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Nitella [24]
3 years ago
8

HURRY PLEASE SOLVE will give brainliest!!!​

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
4 0

2

x-9 = -5

2=x 2+x + 4-2=2 2-6+

-4

2

X

<h2>+2= 4 + 62 =66 +9 = 75</h2>
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Answer:

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3 years ago
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The general equation of a plane is Ax + By + Cz = D, where A, B, C, and D are real numbers and A is nonnegative. Find the equati
yawa3891 [41]
For the answer to the question above,
z=6-(3/4)y-2x 
<span>z+(3/4)y+2x=6 </span>

<span>Just by connecting the three points on the graph, I got this equation by isolating each plane to figure it out. This equation only explains the plane bounded by these three points, though, so to draw it just plot the points and connect them.</span>
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4 years ago
Find an equation of a line passing through the point (8,9) and parallel to the line joining the points (2,7) and (1,5).
TiliK225 [7]

Answer:

2x - y - 7 = 0

Step-by-step explanation:

Since the slope of parallel line are same.

So, we can easily use formula,

y - y₁ = m ( x ₋ x₁)

where, (x₁, y₁) = (8, 9)

and m is a slope of line passing through (x₁, y₁).

and since the slope of parallel lines are same, so here we use slope of parallel line for calculation.

and, Slope = m = \dfrac{y_{b}-y_{a}}{x_{b}-x_{a}}

here, (xₐ, yₐ) = (2, 7)

and, (y_{a},y_{b}) = (1, 5 )

⇒ m = \dfrac{5-7}{1-2}

⇒ m = 2

Putting all values above formula. We get,

y - 9 = 2 ( x ₋ 8)

⇒ y - 9 = 2x - 16

⇒ 2x - y - 7 = 0

which is required equation.

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3 years ago
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See the file attached and let me know whether you understand the explanation.
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Given: Quadrilateral PAST, TX = AX; TP || AS<br> Prove: Quadrilateral PAST is a parallelogram.
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1) Quadrilateral PAST, TX=AX, \overline{TP} \parallel\overline{AS} (given)

2) \angle XPT \cong \angle XSA and \angle XTP \cong \angle XAS (alternate interior angles theorem)

3) \triangle TXP \cong \triangle AXS (AAS)

4) \overline{TP} \cong \overline{AS} (CPCTC)

5) PAST is a parallelogram (a quadrilateral with two pairs of opposite congruent sides is a parallelogram)

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2 years ago
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