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tester [92]
3 years ago
15

Two legs of a right triangle are 4cm, and 3cm. find the length of the hypotenuse of this triangle

Mathematics
1 answer:
sergiy2304 [10]3 years ago
6 0
It'll be 4^2+ 3^2 then the answer you get you take the square root of it which will give you the hypotenuse.
Your answer would be 5 because square root of 25 is 5
If u want the formula it's a^2+b^=c^2
C is the hypotenuse and a and b is the lengths
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The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=
Leno4ka [110]
1 step: S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}, S_{3}=T_{1}+T_{2}+T_{3}, then
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2 step: T_{n}=T_{1}*q^{n-1}, then 
T_{6}=T_{1}*q^{5}
T_{5}=T_{1}*q^{4}
T_{4}=T_{1}*q^{3}
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and \left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right. will have form \left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right..

3 step: Solve this system  \left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right. and dividing first equation on second we obtain \frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}. So, \frac{(q-1)(q+1)}{q+1} = \frac{1}{2} and q-1= \frac{1}{2}, q= \frac{3}{2} - the common ratio.

4 step: Insert q= \frac{3}{2}into equation T_{1}*q^{3}*(q+1)=5 and obtain T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5, from where T_{1}= \frac{16}{27}.




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3 years ago
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Parallelogram as well as a quadrilateral with pairs of equal and parallel opposite sides
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