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ladessa [460]
3 years ago
14

4 math problem

Mathematics
2 answers:
Aneli [31]3 years ago
6 0

Answer:

Step-by-step explanation:

1)y=-6x-14   ---------(i)

-x-3y=-9  ---------- (ii)

Substitution method:

Substitute y  value in equ (i)

-x -3*(-6x-14) = -9

-x - 3*-6x -3*(-14)= -9

-x + 18x + 42 = -9

17x = -9 -42

17x = -51

x = -51/17

x = -3

Substitute x value in equation (i)

y = -6*-3 -14

y =18-14

y = 4

TEA [102]3 years ago
3 0

Step-by-step explanation:

I'm using substitute in the first and something I don't know the name of in the rest.

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The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.​
Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is \dfrac{15}{2}.

To find:

The sum of first six terms.​

Solution:

We have,

S_2=6

S_4=\dfrac{15}{2}

Sum of first n terms of a GP is

S_n=\dfrac{a(1-r^n)}{1-r}              ...(i)

Putting n=2, we get

S_2=\dfrac{a(1-r^2)}{1-r}

6=\dfrac{a(1-r)(1+r)}{1-r}

6=a(1+r)                    ...(ii)

Putting n=4, we get

S_4=\dfrac{a(1-r^4)}{1-r}

\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}

\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}

\dfrac{15}{2}=6(1+r^2)            (Using (ii))

Divide both sides by 6.

\dfrac{15}{12}=(1+r^2)

\dfrac{5}{4}-1=r^2

\dfrac{5-4}{4}=r^2

\dfrac{1}{4}=r^2

Taking square root on both sides, we get

\pm \sqrt{\dfrac{1}{4}}=r

\pm \dfrac{1}{2}=r

\pm 0.5=r

Case 1: If r is positive, then using (ii) we get

6=a(1+0.5)  

6=a(1.5)  

\dfrac{6}{1.5}=a  

4=a

The sum of first 6 terms is

S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}

S_6=\dfrac{4(1-0.015625)}{0.5}

S_6=8(0.984375)

S_6=7.875

Case 2: If r is negative, then using (ii) we get

6=a(1-0.5)  

6=a(0.5)  

\dfrac{6}{0.5}=a  

12=a  

The sum of first 6 terms is

S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}

S_6=\dfrac{12(1-0.015625)}{1.5}

S_6=8(0.984375)

S_6=7.875

Therefore, the sum of the first six terms is 7.875.

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