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2 years ago

Find an equivalent expression to 5+9

Mathematics

1 answer:

Marina86 [1]2 years ago

6 0

Since 5 + 9 is 14 you can use ANY expression that has the same final value/ sum

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From 1994 to 1995 the sales of a book decreased by 80%. If the sales in 1996 for the same as in 1994, By what percent did the in

zheka24 [161]

Answer:

400%

Step-by-step explanation:

Let the sales be "100" in 1994

Since, it decreased 80%, the sales was:

80% = 80/100 = 0.8

0.8 * 100 = 80 (decreased by 80)

So, it was

Sales in 1995: 100 - 80 = 20

Sales in 1996 was same as in 1994, so that's 100

Thus,

Sales in 1994: 100

Sales in 1995: 20

Sales in 1996: 100

We need to find percentage increase form 1995 to 1996, that is what percentage increase is from 20 to 100?

We will use the formula:

$\frac{New-Old}{Old}*100$

Where

New is 100

Old is 20

SO, we have:

$\frac{New-Old}{Old}*100\\=\frac{100-20}{20}*100\\=4*100\\=400$

So, it increased by 400%

3 0

3 years ago

Solve 3x+2y=5 and 7x+2y=9

vekshin1

Answer:

Add the equations in order to solve for the first variable. Plug this value into the other equations in order to solve for the remaining variables.

Point Form:(1,1)

Step-by-step explanation:

6 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 $x \to \frac1x$ :

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