What’s wrong with this picture ?

Fine the equivalent expression 4^-5

Novay_Z [31]

Answer:

1/1024

Step-by-step explanation:

1/(4*4*4*4*4)

1/1024

3 0

2 years ago

Consider the function f(x)=x3+6x2−20x+450.

Damm [24]

Answer:

Therefore reminder = 2802

Step-by-step explanation:

f(x)=x³+6x²-20x+450

x-12)x³+6x²-20x+450(x²+18x+196

x³-12x²

____________________

+18x²-20x+450

18x²-216x

_______________

196x +450

196x-2352

_____________

2802

Therefore reminder = 2802

6 0

3 years ago

Jasmine took a cab home from her office. The cab charged a flat fee of $4, plus $2 per mile. Jasmine paid $32 for the trip. How

vaieri [72.5K]

To solve, make an equation. The equation would be
$$32 = 4 + 2x$
x represents the miles driven
to solve, subtract 4 from each side of the equation.
$32-4 = 4-4 + 2x
The equation is now
$28 = 2x
Now you divide by 2 on each side.
$28 ÷ 2 = 2x ÷ 2
simplify
14 = x
The answer is 14 miles

3 0

3 years ago

Sofia's taco truck lets customers choose 3 ingredients from the list below to add to their tacos. How many groups of 3 different

DIA [1.3K]

Using the combination formula, and considering that the list has 6 ingredients, it is found that 20 groups of 3 different ingredients can go on a taco.

The order in which the ingredients are chosen is not important, hence, the <em>combination formula</em> is used to solve this question.

<h3>Combination formula: </h3>

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem, 3 ingredients are chosen from a set of 6, hence:

$C_{6,3} = \frac{6!}{3!3!} = 20$

20 groups of 3 different ingredients can go on a taco.

To learn more about the combination formula, you can take a look at brainly.com/question/25990169

4 0

2 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