In ΔJKL, k = 430 cm, l = 590 cm and ∠J=117°. Find the length of j, to the nearest centimeter.

In ΔJKL, k = 430 cm, l = 590 cm and ∠J=117°. Find the length of j, to the nearest centimeter.

Step-by-step explanation:

\text{S.A.S.}\rightarrow \text{Law of Cosines}

S.A.S.→Law of Cosines

a^2=b^2+c^2-2bc\cos A

a

2

=b

2

+c

2

−2bccosA

From reference sheet.

j^2 = 430^2+590^2-2(430)(590)\cos 117

j

2

=430

2

+590

2

−2(430)(590)cos117

Plug in values.

j^2 = 184900+348100-2(430)(590)(-0.45399)

j

2

=184900+348100−2(430)(590)(−0.45399)

Square and find cosine.

j^2 = 184900+348100+230354.77957

j

2

=184900+348100+230354.77957

Multiply.

j^2 = 763354.77957

j

2

=763354.77957

Add.

j=\sqrt{763354.77957} \approx873.702 \approx874

j=

763354.77957

​

≈873.702≈874

Square root and round.

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

the answer is (-4,-4)

hope this ! : d