All of the following led to Texas independence EXCEPT: a. San Jacinto c. The Alamo b. Goliad d. The Webster-Ashburton Treaty of 1842

All of the following led to Texas independence EXCEPT: The Webster-Ashburton Treaty of 1842.

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Asked 3/16/2016 11:11:02 AM

Updated 11/17/2018 6:30:43 PM

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Confirmed by jerry06 [11/17/2018 6:30:43 PM]

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Write the following equation in slope-intercept form. -x+5y=3

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Updated 3/12/2016 4:47:42 AM

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Under the Missouri Compromise, what state entered the union as a free state?

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Updated 3/11/2016 7:45:52 AM

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Who failed to become president despite winning most of the popular vote in 1824? **Weegy:** Andrew Jackson failed to become president despite winning most of the popular vote in 1824.
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Updated 12/23/2020 3:57:02 PM

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Write an equation for the line that passes through the points (6,-4) and (-2,-9)

Question|Asked by Finesse_King

Updated 331 days ago|5/26/2023 12:08:57 AM

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y - y1 = m(x - x1),

m = (y2 - y1) / (x2 - x1),

m = (-9 - (-4)) / (-2 - 6)

m = (-9 + 4) / (-2 - 6)

m = -5 / -8

m = 5/8

y - (-4) = (5/8)(x - 6)

y + 4 = (5/8)(x - 6)

Y + 4 = (5/8)x - 15/4

y = (5/8)x - 15/4 - 4

y = (5/8)x - 15/4 - 16/4

Simplifying the expression:

y = (5/8)x - 31/4

Therefore, the equation for the line passing through the points (6, -4) and (-2, -9) is y = (5/8)x - 31/4.

m = (y2 - y1) / (x2 - x1),

m = (-9 - (-4)) / (-2 - 6)

m = (-9 + 4) / (-2 - 6)

m = -5 / -8

m = 5/8

y - (-4) = (5/8)(x - 6)

y + 4 = (5/8)(x - 6)

Y + 4 = (5/8)x - 15/4

y = (5/8)x - 15/4 - 4

y = (5/8)x - 15/4 - 16/4

Simplifying the expression:

y = (5/8)x - 31/4

Therefore, the equation for the line passing through the points (6, -4) and (-2, -9) is y = (5/8)x - 31/4.

Added 331 days ago|5/26/2023 12:08:57 AM

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Write an equation for the line that goes through the point (3,-5) and is parallel to y=4x+6 **Weegy:** 2(4x + 2) = 8x + 4 (More)

Question|Asked by Finesse_King

Updated 331 days ago|5/26/2023 12:10:01 AM

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Using the point-slope form, we substitute the slope (m = 4) and the coordinates (x1 = 3, y1 = -5) into the equation:

y - y1 = m(x - x1)

y - (-5) = 4(x - 3)

y + 5 = 4(x - 3)

y + 5 = 4x - 12

y = 4x - 12 - 5

y = 4x - 17

Therefore, the equation for the line that is parallel to y = 4x + 6 and passes through the point (3, -5) is y = 4x - 17.

y - y1 = m(x - x1)

y - (-5) = 4(x - 3)

y + 5 = 4(x - 3)

y + 5 = 4x - 12

y = 4x - 12 - 5

y = 4x - 17

Therefore, the equation for the line that is parallel to y = 4x + 6 and passes through the point (3, -5) is y = 4x - 17.

Added 331 days ago|5/26/2023 12:09:55 AM

This answer has been confirmed as correct and helpful.

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