If we rewrite it as y=mx+d (which can be taken from here from subtracting ax and c from both sides, then dividing b, resulting in y=(-a/b)(x)-c/b. We can then substitute -a/b for m and -c/b for d), if d=0, then we have m as a constant and as we add a specific number to y (that number being m) every time the x value increases by 1, it therefore forms a straight line. If d is not 0, then we simply add d to every single number - this is still a straight line due to that we still add a specific number to y every time x increases by 1 every single time
The formula for the nth term of a geometric sequence:
a₁ - the first term, r - the common ratio
![54, a_2, a_3, 128 \\ \\ a_1=54 \\ a_4=128 \\ \\ a_n=a_1 \times r^{n-1} \\ a_4=a_1 \times r^3 \\ 128=54 \times r^3 \\ \frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\ \frac{64}{27}=r^3 \\ \sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\ \frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\ r=\frac{4}{3}](https://tex.z-dn.net/?f=54%2C%20a_2%2C%20a_3%2C%20128%20%5C%5C%20%5C%5C%0Aa_1%3D54%20%5C%5C%0Aa_4%3D128%20%5C%5C%20%5C%5C%0Aa_n%3Da_1%20%5Ctimes%20r%5E%7Bn-1%7D%20%5C%5C%0Aa_4%3Da_1%20%5Ctimes%20r%5E3%20%5C%5C%0A128%3D54%20%5Ctimes%20r%5E3%20%5C%5C%0A%5Cfrac%7B128%7D%7B54%7D%3Dr%5E3%20%5C%5C%20%5Cfrac%7B128%20%5Cdiv%202%7D%7B54%20%5Cdiv%202%7D%3Dr%5E3%20%5C%5C%0A%5Cfrac%7B64%7D%7B27%7D%3Dr%5E3%20%5C%5C%0A%5Csqrt%5B3%5D%7B%5Cfrac%7B64%7D%7B27%7D%7D%3D%5Csqrt%5B3%5D%7Br%5E3%7D%20%5C%5C%0A%5Cfrac%7B%5Csqrt%5B3%5D%7B64%7D%7D%7B%5Csqrt%5B3%5D%7B27%7D%7D%3Dr%20%5C%5C%0Ar%3D%5Cfrac%7B4%7D%7B3%7D)
Answer:
nose mate
Step-by-step explanation:
Answer:
B. 1 and -2
Explanation:
First, add up the equations:
x-x+3y+y=4
4y=4
y=1
Now, plug y into any of the original equations above and solve for x:
-x+1=3
-x=2
x=-2
Therefore, the answer is B. 1 and -2
