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Gekata [30.6K]
2 years ago
5

100 points !!!! Please help ASAP: What times 28,000 equals 4,410,483,661

Mathematics
2 answers:
erastova [34]2 years ago
6 0

Answer  it 4410455661

krok68 [10]2 years ago
4 0

Answer:

157517.273607

Step-by-step explanation:

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−3(2+4k)+7(2k−1) eqivalent expression
Alex17521 [72]

Answer:

\large\boxed{-3(2+4k)+7(2k-1)=2k-13}

Step-by-step explanation:

-3(2+4k)+7(2k-1)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\=(-3)(2)+(-3)(4k)+(7)(2k)+(7)(-1)\\\\=-6-12k+14k-7\qquad\text{combine like terms}\\\\=(-12k+14k)+(-6-7)\\\\=2k-13

6 0
3 years ago
Big Mack buys meat for the family reunion cookout. He needs to buy 11 packagesof meat. A package of hotdogs costs $2.59, and a p
IgorC [24]

Hello!

Let's write some important information contained in the exercise:

• hotdogs: $2.59 (,x,)

,

• hamburgers: $5.29 (,y,)

He needs 11 packages. Let's write it:

• x + y = 11

He spent a total of $39.29. We can write it as:

• 2.59x + 5.29y = 39.29

Now, let's solve these two equations as a linear system:

\begin{cases}x+y=11\text{ equation A} \\ 2.59x+5.29y=39.29\text{ equation B}\end{cases}

First, let's isolate x in equation A:

\begin{gathered} x+y=11 \\ x=11-y \end{gathered}

Now, we will replace where's x by 11-y in equation B:

\begin{gathered} 2.59x+5.29y=39.29 \\ 2.59(11-y)+5.29y=39.29 \\ 28.49-2.59y+5.29y=39.29 \\ -2.59y+5.29y=39.29-28.49 \\ 2.7y=10.8 \\ y=\frac{10.8}{2.7} \\  \\ y=4 \end{gathered}

As I called the hamburgers as 'y', we know that he bought 4 packages of hamburgers.

5 0
1 year ago
Please help with this its confusing somehow :(​
RoseWind [281]

Answer:

the answer is 7 because 7 *7 is 49

4 0
2 years ago
A system of two linear inequalities is graphed as shown, where the solution region is shaded.
Zinaida [17]

Answer:

(-3,2) is located in the solution region.

Step-by-step explanation:

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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