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trapecia [35]
3 years ago
6

The base of an isosceles triangle is 7 cm longer than the legs. Find the legs if the perimeter of the triangle is 43 cm.

Mathematics
2 answers:
Archy [21]3 years ago
8 0
An isosceles triangle is a triangle that has one pair of equal sides. The Formula in finding the perimeter of a triangle is PΔ=S₁+S₂+S₃. Since two legs are missing, and the given are the base and perimeter, we will use the formula : S₁+S₂ = PΔ-S₃, where: S₁+S₂ are the legs and <span>S₃ is the base.
Substitute values:
</span><span>S₁+S₂ = 43cm - 7 cm
</span>          = 36 cm
since S₁ and S₂ are equal, divide 36cm by 2. Therefore S₁ and <span>S₂ measures 18cm.
Check: </span><span>PΔ=S₁+S₂+S₃
</span><span>                 = 18cm + 18 cm + 7cm
                 = 43cm

</span>
Nataly_w [17]3 years ago
5 0

Answer:

121

Step-by-step explanation:

legs are x but base is x+7

x+x+x+7 is perimeter = 43

3x+7=43

3x=36

x=12

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3 years ago
last week, Jim worked h hours and earned $9 per hour. if he earned a total of $405, which equation can be used to find out how m
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I hope this helps you




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8 0
3 years ago
-15=-4m+5<br><br>8n+7=31<br><br>-10=10(k-9)
Korolek [52]
-15=-4m+5\ \ \ |both\ sides\ /+4m\\\\4m-15=5\ \ \ \ |both\ sides\ /+15\\\\4m=20\ \ \ \ |both\ sides\ /:4\\\\\underline{\underline{m=5}}

=========================================================

8n+7=31\ \ \ \ |both\ sides\ /-7\\\\8n=24\ \ \ \ |both\ sides\ /:8\\\\\underline{\underline{n=3}}

=========================================================

10(k-9)=-10\ \ \ \ |both\ sides\ /:10\\\\k-9=-1\ \ \ \ |both\ sides\ /+9\\\\\underline{\underline{k=8}}
6 0
3 years ago
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What is the value of 7 cubed (A) 14 (B) 21 (C) 49 (D) 343
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3 years ago
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Figure CDEF is a parallelogram.
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We see that CF and DE are parallel to each other, which means that they had the same length with each other, so:
6n-1=5n+9
Subtract 5n for both side
6n-1-5n=5n+9-5n
n-1=9
Add 1 for both side
n-1+1=9+1
n=10
CF=
6n-1
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=60-1
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DE=
5n+9
=5(10)+9
=50+9
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CD/FE:
4n+2
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True/False:
n=10 True
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CF=59 True
FE=42 True
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3 years ago
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