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Elanso [62]
3 years ago
12

PLEASE HELP,, MARKING BRAINLIEST!!! Solve for the value of x.

Mathematics
1 answer:
Eddi Din [679]3 years ago
5 0

Answer:

37

Step-by-step explanation:

all triangles have a total of 180 degrees

set up equation and solve

180=(2x+4)+(2x-9)+x

180=5x-5

185=5x

37=x

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Solve 7 8 and 9 please
VLD [36.1K]
7) y = 4x
8) y = 7x/8
9) y = 25x/3

I think this are the answers.
7 0
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Qual o valor de 2 3 -1 3 ?
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El valor de 23-13 es 10
7 0
3 years ago
BRAINLIEST ASAP! PLEASE HELP ME :)
nikdorinn [45]

Answer:

m <1 = 61

Step-by-step explanation:

We know that the angle adjacent to 130 is 50 degrees because linear angles are complementary.

Extending line t to make a triangle, we can find the angle adjacent to the angle labeled 111  ( and above it) is 69.

That makes the angle that touches line q 61 to finish the triangle

This angle and angle 1 are corresponding angles, so they are equal

<1 = 61

7 0
3 years ago
Write the equation of a line in slope intercept form that is parallel to the line y= 1/3x + 5 and passes through (-9, 5)
vladimir2022 [97]
Find the slope of line 1
The equation of line 1 is y = (1/3)x + 5
m = 1/3

Find the slope of line 2
Line 2 is parallel to line 1. Parallel lines have the same number of slope. So the slope of line 2 is 1/3

Find the slope-intercept form of equation, with m = 1/3 and (-9,5)
General formula
y - y₁ = m(x - x₁)

Input the number to the formula
y - y₁ = m(x - x₁)
y - 5 = 1/3(x - (-9))
y - 5 = 1/3 (x + 9)
y - 5 = (1/3)x + 3
y = (1/3)x + 3 + 5
y = (1/3)x + 8

The equation is y = (1/3)x + 8
7 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
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