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Alchen [17]
3 years ago
5

M + 5n = p solve for m

Mathematics
2 answers:
atroni [7]3 years ago
8 0
-5n on both sides
M= -5n + p
rusak2 [61]3 years ago
7 0
M+5n=p
m+5n +- 5n= -5n + p
5n + -5n = 0
m+0= -5n + p
m= -5n + p

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One question simple please help
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Answer:

y = \frac{1}{3}x - 9

Step-by-step explanation:

Standard equation of a line is y = mx + b, where m is the slope.

Given line y = - 3x + 78, slope, m₁ = -3

<em><u>To find the line perpendicular to the given line.</u></em>

The lines are perpendicular to each other if the product of their slopes = - 1

That is,

         m_1 \times m_2 = -1

So the slope of new line is

                                                   -3 \times m_2 = -1\\\\m_2 = \frac{-1}{-3} = \frac{1}{3}

Therefore , equation \ of \ new \ line \ \\\\(y - y_1) = m_2(x - x_1) , where \ (x_1 , y_1) = (9, -6 )\\\\(y - (-6))= \frac{1}{3}(x -9)\\\\y + 6 = \frac{1}{3} x - 3\\\\y = \frac{1}{3}x -3-6\\\\y = \frac{1}{3}x -9\\\\

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3 years ago
.Janie is analyzing a quadratic function f(x) and a linear function g(x). Will they intersect?
Alecsey [184]
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Find the value of the variables please please help me out I will mark you the brainly! Please please
RSB [31]

Answer:

m = 54

Step-by-step explanation:

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zhenek [66]

<u>Solving Q1: Given BE = ED and AE = EC.</u>

Considering triangles ΔAED and ΔCEB;

1. AE = EC (given)

2. ∠AED = ∠CEB (vertical angles)

3. ED = EB (given)

ΔAED ≡ ΔCEB (Side-Angle-Side congruency of triangles)

∠DAE = ∠BCE and AD = BC (using CPCTC)

⇒ AD║BC (Converse of Alternate Interior angles theorem)


Similarly consider triangles ΔABE and ΔCDE;

1. AE = CE (given)

2. ∠BEA = ∠DEC (vertical angles)

3. BE = DE (given)

ΔABE ≡ ΔCDE (Side-Angle-Side congruency of triangles)

∠ABE = ∠CDE and AB = CD (using CPCTC)

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Since we have AB║CD, AB = CD and AD║BC, AD = BC.

Therefore, quadrilateral ABCD is a parallelogram.

****************************************************************************************************

<u>Solving Q2: The parallelogram has the angle measures shown in the diagram.</u>

It is clearly visible that both the triangles are Isosceles triangles, so opposite sides in each triangle are equal.

Consider two triangles given in the problem, we have two sets of congruent angles and one included side is common in both triangles.

Using Angle-Side-Angle congruence of triangles, both the triangles would be congruent too.

Using CPCTC, we can say opposite sides of quadrilateral would be congruent.

Therefore, all sides in given quadrilateral are equal.

Hence, Given quadrilateral is a Rhombus.

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