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lyudmila [28]
3 years ago
13

Solve using the quadratic formula or factoring ... 5=6x^2 + 13x

Mathematics
2 answers:
FromTheMoon [43]3 years ago
7 0

Answer:

x = \frac{-5}{2} <u><em>or  </em></u>x = \frac{-1}{3}

Step-by-step explanation:

5 − 6 2 − 1 3 = 0

− 6 2 − 1 3 + 5 = 0 -6x^2-13x+5=0 −6x2−13x+5=0 − 1

( 6 2 + 1 3 − 5 ) = 0

adell [148]3 years ago
4 0

Answer:

- 5/2, 1/3

Step-by-step explanation:

See image below:) I could only show half of the steps. But you can use the app photo math you just take a pic of the problem and it gives you the answer and explains the steps.

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Write <br> 9<br> 5<br> as a mixed number.
Advocard [28]
<span>Well, think of it this way? How many times does 5 go into 9? 

5 * 1 = 5 
5 * 2 = 10 

So 5 goes into 9 one time. Then you have to subtract that 5 from 9 to be left with your fraction value. 

1 and 4/5 is your mixed number.</span>
5 0
3 years ago
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Does the equation 3(2x−1)+5=6(x+1) have one, none, or an infinite amount of solutions?
EleoNora [17]

Answer: No solutions

Step-by-step explanation: If you solve the problem all the way, you get 0 = 4 which is not valid so there is simply no solution

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3 years ago
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Find the amount of time in years. <br><br><br><br> I=$100, P=$200, r=10%
lara31 [8.8K]

Answer:

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4 0
3 years ago
Manny bought 12 pounds of vegetables
finlep [7]

Answer:

Manny bought 3 pounds that were not on sale

Step-by-step explanation:

If 75 % were on sale,  (100% -75% = 25%)  then 25% were not on sale  (The total has to be 100%)

Manny bought 12 pounds of vegetables

To determine how many pounds were not on sale, we take the amount of vegetables purchased and multiply by the percent that were not on sale.

12 * 25%

Change this to decimal form

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3

Manny bought 3 pounds that were not on sale


7 0
3 years ago
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Needhdjdjdjdjdjdnhiskdjdjdjdjdjdjjshdjdjdjdjdjdjdjd<br><br>​
Lilit [14]

9514 1404 393

Answer:

  -(√2)/2

Step-by-step explanation:

The expression evaluated at n=a gives the indeterminate form 0/0, so L'Hopital's rule can be used to find the limit. The second expression comes from differentiating numerator and denominator. Then the form with n=a is no longer indeterminate.

  \displaystyle\lim_{n\to a}{\frac{\sqrt{2n}-\sqrt{3n-a}}{\sqrt{n}-\sqrt{a}}}=\lim_{n\to a}{\frac{\frac{2}{2\sqrt{2n}}-\frac{3}{2\sqrt{3n-a}}}{\frac{1}{2\sqrt{n}}-0}}\\\\=\sqrt{a}\left(\frac{2}{\sqrt{2a}}-\frac{3}{\sqrt{3a-a}}}\right)=\boxed{-\frac{1}{\sqrt{2}}}

5 0
3 years ago
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