All categories

2 years ago

If JL = 8x-3 and LK = 4x + 57, Find JK

Mathematics

2 answers:

lisabon 2012 [21]2 years ago

6 0

JK =8x-3 +4X+57
JK = 12x+ 54

telo118 [61]2 years ago

4 0

Answer: JK = 234

Concept:

Here, we need to know the idea of the segment addition postulate.

The Segment Addition Postulate states that given 2 points A and C, a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation AB + BC = AC.

Solve:

<h2>PART I: Find the value of x</h2>

Given information

JL = 8x - 3

LK = 4x + 57

Given expression deducted from the given figure

JL = LK

Substitute values into the expression

8x - 3 = 4x + 57

Add 3 on both sides

8x - 3 + 3 = 4x + 57 + 3

8x = 4x + 60

Subtract 4x on both sides

8x - 4x = 4x + 60 - 4x

4x = 60

Divide 4 on both sides

4x / 4 = 60 / 4

x = 15

-------------------------------------------------------------------------------

<h2>PART II: Find the value of JK</h2>

Given information

JL = 8x - 3

LK = 4x + 57

x = 15

Given expression deducted from the segment addition postulate

JK = JL + LK

Substitute values into the expression

JK = 8x - 3 + 4x + 57

JK = 8(15) - 3 + 4(15) + 57

Simplify by multiplication

JK = 120 - 3 + 60 + 57

Simplify by simple operation (addition/subtraction)

JK = 117 + 60 + 57

JK = 177 + 57

$\boxed{JK=234}$

Hope this helps!! :)

Please let me know if you have any questions

You might be interested in

Wilson made punch from a recipe that called for 1 and 1/2 cups of orange juice and 3/4 Cup lemonade. Wilson used 2 and 1/4 cups

ryzh [129]

If you don't understand I am really sorry

3 0

3 years ago

In triangle ABC, ∠ABC=70° and ∠ACB=50°. Points M and N lie on sides AB and AC respectively such that ∠MCB=40° and ∠NBC=50°. Find

Sever21 [200]

Answer:

∠NMC = 50°

Step-by-step explanation:

The interpretation of the information given in the question can be seen in the attached images below.

In ΔABC;

∠ A + ∠ B + ∠ C = 180° (sum of angles in a triangle)

∠ A + 70° + 50° = 180°

∠ A = 180° - 70° - 50°

∠ A = 180° - 120°

∠ A = 60°

In ΔAMN ; the base angle are equal , let the base angles be x and y

So; x = y (base angle of an equilateral triangle)

Then;

x + x + 60° = 180°

2x + 60° = 180°

2x = 180° - 60°

2x = 120°

x = 120°/2

x = 60°

∴ x = 60° , y = 60°

In ΔBQC

∠a + ∠e + ∠b = 180°

50° + ∠e + 40° = 180°

∠e = 180° - 50° - 40°

∠e = 180° - 90°

∠e = 90°

At point Q , ∠e = ∠f = ∠g = ∠h = 90° (angles at a point)

∠i = 50° - 40° = 10°

In ΔNQC

∠f + ∠i + ∠j = 180°

90° + 10° + ∠j = 180°

∠j = 180° - 90°-10°

∠j = 180° - 100°

∠j = 80°

From line AC , at point N , ∠y + ∠c + ∠j = 180° (sum of angles on a straight line)

60° + ∠c + ∠80° = 180°

∠c = 180° - 60°-80°

∠c = 180° - 140°

∠c = 40°

Recall that :

At point Q , ∠e = ∠f = ∠g = ∠h = 90° (angles at a point)

Then In Δ NMC ;

∠d + ∠h + ∠c = 180° (sum of angles in a triangle)

∠d + 90° + 40° = 180°

∠d = 180° - 90° -40°

∠d = 180° - 130°

∠d = 50°

Therefore, ∠NMC = ∠d = 50°

3 0

3 years ago

Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\$

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

$Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}$

put this value into equation 1

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}$

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

$Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\$

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

$length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm$