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3 years ago

Solve 16(3x-5) -10(4x-8) = 40

Mathematics

2 answers:

nydimaria [60]3 years ago

6 0

Answer:

x = 5

Step-by-step explanation:

Step 1: Multiply out the brackets

48x - 80 - 40x + 80 = 40 → negative + negative = positive [-10 x -8]

Step 2: Simplify

8x = 40 → collect like terms

Step 3: Solve

x = (40/8)

x = 5

Mashcka [7]3 years ago

3 0

Answer:

x = 5

Step-by-step explanation:

Step 1: Multiply out the brackets

48x - 80 - 40x + 80 = 40 → negative + negative = positive [-10 x -8]

Step 2: Simplify

8x = 40 → collect like terms

Step 3: Solve

x = $\frac{40}{8}$

x = 5

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4 years ago

How would I solve this

vova2212 [387]

$A=\frac{1}{2}\ln17 = 1.417$

Step-by-step explanation:

The area A under the curve can be written as

$\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2}$

To evaluate the integral, let

$u = 1+4x^2 \Rightarrow du = 8xdx\:\text{or}\:\frac{1}{2}du = 4xdx$

so the integral becomes

$\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\int\!\dfrac{du}{u} = \dfrac{1}{2}\ln |u|$

$\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\ln |1+4x^2|$

Putting in the limits of integration, our area becomes

$\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\left.\ln |1+4x^2|\right|_0^2$

$\;\;\;\;= \frac{1}{2}[\ln (1+16) - \ln (1)]$

$\;\;\;\;=\frac{1}{2}\ln17$

Note: $\ln 1 = 0$

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3 years ago

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gyhbgt

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