2x – 1+ 3x=1+ 2x - 5 I need the steps
1 answer:
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The hyperbola having both foci lying in the same quadrant is <u>(y - 16)²/9² - (x + 1)²/12² = 1</u>, making the <u>4th option</u> a right choice.
For the hyperbola with an equation of the form (x - h)²/a² - (y - k)²/b² = 1, the foci in on the points (h + c, k) and (h - c, k), where c = √(a² + b²).
For the hyperbola with an equation of the form (y - k)²/a² - (x - h)²/b² = 1, the foci in on the points (h, k + c) and (h, k - c), where c = √(a² + b²).
<u>Among the options</u>:
(a) (x - 24)²/24² - (y -1)²/7² = 1.
The equation is of the form (x - h)²/a² - (y - k)²/b².
h = 24, k = 1, a = 24, b = 7.
c = √(a² + b²) = √(24² + 7²) = √(576 + 49) = √625 = 25.
Thus, foci are at the points (h + c, k), (h - c, k) = (24 + 25, 1), (24 - 25, 1) = (49, 1), and (-1, 1), which are in different quadrants.
(b) (y - 12)²/5² - (x - 6)²/12² = 1.
The equation is of the form (y - k)²/a² - (x - h)²/b².
h = 6, k = 12, a = 5, b = 12.
c = √(a² + b²) = √(5² + 12²) = √(25 + 144) = √169 = 13.
Thus, foci are at the points (h, k + c), (h, k - c) = (6, 12 + 13), (6, 12 - 13) = (6, 25), and (6, -1), which are in different quadrants.
(c) (y - 16)²/15² - (x - 2)²/8² = 1.
The equation is of the form (y - k)²/a² - (x - h)²/b².
h = 2, k = 16, a = 15, b = 8.
c = √(a² + b²) = √(15² + 8²) = √(225 + 64) = √289 = 17.
Thus, foci are at the points (h, k + c), (h, k - c) = (2, 16 + 17), (2, 16 - 17) = (2, 33), and (2, -1), which are in different quadrants.
(d) (y - 16)²/9² - (x + 1)²/12² = 1.
The equation is of the form (y - k)²/a² - (x - h)²/b².
h = -1, k = 16, a = 9, b = 12.
c = √(a² + b²) = √(9² + 12²) = √(81 + 144) = √225 = 15.
Thus, foci are at the points (h, k + c), (h, k - c) = (-1, 16 + 15), (-1, 16 - 15) = (-1, 31), and (-1, 1), which are in the same quadrants (QUADRANT II).
Thus, the hyperbola having both foci lying in the same quadrant is <u>(y - 16)²/9² - (x + 1)²/12² = 1</u>, making the <u>4th option</u> a right choice.
Learn more about the foci of a hyperbola at
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For the options, refer the image.
Answer:
errr polynomials ahaha
so basically we can multiply the power outside of the parantheses with the exponents on the inside:
so
this will give us:
-3x^8y^4
we also unfortunately, have to use the power on the coefficient -3 as well and we cant just MULTIPLY -3 by 4, we are going to have to multiply -3 with -3 4 TIMES(anytime we see a coefficient we MUST multiply the coefficient by itself according to the number outside of the parantheses)
so:
-3*-3*-3*-3=81
final expression