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klasskru [66]
3 years ago
12

Question is in picture

Mathematics
1 answer:
Marta_Voda [28]3 years ago
7 0

|2x|\geq1\\\\2x\ge1 \ \ \text{or} \ \ 2x\leq-1\\\\\huge\boxed{x\geq\frac{1}{2} \ \ \text{or} \ \ x\leq-\frac{1}{2}}

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10 tenths in a whole.....
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Find the value of the following expression: (3^8 ⋅ 2^-5 ⋅ 90)−2 ⋅ ⋅ 3^28 (5 points) Write your answer in simplified form. Show a
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18452.8125 use PEMDAS
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3 years ago
Find the angle measures. Justify your responses.
MatroZZZ [7]

Answer:

m∠3 = 2x - 10° and m∠6 = 3x + 20

m∠3 = 58°          ;    m ∠6 = 122°

Step-by-step explanation:

a ║ b

m∠2 = m∠6                     If ║ cut by a transversal, then corresponding angles

                                         are ≅ and =

m∠7 = m∠3                   If ║ cut by a transversal, then corresponding angles

                                       are ≅ and =

m∠2 = 3x + 20              Given

m∠7 = 2x - 10                 Given    

m∠6 + m∠3 = 180°         If ║ cut by a transversal, then each pair of same-side

                                       interior angles (also called consecutive interior angles

                                       are supplementary (sum = 180°) .    

3x + 20 + 2x - 10 = 180°

5x + 10 = 180°

5x = 170°

x = 34°

Answer:   m∠3 = 2x - 10° and m∠6 = 3x + 20

                m∠3 = 2(34) - 10  ;    m∠6 = 3(34) + 20

                 m∠3 = 58°          ;    m ∠6 = 122°

                                 

5 0
2 years ago
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Answer

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Step-by-step explanation:

First we turn the equation into y = mx+b format then we we can plug in the value y= 4 and get 4 = -2x + 2

subtract two from both sides so 2= -2x divide btoh sides by 2 and get x = -1

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3 years ago
Line LJ is a diameter of circle K. If tangents to circle K are constructed through points L and J, what relationship would exist
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A tangent line to a circle is perpendicular to the radius drawn to the tangent point.

If tangents are constructed through points L and J, then<span> the tangents are perpendicular to the diameter and parallel to each other.</span>
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