Can I plz get help someone plzzz

7/8 - 5/12 ayudaaa por favor es para mañana

RUDIKE [14]

7/8 = 21/24
5/12 = 10/24

21/24-10/24 = 11/24

3 0

3 years ago

Help please help me please help please please help me please

GarryVolchara [31]

The answer is no. the first equation is not right

8 0

3 years ago

Read 2 more answers

A farmer needs to ship 71 pumpkins to a grocery store if each crates can hold 19 pumpkins how many crates Will the farmer need ?

Lesechka [4]

Answer:

The farmer will need 7 crates.

Step-by-step explanation:

It is given that there are 71 pumpkins in total.

And each crate holds 19 pumpkins.

Let the number of crates required be "n".

The total number of pumpkins can be calculated by multiplying number of pumpkins in each crate with total number of crates.

Thus, the equation is

$19(n) = 71$

$n = \frac{71}{19} = 7$

Thus, the farmer needs 7 crates to hold total of 71 pumpkins.

3 0

3 years ago

A sector of a circle has a central angle of 45 . Find the area of the sector if the radius of the circle is 15 cm.

prohojiy [21]

Answer:

A = 28.125 * pi cm^2

A = 88.357 cm^2

Step-by-step explanation:

Area of a sector is 1/2 r^2 theta where theta is in radians

convert 45 degrees to radians

theta = 45 * pi/180 = pi/4

A = 1/2 * 15^2 * pi/4

A =1/2 * 225 * pi/4

A = 28.125 * pi cm^2

A = 88.357 cm^2

6 0

3 years ago

Find the differential coefficient of <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