Answer:
x=5, y=4. (5, 4).
Step-by-step explanation:
2x-3y=-2
4x+y=24
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-2(2x-3y)=-2(-2)
4x+y=24
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-4x+6y=4
4x+y=24
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7y=28
y=28/7
y=4
4x+4=24
4x=24-4
4x=20
x=20/4
x=5
Answer:
The method of separation is chromatography.
Step-by-step explanation:
It is given that a drawing of a circle made in the center of a piece of paper with a black marker gets wet, the marker bleeds, and as the water moves through the paper the ink separates into several colors. Here, the method of separation used is chromatography. It is a laboratory technique for the separation of a mixture. When water or dye passes through a medium and changes color, then that means chromatography is used there.
Answer:
seven and twenty three hundredths
Step-by-step explanation:
i hate doing this dont you?
Knowing that the area of a square is equivalent the square of its side length, the length of one side can be calculated by taking the square root of the area. The square root of the expression 9q^4r^8s^8 units is equivalent to <span>3q2r4|s3| units. When taking the square root of variables, the power they are raised to is divided by 2.</span>
Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0