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mihalych1998 [28]
4 years ago
14

PLEASE HELP WILL GIVE BRAINLIEST ANSWER

Mathematics
1 answer:
zavuch27 [327]4 years ago
6 0
Corresponding angles are congruent and appear on the same side of the transversal (the dividing line). Out of the answer choices, letter A (angles 1 and 5) fit that description. Hope this helped
You might be interested in
(+1) + (-5) - (+5) + (+9)
olchik [2.2K]

Answer:

The answer is 0

Hope that helps!

5 0
2 years ago
Read 2 more answers
Cual es el resultado?
Korolek [52]
I hope this helps you


x^2/3 [x^2/3.x^-1/4)^6.1/3.2


x^2/3[x^8-3/12]^4


x^2/3 [x^5/12]^4


x^2/3.x^5/12.4


x^2/3.x^5/3


x^2+5/3


x^7/3
5 0
3 years ago
What is the area of this figure?
abruzzese [7]

Answer:

204.27 inches squared

Step-by-step explanation:

The radius is 6 meaning that the area of the quarter-circle is 9pi or 28.274

the rectangle is 176.

3 0
3 years ago
Which of the following is the solution of One-fourth x + 5 = 7? x = 2 x = 6 x = 8 x = 12
salantis [7]
Changes made to your input should not affect the solution:

(1): "x1" was replaced by "x^1". 3 more similar replacement(s).

Step by step solution :

Step 1 :

Equation at the end of step 1 :

x + ((((3•19x2) • x6) • x8) • x12)
Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

57x28 + x = x • (57x27 + 1)

Trying to factor as a Sum of Cubes :

3.2 Factoring: 57x27 + 1

Theory : A sum of two perfect cubes, a3 + b3 can be factored into :
(a+b) • (a2-ab+b2)
Proof : (a+b) • (a2-ab+b2) =
a3-a2b+ab2+ba2-b2a+b3 =
a3+(a2b-ba2)+(ab2-b2a)+b3=
a3+0+0+b3=
a3+b3

Check : 57 is not a cube !!


Final result :

x • (57x27 + 1)

Processing ends successfully
5 0
3 years ago
Find the measure of angle KMJ
Semmy [17]

Answer:

54 Degrees

Step-by-step explanation:

We know that the interior angles of a triangle add up to 180 degrees.

we also know that angle KLM = 48 degrees

and angle JKM = 78 degrees

since these triangles are on parallel lines we know that their angles must be equal (just inverted)

and thus:

180 - (48+78) = 54

5 0
3 years ago
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