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3 years ago

Hi pls answer I'll mark you brainliest

Mathematics

2 answers:

OLEGan [10]3 years ago

7 0

A) y= -2, 3, 7

B) y= 8, 0, -6

C) y= -8, 1, 7

d) y= 1, -2, -6

AVprozaik [17]3 years ago

3 0

A.) y = -2, 3, 7
B.) y = 8, 0, -6
C.) y = -8, 1, 7
D.) y = 1, -2, -6
E.) y = 1, 2, 3, 4, 5

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In ΔJKL, k = 9.6 cm, l = 2.7 cm and ∠J=43°. Find ∠L, to the nearest 10th of a degree.

Bogdan [553]

Answer:

L = 10.64°

Step-by-step explanation:

From the given information:

In triangle JKL;

line k = 9.6 cm

line l = 2.7 cm; &

angle J = 43°

we are to find angle L = ???

We can use the sine rule to determine angle L:

i.e

$\dfrac{j}{SIn \ J} = \dfrac{l}{ SIn \ L}$

Using Pythagoras rule to find j

i,e

j² = k² + l²

j² = 9.6²+ 2.7²

j² = 92.16 + 7.29

j² = 99.45

$j = \sqrt{99.45}$

j = 9.97

∴

$\dfrac{9.97}{Sin \ 43} = \dfrac{2.7}{ Sin \ L}$

${9.97 \times Sin (L ) = (2.7 \times Sin \ 43)$

$= Sin \ L = \dfrac{ (2.7 \times Sin \ 43)}{9.97 } \\ \\ = Sin \ L = \dfrac{ (2.7 \times 0.6819)}{9.97 } \\ \\ = Sin \ L = 0.18466 \\ \\ L = Sin^{-1} (0.18466) \\ \\ L = 10.64 ^0$

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3 years ago

Hi pls answer I'll mark you brainliest​

Hi pls answer I'll mark you brainliest