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nignag [31]
3 years ago
5

The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) us

ing a point and the slope. Which point did Harold use? When Harold wrote his equation, the point he used was (7, 3). When Harold wrote his equation, the point he used was (0, 7). When Harold wrote his equation, the point he used was (7, 0). When Harold wrote his equation, the point he used was (3, 7).
Mathematics
2 answers:
xz_007 [3.2K]3 years ago
8 0

For this case we must find the point that Harold used to arrive at the following equation:

y = 3 (x-7)

Starting from the fact that the equation of the point-slope form of a line is given by:

(y-y_ {1}) = m (x-x_ {1})

If we compare the standard equation with Harold's, we see that the slope of the line is m = 3.

In addition, it is observed that x_ {1} = 7and y_ {1} = 0.

Then, the correct option is: Harold used the point (7,0)

ANswer:

When Harold wrote his equation, the point was used (7,0).

Kamila [148]3 years ago
8 0

Answer:

c

Step-by-step explanation:

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What is the exact perimeter of kite ? Show your work.
Afina-wow [57]

Answer:

<u>The perimeter of the kite is 29 units.</u>

Step-by-step explanation:

1. Let's use the Pythagorean Theorem to find the perimeter of the kite:

With the information given, we have four right triangles with a new point A, intersecting UW and VX. Those triangles are:

Δ UVA

Δ VWA

Δ UXA

Δ XWA

As we can see in the figure, length of UV and VW is the same and the length of UX and WX is also the same.

2. Let's find the value of UV and VW

UV is the hypotenuse of Δ UVA and it sides are 3 and 4 units length. So, we can calculate the length of UV this way:

Length of UV ² = UA ² + VA ²

Replacing with the real values:

Length of UV ² = 3 ² + 4 ²

Length of UV ² = 9 + 16

Length of UV ² = 25

√Length of UV ² = √25

<u>Length of UV  = 5 units ⇒ Length of VW = 5 units</u>

3. Let's find the value of UX and WX

UX is the hypotenuse of Δ UXA and it sides are 3 and 9 units length. So, we can calculate the length of UX this way:

Length of UX ² = UA ² + XA ²

Replacing with the real values:

Length of UX ² = 3 ² + 9 ²

Length of UX ² = 9 + 81

Length of UV ² = 90

√Length of UV ² = √90

<u>Length of UV  = 9.5 units (rounding to the nearest tenth) ⇒ Length of WX= 9.5 units</u>

<u>4. </u>Let's calculate the perimeter of the kite:

Perimeter of the kite = UV + VW + WX + UX

Perimeter of the kite = 5 + 5 + 9.5 + 9.5

<u>Perimeter of the kite = 29 units</u>

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