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olga nikolaevna [1]
3 years ago
9

How do you solve this equation ( ASAP PLS )

Mathematics
1 answer:
makvit [3.9K]3 years ago
8 0
<span>Solve this by substitution:

</span>Solve 5x - 4y = -13 for x:<span><span><span><span>5x </span>− <span>4y </span></span>+ <span>4y </span></span>= -<span>13 + <span>4y     </span></span></span>(Add 4y to both sides)
<span><span><u>5x  </u></span>= <span><span><u>4y - </u><u>13 </u></span><u> </u>  </span></span>(Divide both sides by 5)
 5          5

<span>x = <span><span><span><u>4</u></span> y </span>+ -<span><u>13</u><span>
</span></span></span></span>      5        5
You can use https://www.mathpapa.com/algebra-calculator.html
5x - 4y = -13, 8x +10y = 12 (enter both equations with a comma between them and it will solve for you and show you step by step.


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Consider the provided information.

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AleksAgata [21]

Answer:

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h =  - 16 {t}^{2}  + 120t \\  \because \:  we \: have \: to \: find \:  the  \:  time  \: required  \:  \\ for \:  the \:  object \:  to \:  return to \:  its  \:  \\ point \:  of  departure. \\  \therefore \: plug \: h = 0 \: in \: the \: given \: euation. \\  \therefore \: 0 =   - 16 {t}^{2}  + 120t \\ \therefore \:16 {t}^{2}  - 120t = 0 \\ \therefore \:8t(2t - 15) = 0 \\ \therefore \:8t = 0 \:  \: or \:  \: (2t - 15) = 0 \\ \therefore \:t =  \frac{0}{8}  \: or \: t =  \frac{15}{2}  \\ \therefore \:t = 0 \: or \: t = 7.5 \\  \because \: t = 0 \: is \: not \: possible \\ \huge \purple{ \boxed{  \therefore \: t = 7.5 \: seconds}}

Thus, the time required for the object to return to its point of departure is 7.5 seconds.

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