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lapo4ka [179]
3 years ago
8

Question 1 (1 point)

Mathematics
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer:

The answer would be (B) .

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Paul is going to have the exterior of his home painted. Painting USA charges $75.00 to come out and evaluate the house plus $25.
user100 [1]
Given:$75
25 hours

Painting USA : Fixed cost: $75 ; Variable Cost: $25x
Total Cost = 75 + 25(25) = 75 + 625 = 700

Upscale Home Painting: No Fixed Cost ; Variable Cost : $40x
Total Cost = 40(25) = 1,000

The most cost effective company will be Painting USA whose charge will amount to $700.
3 0
3 years ago
What is 1 and 5/6 + 4 and 3/6 ​
Simora [160]
The answer is 6 1/3 because 1+4= 5 and 3/6 + 5/6 is 8/6 which simplify a to 1 and 1/3 and 5+ 1 1/3= 6 1/3
3 0
3 years ago
If a polynomial function f(x) has roots 6 and square root of 5, what must also be a root of f(x)?
ivolga24 [154]

Answer:

-\sqrt{5}

Step-by-step explanation:

A root with square root or under root is only obtained when we take the square root of both sides. Remember that when we take a square root, there are two possible answers:

  • One answer with positive square root
  • One answer with negative square root

For example, for the equation:

x^{2}=3

If we take the square root of both sides, the answers will be:

x=\sqrt{3} \text{ or } x= -\sqrt{3}

Only getting one solution with square root is not possible. Solutions with square root always occur in pairs.

For given case, the roots are 6 and \sqrt{5}. Therefore, the 3rd root of the polynomial function f(x) had to be -\sqrt{5}

It seems you made error while writing option B, it should be - square root of 5.

4 0
3 years ago
Please answer quickly thanks
Anastaziya [24]
8/3 = 2 2/3 and -11/3 = -3 2/3
so the middle line is 0 and each big line is 1.
Put the first dot on 2 2/3 or right before the 3rd big line.
Put the second dot on -3 2/3 or the line right after the 4th big line going back from 0
3 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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