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vladimir1956 [14]
3 years ago
15

Point (7,2)is translated vertically 2 units and horizontally -5 units. Where is the new

Mathematics
1 answer:
Sonja [21]3 years ago
3 0

Answer:(2,4)

Step-by-step explanation:

you can make a grid to check

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Multiply both sides of the original equation by x, and find the solution set of the new equation.
prisoha [69]

Answer: (0, 1) Usually, we put the solutions in parenthesis.

Step-by-step explanation: x(x+1) = x(2) which = x²+ x = 2x = x² + x - 2x = 0

x² - x = 0. Now we divide by x which = x(x - 1) = 0 and the new solution set is x = 0 and x -1 = 0 which equals x = 1

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Prove the quadratic formula
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ax² + bx + c = 0

x = (-b ± √(b² - 4ac))/2a


First, rewrite the first equation so that the first coefficient is 1. Divide everything by a.

(ax² + bx + c = 0)/a =

x² + (b/a)x + (c/a) = 0

Isolate (c/a) by subtracting (c/a) from both sides

x² + (b/a)x + (c/a) (-(c/a) = 0 (- (c/a)

x² + (b/a)x = 0 - (c/a)

Add spaces

x² + (b/a)x          =  -c/a

Take 1/2 of the middle term's coefficient and square it. Remember that what you add to one side, you add to the other.

x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²

Simplify the left side of the equation.

x² + (b/a)x + (b/2a)² = (x + (b/2a))²

(x + b/2a))² = ((b²/4a²) - (4ac/4a²)) -> ((b² - 4ac)/(4a²))

Take the square root of both sides of the equation

√(x + b/2a))² = √((b²/4a²) - (4ac/4a²))

x + b/(2a) = (±√(b² - 4ac)/2a

Simplify. Isolate the x.

x = -(b/2a) ± (∛b² - 4ac)/2a = (-b ± √(b² - 4ac))/2a

~

3 0
3 years ago
then we can walk 1/3 miles in 5 minutes if Tammy walks a constant Pace how far does he walk in 1 minute ​
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3 years ago
Covert 3.92 into a mixed number in simplest from
Kobotan [32]
First, take three off it, because that will form the whole-number part:
3.92 = 3 + 0.92

Next, read off the place value of the last digit. The 2 is in the 'hundredths' column, which means that 0.92 = 92/100:
3.92 = 3 + 92/100

Finally, simplify 92/100 by dividing top and bottom by 4 to get 23/25. Then, shove it all together:
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Answer: B: copernicus....

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