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3 years ago

Find a fraction that is more than 7/4 but less than 9/5. Explain your answer.

Mathematics

1 answer:

Maksim231197 [3]3 years ago

8 0

Step-by-step explanation:

for such problems it is always the best approach to being both fractions to the same denominator.

that means for one or both of them we need to multiply them with the same factor on the top and the bottom (so that the fraction value remains unchanged), so that we get 2 fractions with the same denominators (bottom part).

the the least common multiple for 4 and 5 is 20.

so, we need to bring both fractions to .../20 :

7/4 × 5/5 = 35/20

9/5 × 4/4 = 36/20

now, to find a fraction between them, we need to actually refine the denominator even further, as 35 and 36 are consecutive integers.

in such a case I suggest doubling things.

35/20 × 2/2 = 70/40

36/20 × 2/2 = 72/40

now we see, a fraction between 7/4 and 9/5 would be

71/40

as 7/4 = 70/40 is smaller, and 9/5 = 72/40 is larger.

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What are the women's right included by UN

fomenos

Answer:

Step-by-step explanation:

Women's rights are the fundamental human rights that were enshrined by the United Nations for every human being on the planet nearly 70 years ago. These rights include the right to live free from violence, slavery, and discrimination; to be educated; to own property; to vote; and to earn a fair and equal wage.

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4 years ago

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0

2 years ago

Help me with this please

Pavel [41]

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Step-by-step explanation:

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3 years ago

Find the inverse function of the function f(x) = 5x – 2.

kirill115 [55]

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