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schepotkina [342]
3 years ago
8

a triangle has a height of 34 inches and an area of 391 square inches. Which equation can be used to find the base of the triang

le, x?​
Mathematics
1 answer:
tester [92]3 years ago
3 0

Answer:

The area of a triangle is the space contained within its 3 sides. To find out the area of a triangle, we need to know the length of its three sides. The sides should be measured in feet (ft) for square footage calculations and if needed, converted to inches (in), yards (yd), centimetres (cm), millimetres (mm) and metres (m). The formula:

Square feet to square meters: multiply ft² by 0.092903 to get m²

Square yards to square feet: multiply yd² by 9 to get ft²

Square feet to square yards: multiply ft² by 0.11111 to get yd²

Square yards to square meters: multiply yd² by 0.8

Step-by-step explanation:

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Step-by-step explanation:

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The  value is   T  =  51 \  inches

Step-by-step explanation:

 From the diagram we can obtain the the height of T  as follows

      T  =  \frac{60 + 42 }{2}

     

3 0
3 years ago
Evaluate it..........​
trasher [3.6K]

Answer:

the answer is \frac{1}{41} (23x-2ln(5sinx+4cosx))

Step-by-step explanation:

\int {\frac{3sinx+2cosx}{5sinx+4cosx} } \, dx

here,

a=3, b=2, c=5, d=4

M=\frac{ac+bd}{c^2+d^2 }=\frac{3*5+2*4}{5^2+4^2}=\frac{23}{41}

N=\frac{bc-ad}{c^2+d^2 }=\frac{2*5-3*4}{5^2+4^2}=\frac{-2}{41}

\int {\frac{3sinx+2cosx}{5sinx+4cosx} } \, dx=\int{\frac{\frac{23}{41}(5sinx+4cosx)-\frac{2}{41}(5cosx-4sinx)  }{5sinx+4cosx} }\,dx

=\frac{23}{41}\int\, dx -\frac{2}{41} \int{\frac{5cosx-4sinx}{5sinx+4cosx} }\, dx

=\frac{1}{41} (23x-2ln(5sinx+4cosx))

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2 years ago
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