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deff fn [24]
3 years ago
15

Solve by using substitution-7x-2y=-13   and   x-2y=11

Mathematics
1 answer:
netineya [11]3 years ago
7 0
\left \{ {{-7x-2y=-13} \atop {x-2y=11}} \right. \\\\ \left \{ {{-7x-2y=-13} \atop {x=11+2y}} \right.\\\\
Substitute\ second\ equation\ into\ the\ first\ one:\\\\
-7(11+2y)-2y=-13\\
-77-14y-2y=-13\\
-77-16y=-13\ \ \ |Add\ 77\\
-16y= 64\ \ \ |Divide\ by\ -16\\
y=-4\\\\x=11+2y=11+2*(-4)=11-8=3\\\\ \left \{ {{y=-4} \atop {x=3}} \right.
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Carla is saving money for a trip. She already has some money in her savings account and will add the same amount to her account
Bingel [31]

Answer:

280=8x + 130

Step-by-step explanation:

Im pretty sure this is correct it has been a while since I did this but y is the total amount so y is 280. M is the slope so it would be how often the amount changes so m is 8. x is unknown and what needs to be solved for. b is the y intercept which is just the initial amount of 130.

4 0
3 years ago
Two sixth-grade classes are collecting money to buy a present for one of their teachers. One class collected $30.25. About how m
zavuch27 [327]
If they only have 30.25, then they still need 39.50
6 0
3 years ago
All changes saved
slamgirl [31]
Your answer would be :A


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4 0
3 years ago
If p is the incenter of triangle jkl, find each measure
Hunter-Best [27]

Answer:

PO = 7

PM = 7

MJ = 11

∠PJO = 32°

∠KJL = 64°

PL ≈ 18.385

OL = 17

∠PLO = 22°

∠NLO = 44°

∠JKL = 72°

∠MKP = 36°

∠NKP = 36°

∠PKN = 36°

KN = 10

PL ≈ 13.04

PK ≈ 12.21

JL = 28

JK = 21

LK = 27

Step-by-step explanation:

The given parameters are;

The point representing the incenter of the triangle = P

Therefore PO = PM = PN = 7

tan(32°) = PM/JM = 7/JM

∴JM = 7/(tan(32°)) ≈ 11.2

∠PJO = tan⁻¹(7/11)≈ 32.47°

∠PJO = ∠PJM = 32° similar triangles

∠KJL = ∠KJP + ∠PJO = 32 + 32 = 64°

∠KJL ≈ 64°

PL = √(7² + 17²) ≈ 18.385

OL = NL = 17 similar triangles

∠PLO = sin⁻¹(7/18.385) ≈ 22.380°

∠PLO = ∠PLN = 22°

∠NLO = ∠PNL + ∠OLP ≈ 22° + 22° ≈ 44°

∠NLO ≈ 44.380°

∠JKL = 180 - (∠KJL + ∠NLO)

∠JKL = 180° - (64° + 44°) ≈ 72°

∠JKL  ≈ 72°

∠MKP = ∠NKP = 72°/2 = 36°

∠MKP = 36°

∠NKP = 36°

∠PKN = ∠JKL - ∠MKP = 72° - 36° ≈ 36°

∠PKN  ≈ 36°

KN = KM = 10

MJ = OJ = 11

PL = √(7² + 11²) ≈ 13.04

PK = √(7² + 10²) ≈ 12.21

JL = JO + OL = 11 + 17 = 28

JK = JM + MK = 11 + 10 = 21

LK = LN + NK = 17 + 10 = 27

6 0
2 years ago
One sixth of a right angle
QveST [7]
The answer will be fifteen degrees because one sixth is 1/6 and a right angle is 90 degrees so it will be 1/6 * 90/6 which will be 90 divided by 6 and equal fifteen .
7 0
3 years ago
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