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kari74 [83]
3 years ago
11

HELP! WILL GIVE BRAINLIEST

Mathematics
1 answer:
likoan [24]3 years ago
3 0

Answer: area 24 perimeter/circumstance 32 meters

Step-by-step explanation:

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How do you find the area of a right triangle if you only have the base
valentina_108 [34]

Step-by-step explanation:

You can find the area of a right triangle the same as you would any other triangle by using the following formula:

A = (1/2)bh, where A is the area of the triangle, b is the length of the base and h is the height of the triangle; However, with a right triangle, it's much more convenient in finding its area if we utilize the lengths of the two legs (the two sides that are shorter than the longest side, the hypotenuse and that are perpendicular to each other and thus form the right angle of the right triangle), that is, since the two legs of a right triangle are perpendicular to each other, when we treat one leg as the base, then, consequently, we can automatically treat the length of the other leg as the height, and if we initially know the lengths of both legs, then we can then just plug this information directly into the area formula for a triangle to find the area A of the right triangle.

For example: Find the area of a right triangle whose legs have lengths of 3 in. and 4 in.

Make the 4 in. leg the base. Since the two legs of a right triangle are perpendicular to each other, then the length of the other leg is automatically the height of the triangle; therefore, plugging this information into the formula for the area of a triangle, we have:

A = (1/2)bh

= (1/2)(4 in.)(3 in.)

= (1/2)(12 in.²)

A = 6 in.² (note: in.² means square inches)

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3 years ago
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USPshnik [31]
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a master roof can cover a garage in 1 hour less than her new assitant. if they work together, they can complete the job in 7.75
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Please help please please ASAP please please
Lelu [443]

Step-by-step explanation:

ac is the answer

.............

4 0
3 years ago
Help. please read the instructions and Answer the questions​
Dominik [7]

9514 1404 393

Explanation:

We refer to the equations as [1] and [2]. We refer to the items as (1) – (4).

__

1. The terms to be eliminated have matching coefficients in (1) and (2). They can be eliminated by subtracting one equation from the other.

In (3) and (4), putting the equations in standard form* results in terms with opposite coefficients. Those terms can be eliminated by adding the equations.

__

2. Terms to be eliminated will have matching or opposite coefficients.

__

3. In (1) and (2), the variable x can be eliminated by subtracting one equation from the other. In the attachment, we have indicated the subtraction that will result in the remaining variable having a positive coefficient.

__

4. In (3) and (4), the coefficients of the variables are not equal or opposite in the two equations, so no variable can be eliminated directly.

__

5. As suggested by the answer to Q4, an equivalent equation must be found that has an equal or opposite variable coefficient with respect to the other equation. The new equations are ...

  (3) [2] ⇒ x -y = 2

  (4) [1] ⇒ 2x +2y = 3

_____

Here are the solutions:

(1) [1] -[2]  ⇒  (x +y) -(x -y) = (-1) -(3)

  2y = -4  ⇒  y = -2

  x = y +3 = 1 . . . . from [2]

  (x, y) = (1, -2)

__

(2) [2] -[1]  ⇒  (x +2y) -(x +y) = (8) -(5)

  y = 3

  x = 5 -y = 2 . . . . from [1]

  (x, y) = (2, 3)

__

(3) [1] +[2]/2  ⇒  (x +y) +(x -y) = (1) +(2)

  2x = 3  ⇒  x = 3/2

  y = 1 -x = -1/2 . . . . from [1]

  (x, y) = (3/2, -1/2)

__

(4) [2] +[1]/2  ⇒  (5x -2y) +(2x +2y) = (4) +(3)

  7x = 7  ⇒  x = 1

  y = (3 -2x)/2 = 1/2

  (x, y) = (1, 1/2)

_____

* Equations in standard form have mutually prime coefficients. In (3) a factor of 2 can be removed from equation [2]. In (4), a factor of 2 can be removed from equation [1].

7 0
3 years ago
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