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Ilia_Sergeevich [38]
3 years ago
15

VT is the diagonal of a rhombus. What is the slope of the other diagonal?

Mathematics
1 answer:
noname [10]3 years ago
6 0

Answer:

C

Step-by-step explanation:

Remark

The slope of the given segment is found by using the 2 end points.

V(3,2) and T(5,7)

Slope of VT

<u>Givens</u>

  • y2 = 7
  • y1 = 2
  • x2 = 5
  • x1 = 3

<u>Formula</u>

m = (y2 - y1)/(x2 - x1)

<u>Solution</u>

m = (7 - 2)/(5 -3)

m = 5/2

Slope of other diagonal

m * m1 = - 1

5/2 * m1 = - 1

5m1 = -1*2 = -2

m1 = -2/5

C

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Inga [223]

First, we have to calculate the length of the sides of the pool, we are told that the scale factor of the backyard to the pool size equals 1/2, then we can find the length of the sides of the pool by multiplying the lengths of the sides of the backyard by 1/2, like this:

length of the pool = length of the yard * 1/2

width of the pool = width of the yard * 1/2

By replacing the 60 m for the length of the yard and 50 m for the width, we get:

length of the pool = 60 * 1/2 = 30

width of the pool = 50 * 1/2 = 25

Let's call x1 to the distance from the base of the pool to the bottom side of the yard and x2 to the distance from the top side of the pool to the top side of the yard, then we can formulate the following equation:

width of the yard = width of the pool + x1 + x2

Since we want the edges to be at an equal distance, x1 and x2 are the same, then we can rewrite them as x:

width of the yard = width of the pool + x + x

width of the yard = width of the pool + 2x

Replacing the known values:

60 = 30 + 2x

From this equation, we can solve for x to get:

60 - 30 = 30 - 30 + 2x

30 = 2x

30/2 = 2x/2

15 = x

x = 15

Now, let's call y1 to the distance from the right side of the corresponding side of the yard and y2 to the distance from the left side of the pool to the left side of the yard, with this, we can formulate the following equation:

length of the yard = length of the pool + y1 + y2

Since we want the edges to be at an equal distance, y1 and y2 are the same, then we can rewrite them as y:

length of the yard = length of the pool + y + y

length of the yard = length of the pool + 2y

Replacing the known values:

50 = 25 + 2y

50 - 25 = 25 - 25 + 2y

25 = 2y

25/2 = 2y/2

12.5 = y

y = 12.5

Now, we know that the pool must be at a distance of 15 m from the horizontal sides of the pool to the horizontal sides of the yard and that it must be at a distance of 12.5 m from the vertical sides of the pool to the vertical sides of the yards.

Here is a figure that depicts the results:

5 0
1 year ago
The following is a student’s work when solving for y. There is a mistake in the work. What is the mistake? What is the correct a
laila [671]

Answer:

Part 1

The mistake is Step 2: P + 2·x = 2·y

Part 2

The correct answer is

Step 2 correction: P - 2·x = 2·y

(P - 2·x)/2 = y

Step-by-step explanation:

Part 1

The student's steps are;

Step 1; P = 2·x + 2·y

Step 2: P + 2·x = 2·y

Step 3: P + 2·x/2 = y

The mistake in the work is in Step 2

The mistake is moving 2·x to the left hand side of the equation by adding 2·x to <em>P </em>to get; P + 2·x = 2·y

Part 2

To correct method to move 2·x to the left hand side of the equation, leaving only 2·y on the right hand side is to subtract 2·x from both sides of the equation as follows;

Step 2 correction: P - 2·x = 2·x + 2·y - 2·x = 2·x - 2·x + 2·y = 2·y

∴ P - 2·x = 2·y

(P - 2·x)/2 = y

y = (P - 2·x)/2

7 0
3 years ago
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nadya68 [22]

Answer:

The cost of the diamond last year is x = $ 830

Step-by-step explanation:

Let the cost of diamond last year = x

the cost of diamond present year = y

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⇒ y = 2 x + 10  ------ (1)

The sum of the cost (last year and this year) is  = $ 2500

⇒ x + y = $ 2500 -------- (2)

Put the value of y in equation (2) from equation (1), we get

⇒ 2 x + 10 + x = $ 2500

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⇒ y = $ 1670

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Answer:

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8 0
3 years ago
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