Dan had 15 cups Of apple juice he wanted to add 6 oz how much would he have in the end?

1 answer:

6 0

If you want to know the total, it would be 15 cups and 6 oz. One cup is 8 oz, so 6 oz would have to stand next to 15 cups.
If you want to know the total amount of ounces, it would be 126 ounces.
This information is only accurate if you are using USA measurements.
Hope this helps! Have a happy holiday!

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Hugo works in a shop.

RSB [31]

Multiply hourly rate by hours worked:

Saturday 7 hours x 9.00 = 63.00

Sunday: he gets paid 1 times his rate: 9.00 x 1 = 9.00

Multiply by hours : 9.00 x 4 = 36.00

Add the days together: 63 + 36 = 99.00

He earned 99.00

7 0

3 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 $x \to \frac1x$ :

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