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aliina [53]
3 years ago
7

in an isosceles triangle there are two sided, called legs, that are the same length. the 3rd side is the base. if an isosceles t

riangle has a perimeter 345 cm and the base length 85 cm. what is the length of each leg
Mathematics
1 answer:
N76 [4]3 years ago
4 0
130. is the answer.....
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The area of a field shaped like a right triangle is 750yd^2.One leg of the field is fenced with wood costing $5/yd. the remainde
Elenna [48]
Assume that
 a and b = the two legs of the right triangle. 
 c =  the hypotenuse.

The area of the right triangle is 750 yd², therefore
(1/2)*a*b = 750
ab = 1500               (1)

The perimeter is 150yd, therefore
a + b + c = 150         (2)

Let the side fenced with wood be a, at $5/yd.  Sides b and c are fenced with steel at $10/y. The total cost is $1200, therefore
 5a + 10b + 10c = 1200            
or
a + 2b + 2c = 240         (3)

From (2), obtain 
c = 150 - a - b              (4)

Substitute (4) into (3)
a + 2b + 2(150 - a - b) = 240
-a + 300 = 240
a = 60

From (1), obtain
60b = 1500
b = 25

From (4), obtain
c = 150 - 60 - 25 = 65

Answer:
A. The length of the leg fenced with wood is 60 yd.
B.  The length of the leg fenced with steel is 25 yd.


5 0
4 years ago
Write the point-slope form of the equation of the line through the given point with the given
Marina86 [1]

Question 3)

Given

The point (1, -5)

The slope m = -5/6

Using the point-slope form of the equation of a line

y-y_1=m\left(x-x_1\right)

where

  • m is the slope of the line
  • (x₁, y₁) is the point

In our case:

  • m = -5/6
  • (x₁, y₁) = (1, -5)

substituting the values m = -5/6 and the point (1, -5) in the point-slope form of the equation of the line

y-y_1=m\left(x-x_1\right)

y-\left(-5\right)=-\frac{5}{6}\left(x-1\right)

y+5=-\frac{5}{6}\left(x-1\right)

Thus, the point-slope form of the equation of the line is:

y+5=-\frac{5}{6}\left(x-1\right)

Question 4)

Given

The point (-1, 5)

The slope m = -7/2

In our case:

  • m = -7/2
  • (x₁, y₁) = (-1, 5)

substituting the values m = -7/2 and the point (-1, 5) in the point-slope form of the equation of the line

y-y_1=m\left(x-x_1\right)

y-5=-\frac{7}{2}\left(x-\left(-1\right)\right)

y-5=-\frac{7}{2}\left(x+1\right)

Thus, the point-slope form of the equation of the line is:

y-5=-\frac{7}{2}\left(x+1\right)

7 0
3 years ago
Graph y = 2x - 5 using the
borishaifa [10]

O The line increases from the left to right

O The line crosses the x-axis at 2.5

O The line does not enter quadrant II at all...

...Are your answers

7 0
3 years ago
The following argument claims to prove that the requirement that an equivalence relation be reflexive is redundant. In other wor
sergij07 [2.7K]

Answer:

2

If R is a relation that is transitive and symmetric, then R is reflexive on dom(R)={a∣(∃b)aRb}: if a∈dom(R), then there is b such that aRb, thus bRa by symmetry, so aRa by transitivity.

Note that if R is symmetric, then dom(R)=range(R)={b∣(∃a)aRb}.

Hence, to get an example of a relation R on a set A that is transitive and symmetric but not reflexive (on A), there has to be some a∈A which is not R-related to any b∈A. There are many examples of this:

A={0,1} and R={(0,0)},

not reflexive on A because (1,1)∉R,

A={0,1,2} and R={(0,0),(0,1),(1,0),(1,1)},

not reflexive on A because (2,2)∉R.

Step-by-step explanation:

6 0
3 years ago
Express this number in standard form<br> 5.301 x 107 = ?
defon

Answer:

The answer is 567.207

Step-by-step explanation:

I got the answer by multiplying 5.301 by 107.So then I got 567.207.

7 0
3 years ago
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