<span>-2x^2-x+7=0
Variable with the highest degree's (exponent) constant, -2 is a, next variable's constant, -1 is b, the constant or number without a variable, 7 is c
using substitution put the numbers into the formula
</span>(-b±√(b^(2)-4ac))/(a^(2))
(-(-1)±√((-1)^(2)-4(-2)(7))/((-2)^(2)) simplify
(1±√(1+56))/4
1±√(57)/4 is your answer
Answer:
w < 7
Step-by-step explanation:
Since the inequality is an OPEN circle, that means the sign is < or >
Since the sign is going to the left that means it's < or ≤
The one that has both of these effects is <
Since the sign starts at 7, the inequality is w < 7
Tbh I would say ounces or grams. I wouldn't make a final decision on my answer though.
._.
Answer:
Technically neither, but Mia is correct.
Step-by-step explanation:
Slope-Intercept Form: y = mx + b
Since our slope <em>m</em> is equal to 6, it is a positive number. That means that the slope/line will be positive, so the line should be going up, not down. That means the blue line would be correct. Therefore, Mia is correct.
Technicality:
Since the blue line is dotted, the equation should be a <em>inequality </em>and not a linear equation. No solutions on the line would technically work. If the equation was y < 6x + 4 or something with an inequality sign, then it would be correct.
Answer:
D. $31,337.27
Step-by-step explanation:
We have that the initial amount of the loan is $5500.
Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.
The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.
Also, it is given that the loan was compounded annually.
We have the formula as,
i.e. ![PV=\frac{P\times [1-(1+\frac{r}{n})^{-t\times n}]}{\frac{r}{n}}](https://tex.z-dn.net/?f=PV%3D%5Cfrac%7BP%5Ctimes%20%5B1-%281%2B%5Cfrac%7Br%7D%7Bn%7D%29%5E%7B-t%5Ctimes%20n%7D%5D%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D)
Substituting the values, we get,
i.e. ![PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}](https://tex.z-dn.net/?f=PV%3D%5Cfrac%7BP%5Ctimes%20%5B1-%281%2B%5Cfrac%7B0.075%7D%7B12%7D%29%5E%7B-10%5Ctimes%2012%7D%5D%7D%7B%5Cfrac%7B0.075%7D%7B12%7D%7D)
i.e. ![22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}](https://tex.z-dn.net/?f=22000%3D%5Cfrac%7BP%5Ctimes%20%5B1-%281%2B0.00625%29%5E%7B-120%7D%5D%7D%7B0.00625%7D)
i.e. ![22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}](https://tex.z-dn.net/?f=22000%3D%5Cfrac%7BP%5Ctimes%20%5B1-%281.00625%29%5E%7B-120%7D%5D%7D%7B0.00625%7D)
i.e. ![22000=\frac{P\times [1-0.4735]}{0.00625}](https://tex.z-dn.net/?f=22000%3D%5Cfrac%7BP%5Ctimes%20%5B1-0.4735%5D%7D%7B0.00625%7D)
i.e. 
i.e. 
i.e. 
i.e. 
Thus, the total lifetime cost to pay of the loans compounded annually = 261.16 × 120 = $31,339.2
Hence, the total cost close to the answer is $31,337.27
