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andreev551 [17]
3 years ago
5

Write the point-slope form of the line that passes through (5, 5) and is PARALLEL to a line with a slope of 1/4. Include all of

your work in your final answer.
Write the point-slope form of the line that passes through (5, 5) and is PERPENDICULAR to a line with a slope of 1/4.

PLEASE HELP!
Mathematics
1 answer:
lesya692 [45]3 years ago
4 0
Parallel lines will have the same slope

y - y1 = m(x - 1)
slope(m) = 1/4
(5,5)...x1 = 5 and y1 = 5
now we sub
y - 5 = 1/4(x - 5) <==== ur parallel line
==================
perpendiculr lines will have a negative reciprocal slope. To find the negative reciprocal, u flip the number and change the sign. So the negative reciprocal of 1/4 is -4....see how I flipped 1/4 and made it 4/1...then changed the sign making it -4/1 or just -4. That will be ur slope of the perpendicular line.

y - y1 = m(x - x1)
slope(m) = -4
(5,5)...x1 = 5 and y1 = 5
now we sub
y - 5 = -4(x - 5) <=== ur perpendicular line
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xxTIMURxx [149]

Let's see

\\ \rm\rightarrowtail y=6x^2-25x+11

Graph attached

  • vertex(-2,-13)

Axis of symmetry

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7 0
2 years ago
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What’s the answer to this question?
denis23 [38]
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8 0
3 years ago
Round 82736 to 1 significant figure
hram777 [196]

Answer:

80000

Step-by-step explanation:

Significant figures are any number but zeroes.

However, in some cases, zeroes also count as significant figures.

  • Zeroes are significant figures when between other non-zero numbers.
  • Zeroes are significant when you see a dot on the right indicating that zero is not a place holder.
  • They are <u>not</u> significant when serving as place holders, such as zeroes on the right side of a number. For example, .004 has 1 sig fig. The 2 zeroes on the left aren't sig figs because they're just place holders.
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5 0
3 years ago
How do I solve this? How do you get this answer?I need this by tomorrow someone please help!!
rusak2 [61]
M(x) = x² - 4x + 3
m(x) = x² - x - 3x + 3
m(x) = x(x - 1) - 3(x - 1)
m(x) = (x - 3)(x - 1)

(x - 3)(x - 1) = 0
x - 3 = 0 ∧ x - 1 = 0
<u>x = 3 ∧ x = 1</u>

or 

m(x) = x² - 4x + 3
Δ = (-4)² - 4 · 1 · 3
Δ = 16 - 12 = 4
√Δ = √4
√Δ = 2

x = (4 - 2)/2 ∧ x = (4 + 2)/2
x = 2/2 ∧ x = 6/2
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topjm [15]

Answer:

(y+7)²/364 - (x+8)²/364 = 1

Step-by-step explanation:

This is the standard form of a vertical hyperbola (y-k)²/a² - (x-h)²/b² = 1

so, (y²+14y+?) - (x²+16x+?) = 251

(y²+14+49)-49 - (x²+16x+64)-64 = 251

(y+7)² - (x+8)² = 251 + 49 + 64

(y+7)² - (x+8)² = 364

(y+7)²/364 - (x+8)²/364 = 1

7 0
4 years ago
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