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Fittoniya [83]
3 years ago
11

Point J is on line segment IK. Given JK = 2x, IJ = 5x, and IK = x + 6, determine the numerical length of JK.

Mathematics
1 answer:
Andre45 [30]3 years ago
8 0

Answer:

JK = 2

Step-by-step explanation:

Given:

JK = 2x

IJ = 5x

IK = x + 6

Required

Solve for JK.

Since J is on IK, we have:

IK = IJ + JK

x + 6 = 5x + 2x

x + 6 = 7x

Collect Like Terms

7x - x = 6

6x = 6

Solve for x

x = 6/6

x = 1

Substitute 1 for x in JK = 2x

JK = 2 * 1

JK = 2

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Find x if AB = 9, BC = 2x - 5,<br> and AC = x + 9
Oxana [17]

Answer:

x = 5

Step-by-step explanation:

Using the fact that line segments AB and BC are parts of the whole line segment AC, we can write the following equation:

AB + BC = AC

Now, using the given values, we can substitute in for the equation and solve for x:

AB + BC = AC

9 + 2x - 5 = x + 9

2x + 4 = x + 9

x = 5

Thus, we have found that for these sets of equations for these line segments, our value for x should be 5.

Cheers.

5 0
3 years ago
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98 x 24 = (100 - 2 ) x 24 = ?
Ray Of Light [21]

Answer:

2352

Step-by-step explanation:

(100 - 2) x 24 = 2400 - 48 = 2352

3 0
3 years ago
HELPP ASaPP I’ll mark you as brainlister
nika2105 [10]

Answer:

BC ≈ 8.9 units

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos36° = \frac{adjacent}{hypotenuse} = \frac{BC}{AB} = \frac{BC}{11} ( multiply both sides by 11 )

11 × cos36° = BC , then

BC ≈ 8.9 ( to the nearest tenth )

5 0
3 years ago
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Of the 250 passengers on the plane, 177 checked their bags rather than carry-on. What *percent* opted for carry-on?
Harrizon [31]

Answer:

29.2%

Step-by-step explanation:

250-177=73

250x0.292=73

0.292 is 29.2% in decimal form.

7 0
3 years ago
An equilateral triangle has a side length of 6. What is the height of the triangle?
vlabodo [156]
Since the height of an equilateral triangle in terms of its side s is s√3/2, the height of the triangle is 6√3/2 = 3√3 and so the area is (1/2)(6)(3√3) = 9√3. 

<span>If we draw a horizontal line a height of h from the base of the triangle, the region is split into two regions: the lower region consisting of a trapezoid of height h and the upper region consisting of a triangle of height 3√3 - h. </span>

<span>Since the upper triangle and the triangle itself are similar triangles, the base and height are proportional. If we let x denote the base of the length of the upper triangle, we have: </span>

<span>(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle) </span>
<span>==> x/6 = (3√3 - h)/(3√3) </span>
<span>==> x = (6√3 - 2h)/√3 </span>

<span>Thus, the area of the upper triangle is: </span>

<span>A = (1/2)[(6√3 - 2h)/√3](3√3 - h) = [(6√3 - 2h)(3√3 - h)]/(2√3). </span>
<span>(Made a dumb mistake about the height here for some reason) </span>

<span>Since we require that the area of this triangle is to be half of the total area (9√3/2), we need to solve: </span>

<span>[(6√3 - 2h)(3√3 - h)]/(2√3) = 9√3/2 </span>
<span>==> (6√3 - 2h)(3√3 - h) = 27 </span>
<span>==> 54 - 6h√3 - 6h√3 + 2h^2 = 27 </span>
<span>==> 2h^2 - 12h√3 + 27 = 0. </span>

<span>Solving with the Quadratic Formula gives: </span>

<span>h = (6√3 + 3√6)/2 ≈ 8.87 units and h = (6√3 - 3√6)/2 ≈ 1.52 units. </span>

<span>Since h = (6√3 + 3√6)/2 would place the line outside of the triangle, we pick h = (6√3 - 3√6)/2. </span>

<span>Therefore, the line should be ==> (6√3 - 3√6)/2 units from the base. </span>

<span>I hope this helps! ^^ Brainliest Please?</span><span>
</span>
5 0
3 years ago
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