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3 years ago

Solve 4(6k -12) = 3 (8k - 16)

Mathematics

2 answers:

Svetradugi [14.3K]3 years ago

8 0

Answer:

or,24k-48=24k-48

or,24k-24k= -48+48

0=0

schepotkina [342]3 years ago

5 0

Answer:

i think the answer is 0=0

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Share £40 in the ratio 1:4 between Tim and Sam

lesantik [10]

Answer:

$\£8:\£32$

Step-by-step explanation:

$\frac{40}{1+4}$

$\frac{40}{5}$

$=8$

$1:4$

$1 \times 8:4 \times 8$

$8:32$

7 0

3 years ago

Read 2 more answers

When constructing inscribed polygons, how can you be sure the figure inscribed is a regular polygon

alina1380 [7]

You can just draw arcs....

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3 years ago

Find the measures of the angles of a triangle whose angles have a measure of x, 1/2x, and 1/6x. Also, what kind of triangle is i

Alexandra [31]

the sum of the angles in a triangle = 180°, thus

x + $\frac{1}{2}$ x + $\frac{1}{6}$ x = 180

multiply through by 6

6x + 3x + x = 1080

10x = 1080 ( divide both sides by 10 )

x = 108

the angles are 108°, 54° and 18°

Since all the angles are different and the largest is 108°

The triangle is an obtuse scalene triangle

5 0

3 years ago

Heather's class took a field trip to the science museum. They left school at 7:30 A.M. It took

vodka [1.7K]

Answer:

10:30 A.M

Step-by-step explanation:

5 0

3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago