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3 years ago

1 inch=2.5 cm how many inches are there in 16.4cm

1 answer:

4 0

Answer: 6,5 inch $\displaystyle\ 1 \ inch=2,5=\frac{10}{4} \\\\ x \ inch=16,4 \\\\x=\frac{16,4}{\frac{10}{4} }= 6,56 inch$

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Hi! Can you please help me with problems? If you can, please show work. Thank you!

Anvisha [2.4K]

1.)
x + (x + 1) = 145
2x + 1 = 145
2x = 145 - 1
2x = 144
x = 72

if x is equal to 72
72 + 1 = 73

therefore the 2 integers is 72 and 73

2.
x + (x + 1) + (x + 2) = 54
3x + 3 = 54
3x = 54 - 3
3x = 51
x = 17

numbers are, 17, 18 and 19

5 0

3 years ago

The line of symmetry for the quadratic equation y = ax^2 + 8x - 3 is x = 4. What is the value of "a"?

alexdok [17]

Answer:

a = - 1

Step-by-step explanation:

given a quadratic in standard form : y = ax² + bx + c : a ≠ 0

Then the x-coordinate of the vertex is

• $x_{vertex}$ = - $\frac{b}{2a}$

The line of symmetry passes through the parabola, hence x-coordinate is 4

y = ax² + 8x - 3 is in standard form

with a = a, b = 8 and c = - 3, hence

- $\frac{8}{2a}$ = 4 ( multiply both sides by 2a

- 8 = 8a ( divide both sides by 8 )

a = - 1

6 0

3 years ago

What is 9/2 in a percentage

Tpy6a [65]

Answer:

450%

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

The price have changed from $120 to $75. By how many

Kitty [74]

Answer:

35.7%

Step-by-step explanation:

The change is 45 dollars. (120-75)

The percent of change is the dollars divided by the original price.

45/120=0.357

Or 35.7%

3 0

3 years ago

Read 2 more answers