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3 years ago

Please HELPP!!! Solve the system of equations. y= 2x y= x^2 - 15

1 answer:

7 0

$y=2x\\y=x^2-15\\\\x^2-15=2x\\x^2-2x-15=0\\x^2-5x+3x-15=0\\x(x-5)+3(x-5)=0\\(x+3)(x-5)=0\\x=-3 \vee x=5\\\\y=2\cdot(-3) \vee y=2\cdot5\\y=-6 \vee y=10\\\\(x,y)\in\{(-3,-6),(5,10)\}$

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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.

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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:

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Using Pythagoras rule to find j

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$j = \sqrt{99.45}$

j = 9.97

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

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

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Answer:

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