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Furkat [3]
3 years ago
8

Help i'm in a hurry!!! first gets brainliest!!!!!!!!!!!!!!!!!!! Help on the second one pls!

Mathematics
1 answer:
mina [271]3 years ago
5 0

Answer:

THX FOR AWNSERING

Step-by-step explanation:

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Is number 4 correct?
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sure

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Which triangle is a rotation of triangle 3?<br><br><br><br> Δ1<br> Δ2<br> Δ3<br> Δ4
nikdorinn [45]

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1

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6 0
2 years ago
So ima ask a question am i racist and trying to act black
vfiekz [6]

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8 0
2 years ago
In the figure Line “P” is the ____?
Stella [2.4K]

Answer:

  "line of reflection."

Step-by-step explanation:

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6 0
3 years ago
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
 S_{5}-S_{3}=T_{4}+T_{5}=5.

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}
T_{3}=T_{1}*q^{2}
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}.




5 0
3 years ago
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