C- = C(x) /x Average cost of producing x units x > 0

Cannon Precision Instrument

C (x) Total monthly cost for producing x units of M1 cameras

is C(x) = 0.0025x² + 80x + 10000

Then average cost of producing x cameras M1 is

C-(x) = ( 0.0025x² + 80x + 10000) /x

C-(x) = 0.0025x + 80 + 10000/x

Taking derivatives on both sides of the equation

C-´(x) = 0.0025 - 10000/x²

Then

C-´(x) = 0

( 0.0025x² - 10000 ) / x² = 0

0.0025x² - 10000 = 0

x² = 10000 /0.0025 x² = 4000000

x = 2000 cameras

8 0

3 years ago