Please say brainliest.
The answer is yes.
Plug 4 in for the value of n so you have 4+8 is less than it equal to 13. And since 12 is less than 13 it is true, and a solution to the inequality.
Answer:
s = 2x²+3x-23
Step-by-step explanation:
Given that,
First side of a triangle = 3x+8
Second side of a triangle = 2x²-4x+5
The perimeter of the triangle = 4x²+2x-10
We need to find the expression for the third side. Let the third side b s.
Perimeter = Sum of all sides
4x²+2x-10 = 3x+8 + 2x²-4x+5 +s
4x²+2x-10 = 2x²-x+13+s
s = 4x²+2x-10- (2x²-x+13)
= 2x²+3x-23
So, the third side will be 2x²+3x-23.
Answer:
1
- Two over three
2
Step-by-step explanation:
Answer:
∠JKL = 38°
Step-by-step explanation:
PQRS, JQK and LRK are straight lines
Let's take the straight lines in the diagrams one after the other to find what they consist.
The related diagram can be found at brainly (question ID: 18713345)
Find attached the diagram used for solving the question.
For straight line PQRS,
2x°+y°+x°+2y° = 180°
(Sum of angles on a Straight line = 180°)
Collect like terms
3x° + 3y° = 180°
Also straight line PQRS = straight line PQR + straight line SRQ
For straight line PQR,
2y + x + ∠RQM = 180° ....equation 1
For straight line SRQ,
2x + y + ∠MRQ = 180° ....equation 2
Straight line PQRS = addition of equation 1 and 2
By collecting like times
3x +3y + ∠RQM + ∠MRQ = 360°....equation 3
Given ∠QMR = 33°
∠RQM + ∠MRQ + ∠QMR = 180° (sum of angles in a triangle)
∠RQM + ∠MRQ + 33° = 180°
∠RQM + ∠MRQ = 180-33
∠RQM + ∠MRQ = 147° ...equation 4
Insert equation 4 in 3
3x° +3y° + 147° = 360°
3x +3y = 360 - 147
3x +3y = 213
3(x+y) = 3(71)
x+y = 71°
∠JQP = ∠RQK = 2y° (vertical angles are equal)
∠LRS = ∠QRK = 2x° (vertical angles are equal)
∠QRK + ∠RQK + ∠QKR = 180° (sum of angles in a triangle)
2x+2y + ∠QKR = 180
2(x+y) + ∠QKR = 180
2(71) + ∠QKR = 180
142 + ∠QKR = 180
∠QKR = 180 - 142
∠QKR = 38°
∠JKL = ∠QKR = 38°
Answer:
142
Step-by-step explanation:
50:284*100 =
(50*100):284 =
5000:284 = 142
Definition = the word OF means to multiply
Hope this helps!
- <em>Tobie the dog</em>