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vova2212 [387]
3 years ago
6

In Central City, Elm Street and Maple Street are parallel to one another. Oak Street crosses both Elm Street and Maple Street as

shown.Tell whether each statement is True or False.

Mathematics
2 answers:
lesya692 [45]3 years ago
8 0

Answer:

a. True.

b. False.

c. True.

d. True.

e. True.

Step-by-step explanation:

a. This statement is true because angles 6 and 8 are vertical angles. That means they are congruent.

b. This statement is false because Elm Street is a straight line, which means that angles 1 and 2 are supplementary angles and their measures will add up to be 180 degrees. 65 + 125 = 190, which is not equal to 180.

c. This statement is true because angles 5 and 1 correspond to each other.

d. This statement is true because angles 7 and 8 form a straight line.

e. This statement is true because they are interior angles that are alternate.

Hope this helps!

mestny [16]3 years ago
4 0

Answer:

a. True, because they are corresponding angles.

b. False, because angles 1 and 2 are supplementary, meaning that they add up to 180 degrees, but 125 + 65 = 190 degrees.

c. True, because they are corresponding angles.

d. True, becuase they are two angles that add up to 180 degrees.

e. True, because they are both interior angles on opposite sides of the transversal.

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3 years ago
PLEASE LOOK AT PHOTO! WHOEVER ANSWERS CORRECTLY I WILL MARK BRAINIEST!!
8_murik_8 [283]

Step-by-step explanation:

let m2 be x

then m1 = 4x

Since, m1+m2 = 115

therefore x + 4x = 115

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x = 23

so , m2 = 23°

3 0
2 years ago
Read 2 more answers
Refer to the figure to the right.
Law Incorporation [45]

Step-by-step explanation:

Use arc length:

s = rθ

where r is the radius and θ is the angle in radians.

a) First, convert the angle to radians (1 degree = 60 minutes).

77 ⁵⁰/₆₀° × (π radian / 180°) = 1.358 radians

s = rθ

s = (9.67 in) (1.358 rad)

s = 13.1 in

b) s = rθ

4 in = (9.67 in) θ

θ = 0.414 radians

θ = 23.7°

θ = 23° 42'

6 0
3 years ago
Solve the following equation. X cubed minus 6X squared plus 6X equals zero
anyanavicka [17]

We have to solve this equation:

x^3-6x^2+6x=0

Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:

\begin{gathered} x^3-6x^2+6x=0 \\ x(x^2-6x+6)=0 \end{gathered}

Now we can find the roots of the quadratic polynomial as:

\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{6\pm\sqrt[]{36-24}}{2} \\ x=\frac{6\pm\sqrt[]{12}}{2} \\ x=\frac{6\pm\sqrt[]{4\cdot3}}{2} \\ x=\frac{6\pm2\sqrt[]{3}}{2} \\ x=3\pm\sqrt[]{3} \\ x_1=3-\sqrt[]{3} \\ x_2=3+\sqrt[]{3} \end{gathered}

Then, the solutions to the equation are:

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4 0
1 year ago
What are the solutions of the system?
siniylev [52]
y=x^2-3x+2 \\y=4x-4

A. (1, 0) and (–6, –28) 

B. (1, 20) and (6, 0) 

C. (1, 0) and (6, 20) 

D. no solution


Correct answer is C.


y=x^2-3x+2 \\y=4x-4\\ \\4x-4=x^2-3x+2 \\x^2-7x+6=0 \\x^2-6x-x+6=0 \\x(x-6)-(x-6)=0 \\(x-6)(x-1)=0 \\x=6\,\text{ or }x=1 \\y=4\times 6-4=20 \,\text{ or }y=4\times 1-4=0 \\ \\(6,20)\,\text{ and }(1,0)

8 0
3 years ago
Read 2 more answers
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