Answer:
nah
Step-by-step explanation:
Considerando las fórmulas para el perímetro y el área de un rectángulo, hay que se chega en una <u>eccuación cuadrática sin solución</u>, o sea, las medidas no son posibles y la persona estaba mintiendo.
<h3>¿Cuál es la fórmula para el perímetro y el área de un rectángulo?</h3>
Considerando que las dimensiones son l y w, hay que:
- El perímetro es: P = 2(l + w).
El <u>perímetro es de 18 m</u>, o sea:
2(l + w) = 18
l + w = 9
l = 9- w.
El <u>área es de 21 m²</u>, o sea:
lw = 21
(9- w)w = 21
-w² + 9 - 21 = 0
w² - 9w + 21 = 0
El discriminante es dado por:
D = 9² - 4 x 1 x 21 = -3.
El discriminante negativo implica que la <u>eccuación cuadrática no tiene solución</u>, o sea, las medidas no son posibles y la persona estaba mintiendo.
Puede-se aprender más a cerca de el perímetro y el área de un rectángulo en brainly.com/question/26475963
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Answer:
1.38 * 10^ -2
Step-by-step explanation:
.0138
We need 1 number to the left of the decimal
1.38 * 10^ something
We moved it 2 places to the right, that means our exponent is -2. If we had moved it to the left, our exponent would have been positive
1.38 * 10^ -2
Answer:
Use the slope formula to find the slope of a line given the coordinates of two points on the line. The slope formula is m=(y2-y1)/(x2-x1), or the change in the y values over the change in the x values. The coordinates of the first point represent x1 and y1. The coordinates of the second points are x2, y2.
Step-by-step explanation:
now, there are 12 months in a year, so 18 months is really 18/12 of a year, thus
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\to \frac{18}{12}\dotfill &\frac{3}{2} \end{cases} \\\\\\ 4000=P[1+(0.05)(\frac{3}{2})]\implies 4000=P(1.075) \\\\\\ \cfrac{4000}{1.075}=P\implies 3720.93\approx P](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B18%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%204000%3DP%5B1%2B%280.05%29%28%5Cfrac%7B3%7D%7B2%7D%29%5D%5Cimplies%204000%3DP%281.075%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4000%7D%7B1.075%7D%3DP%5Cimplies%203720.93%5Capprox%20P)