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3 years ago

Rewrite 2/5 and 3/4 so that they have common denominators. Then add the fractions and simplify the sum

Mathematics

1 answer:

leonid [27]3 years ago

6 0

Answer:

8/20+15/20=23/20

23/20=1 3/20

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2 2 6 6 7 8 9 10 TIME REN 36: A baseball player had 4 hits in 8 games. At this rate, how

Nostrana [21]

Answer:

14 hits in 28 games

Step-by-step explanation:

every 2 games there is 1 hit

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4 years ago

In order to join a dancing club, there is a $30 startup fee and a $4 monthly fee. Write an equation in slope-intercept form that

sveta [45]

Answer:

Step-by-step explanation:

$30 is your startup fee. You automatically pay that to join. When is is also $4 per month, you pay that much times however many months there are.

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3 years ago

In two or more complete sentences, describe the transformation(s) that take place on the parent function, f(x)=log(x), to achiev

matrenka [14]

Transforming $f(x) = log(x)$ to $f(x) = log (2x)$ gives the effect of squashing the graph horizontally (by halving the x-coordinate)

Then from $f(x) = log(2x)$ to $f(x) = log (-2x)$ is reflecting on the y-axis

Then from $log(-2x)$ to $log(-2x-4)$ is to translate by 4 units to the right

Finally from $log(-2x-4)$ to $log(-2x-4)+5$ is translating the graph up by 5 units

5 0

4 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%20%5Cinfty%20%20%5Cfrac%7B%20%5Csqrt%5B%20%20%5Cscriptsize%5Cphi%

Rasek [7]

With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

$I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx$

Replace $x \to x^{\frac1\phi} = x^{\phi-1}$ :

$I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

Split the integral at x = 1. For the integral over [1, ∞), substitute $x \to \frac1x$ :

$\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

The integrals involving tan⁻¹ disappear, and we're left with

$I = \displaystyle \frac\pi{2\phi} \int_0^1 \frac{dx}{(1+x)^2} = \boxed{\frac\pi{4\phi}}$

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2 years ago

45 students will be grouped in pairs (groups of 2). How many different combinations of groups

Sindrei [870]

Answer:

its 21

Step-by-step explanation:

im just really smart .

7 0

3 years ago