Answer:
a picture would help
Step-by-step explanation:
you need a picture to further understand and answer the question right
Primeiro Você IRA Dividir o c (APENAS Colocar 5 + 2 embaixo do c). Depois faça a soma 5 + 2 = 7. Então faça o mmc (mínimo múltiplo comum) entre 7 e 1(o 1 é invisível, mas continua estando embaixo do 3). Deu 7 o mmc. Transforme seus números em frações, com o denominador 7. Transforme os números em frações o 3 virá 21, pois você divide em baixo e multiplica em baixo, e o c continua normal, pois já estava em baixo de 7. Como é para descobrir uma incógnita você tira os denominadores e ficará 21 = c.
3 = c/5 + 2
3 = c/7
21/7 = c/7
21 = c
Answer:
Neither
Step-by-step explanation:
5 x -10 = -50
4 x 9 = 36
-50 + 36 = -14 and therefore can not be orthogonal
sqrt(25+16)=sqrt(41)
sqrt(100+81)=sqrt(181)
sqrt(41) x sqrt(181) is not equal to -14 so therefore it is not parallel
So the answer is then neither
The function which represents the taxi fare in terms of distance d in meter is f(d) = 3.50 ×
+ 40 .
According to the question,
For first 500 meters cost is 40.00 and For each additional 300 meters cost is 3.50 .
Let, the total distance of taxi ride = d meters
Now, representing the equation of taxi fare in terms of distance when distance is less than 500
taxi fare = 3.50 × + 40
∴ f(d) = 3.50 × + 40
To know more about functions refer below link: brainly.com/question/12431044
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Answer:
The value of Car B will become greater than the value of car A during the fifth year.
Step-by-step explanation:
Note: See the attached excel file for calculation of beginning and ending values of Cars A and B.
In the attached excel file, the following are used:
Annual Depreciation expense of Car A = Initial value of Car A * Depreciates rate of Car A = 30,000 * 20% = 6,000
Annual Depreciation expense of Car B from Year 1 to Year 6 = Initial value of Car B * Depreciates rate of Car B = 20,000 * 15% = 3,000
Annual Depreciation expense of Car B in Year 7 = Beginning value of Car B in Year 7 = 2,000
Conclusion
Since the 8,000 Beginning value of Car B in Year 5 is greater than the 6,000 Beginning value of Car A in Year 5, it therefore implies that the value Car B becomes greater than the value of car A during the fifth year.
