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drek231 [11]
3 years ago
5

Ribbon A is 1/3 meters long. It is 2/5 meters shorter than ribbon b. What's the total length of 2 ribbons?

Mathematics
1 answer:
Pavel [41]3 years ago
4 0
1/3+2/5=5/15+10/15
answer is 1
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2x + y = 7<br> -2x - y = 6<br> Solve systems of equations by substitution
ehidna [41]

Answer:

No solution.

Step-by-step explanation:

2x + y = 7

-2x - y = 6

From the first equation:

y = 7 - 2x

Plug this into the second equation:

-2x - (7 - 2x) = 6

-2x  + 2x = 6 + 7

0 = 13

this doesn't make sense so theer is:

No solution.

4 0
2 years ago
Determine whether the polynomial below can be factored into perfect squares. If so, factor the polynomial. Otherwise, select tha
kkurt [141]

Answer:

A. =(9x-12)^2

Step-by-step explanation:

We need to determine whether the polynomial 81x^2-216x+144 can be factored into perfect squares. If so, factor the polynomial. Otherwise, select that it cannot be factored into a perfect square.

81x^2-216x+144

=9(9x^2-24x+12)

=9(9x^2-12x-12x+12)

=9(3x(3x-4)-4(3x-4))

=9(3x-4)(3x-4)

=9(3x-4)^2

=3^2(3x-4)^2

=(3(3x-4))^2

=(9x-12)^2

Hence choice A. =(9x-12)^2 is correct.

3 0
3 years ago
Read 2 more answers
Please help me on this math I have been trying to solve it but I can't please please help me!!!!!!​
ludmilkaskok [199]

Answer:The first table would be a proportional relationship.

Step-by-step explanation:

It's the only table with a constant, which is required in a proportional relationship. The constant would be 1300 which you can find by divided the two numbers in each row.

7 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
What is h−(−2.22)=−7.851 someone please help me
telo118 [61]

Answer:

h=-10.071?

Step-by-step explanation:

7 0
2 years ago
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