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krek1111 [17]
3 years ago
8

Find a number if: 3/7 of it is 2 1/4

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
8 0

Answer:

x=5 1/4

Step-by-step explanation:

2 1/4= \frac{9}{4}---> \frac{9}{4} divided by 3 to find 1/7x----> \frac{3}{4}=1/7x---> x=21/4---> x=5 1/4

Inessa [10]3 years ago
6 0

Answer:

The number is 5¹/₄

Step-by-step explanation:

3/7 of it is 2 ¹/₄

let the it be represented as A

3/ 7 of A = 2 ¹/₄

3A/7 = 4X2+1/4 = 9/4

3A/7= 9/4

cross multiply

3A X4 = 9x7

12A = 63

A = 63/12 = 21/4= 5¹/₄

The number is 5¹/₄

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Vaselesa [24]

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Evaluate the following limit, if it exists : limx→0 (12xe^x−12x) / (cos(5x)−1)
Amanda [17]

Answer:

\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}

Step-by-step explanation:

Notice that \lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=\frac{12(0)e^{0}-12(0)}{cos(5(0))-1}=\frac{0}{0}, which is in indeterminate form, so we must use L'Hôpital's rule which states that \lim_{x \to c} \frac{f(x)}{g(x)}=\lim_{x \to c} \frac{f'(x)}{g'(x)}. Basically, we keep differentiating the numerator and denominator until we can plug the limit in without having any discontinuities:

\frac{12xe^x-12x}{cos(5x)-1}\\\\\frac{12xe^x+12e^x-12}{-5sin(5x)}\\ \\\frac{12xe^x+12e^x+12e^x}{-25cos(5x)}

Now, plug in the limit and evaluate:

\frac{12(0)e^{0}+12e^{0}+12e^{0}}{-25cos(5(0))}\\ \\\frac{12+12}{-25cos(0)}\\ \\\frac{24}{-25}\\ \\-\frac{24}{25}

Thus, \lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}

3 0
2 years ago
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Aleksandr [31]

Answer:

f'(x)=-(x+42)/5

Step-by-step explanation:

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y=-5x-42

Flip x and y and solve for y:

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(x+42)/-5=y

So the inverse function is f'(x)=-(x+42)/5 (note that f' means inverse)

6 0
3 years ago
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