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asambeis [7]
3 years ago
8

1. What is 26% written as a decimal?

Mathematics
2 answers:
Pani-rosa [81]3 years ago
7 0

Answer:

26% = 0.26 in decimal form

Step-by-step explanation:

You just put a 0. in front of it,

take .26

Then 100 = 1.00

and 26 = .26

Hope This Helps:)

Would appreciate brainliest!!:)

sattari [20]3 years ago
5 0

Answer:

.26 = 26%     hope this helps!   :)

Step-by-step explanation:

.26

100 = 1.00

26 = .26

you just move the decimal point over 2

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Round 713,923 to the nearest hundred.
jeyben [28]

Answer:

713,7923 to the nearest hundred is 713,900

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3 years ago
Allyson and Adrian have decided to connect their ankles with a bungie cord; one end is tied to each person's ankle. The cord is
BARSIC [14]

Answer: 5.5 seconds

Step-by-step explanation:

30=5/4 ×(-20) + 10t

30 =-25 +10t

10t=55

t=5.5seconds

The cord will first touch the corner of the building after 5.5secs

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3 years ago
A bedroom door in the house has the same dimension as the front door but the length is 30 inches rather than 36 inches how much
arsen [322]

Answer:

6/5

Step-by-step explanation:

Calculate the volume by multiplying the measured length and width of the space together, then multiply the result by the height of the room. From the example, 10 * 25 feet = 250 square feet, and 5 * 10 feet = 50 square feet.

5 0
2 years ago
The cost y (in dollars) to rent a pickup truck is proportional to the number x of hours that the pickup truck is rented. It cost
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6 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
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