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White raven [17]
3 years ago
10

I need help on this. Will mark branliest

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
4 0
There is no question or picture.
You might be interested in
The first term of an AP is -7 and the ratio of the eight term to the third term is 7:1 calculate the common difference
alukav5142 [94]

Answer:

the common difference is 6.

Step-by-step explanation:

Given;

first term of an AP, a = -7

let the common difference = d

The third term is written as;

T₃ = a + 2d

The eight term is written as;

T₈ = a + 7d

The ratio of the eight term to third term = 7:1

\frac{T_8}{T_3} = \frac{7}{1} = \frac{a+ 7d}{a + 2d} \\\\7 =  \frac{a+ 7d}{a + 2d}\\\\Recall, a = -7\\\\7 =   \frac{-7+ 7d}{-7 + 2d}\\\\7(-7 + 2d) = -7+ 7d\\\\-49 + 14d = -7 + 7d\\\\14d -7d = -7 + 49\\\\7d = 42\\\\d = \frac{42}{7} \\\\d = 6

Therefore, the common difference is 6.

4 0
3 years ago
Help please if you can will give branliest
Yuliya22 [10]

Answer:

i’m thinking it might be 10?

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
An unmanned submarine ascends from the ocean floor at a rate of 2 meters per second. If the ocean floor is 4220 feet below sea l
KengaRu [80]

Answer:

a=4220-2t

This would be if we are finding the altitude of the submarine from the surface of the water(how many meters away from the surface it is). I don't know what it would be if you want to know from the bottom of the ocean.

4 0
2 years ago
Read 2 more answers
In arithmetic sequence tn find: S10, if t2=5 and t8=15
makkiz [27]

Given:

\bold{t_2=5}\\\\\bold{t_8=15}\\\\

To find:

\bold{T_n=?}\\\\\bold{S_{10}=?}

Solution:

\to \bold{t_2=5}\\\\  \bold{a+d=5} \\\\  \bold{a=5-d} ................(i)\\\\ \to \bold{t_8=15}\\\\ \bold{a+ 7d=15}.................(ii)\\\\

Putting the equation (i) value into equation (ii):

\to \bold{5-d+7d=15}\\\\\to \bold{5+6d=15}\\\\\to \bold{6d=15-5}\\\\\to \bold{6d=10}\\\\\to \bold{d=\frac{10}{6}}\\\\\to \bold{d=\frac{5}{3}}\\\\

Putting the value of (d) into equation (i):

\to \bold{a=5-\frac{5}{3}}\\\\\to \bold{a=\frac{15-5}{3}}\\\\\to \bold{a=\frac{10}{3}}\\\\

Calculating the \bold{t_n :}

\to \bold{t_n=a+(n-1)d}\\\\\to \bold{t_n=\frac{10}{3}+(n-1)\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}+\frac{5}{3}n-\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}-\frac{5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{10-5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{5}{3} +\frac{5}{3}n}\\\\ \to \bold{t_n=\frac{5}{3}(1+n)}\\\\

Formula:

\bold{S_n = \frac{n}{2}[2a + (n - 1)d ] }\\

Calculating the \bold{S_{10}} :

\to \bold{S_{10} = \frac{10}{2}[2\frac{10}{3} + (10- 1)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + (9)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 9 \times \frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 3\times 5 ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 15 ] }\\\\\to \bold{S_{10} = 5[\frac{20+ 45}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{65}{3} ] }\\\\\to \bold{S_{10} = 5 \times \frac{65}{3}  }\\\\\to \bold{S_{10} =  \frac{325}{3}  }\\\\\to \bold{S_{10} =108.33 }

Therefore, the final answer is  " \bold{\frac{5}{3}(1+n)\ and\ 108.33}"

Learn more:

brainly.com/question/11853909

7 0
3 years ago
Please help me please!!!!!
sattari [20]
C 63 i guesses dude but can be 50 right or wrong
5 0
3 years ago
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