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lina2011 [118]
2 years ago
15

WILL GIVE BRAINLIEST

Mathematics
1 answer:
vesna_86 [32]2 years ago
7 0

The simplified forms of the given expressions are

a. 32x

b. 0

c. -26t + 14x

d. 3t

<h3>Simplifying an expression </h3>

From the question, we are to simplify the given expressions by adding or subtracting

a. (10x) + (22x)

= 32x

b. (-6x) - (-6x)

= -6x + 6x

= 0

c. (-13t) + (14x) - (+13t)

= -13t + 14x - 13t

Collect like terms

= -13t -13t + 14x

= -26t + 14x

d. (-4t) + (+4t) - (-3t)

= -4t + 4t + 3t

= 3t

Hence, the simplified forms of the given expressions are

a. 32x

b. 0

c. -26t + 14x

d. 3t

Learn more on Simplifying an expression here: brainly.com/question/723406

#SPJ1

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Elodia [21]
We have 2x + 2x + 32 = 256;
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3 years ago
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Aleksandr-060686 [28]

-4x-2y=-12

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--------------------------add

6y = -36

y = -6

Substitute y = -6 into -4x-2y=-12

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Answer

x = 6 and y = -6

3 0
4 years ago
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<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%
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With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

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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}}

8 0
2 years ago
What’s 1/6 plus 3/8 in simplest form
gregori [183]

13/24 is the answer to your problem.


7 0
3 years ago
10+4-3/2-9/2 please show work
nexus9112 [7]
Here is the answer
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