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tino4ka555 [31]
3 years ago
14

In ΔJKL, l = 380 cm, j = 620 cm and ∠K=30°. Find the area of ΔJKL, to the nearest square centimeter.

Mathematics
1 answer:
irina1246 [14]3 years ago
3 0

Answer:

58,900cm²

Step-by-step explanation:

Area of the triangle = 1/2absin thets

Area of the triangle  = 1/2* i * j sin 30

Area of the triangle = 1/2 * 380 * 620 sin 30

Area of the triangle = 380 * 310 * 0.5

Area of the triangle = 380*155

Area of the triangle  = 58,900cm²

hence the area of the triangle is 58,900cm²

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Answer:

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Step-by-step explanation:

1) all straight lines sum to 180°

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the other angle for b is 25°, while the other angle for m is 155°

so we can see that the angles for both lines are the same, hence they are parallel.

2) ∠8 should be the same as ∠6, ∠10 should be the same as ∠3, ∠7 same as ∠5 and ∠9 same as ∠4

in the options we are only given '∠8 should be the same as ∠6' as the correct answer, so we take that.

3) from the image we can see that both horizontal lines are parallel to each other, so both angles on the lines should be same, so ∠CET would be (2x-16)°

(2x-16)°+(7x+20)°=180°

we get x=20(nearest whole number)

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4) since we need to show that they are parallel,

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we then plug the x value into the two equations, in which we get 150° for both the angles [2(60)+30=4(60)-90] ⇒ (150=150)

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Step-by-step explanation:

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Find y when x=77, if y varies inversly as x, and y = 23 when x = 115.
Eduardwww [97]

The required value of y is 2645 / 77 when x = 77.

Step-by-step explanation:

1. Let's check the information given to answer:

if y varies inversely as x, and y = 23 when x = 115. Find y when x=77

2. What is y when x = 77?

As the statement "y varies inversely as x" translates into y = k/x

If we let y = 23 and x = 115, the constant of variation becomes

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        k = 23 × 115

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Thus the specification equation is y = 2645  ÷ x . Now, letting x = 77, we obtain

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