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Alexxx [7]
4 years ago
11

xFormula1" title="x {}^{2} + 140 \times + 2400 = 6 \times " alt="x {}^{2} + 140 \times + 2400 = 6 \times " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ipn [44]4 years ago
3 0
Can you write it normally lol

Fr just use photomath
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Shkiper50 [21]
You would be able to make about 120 shots
7 0
3 years ago
Read 2 more answers
PLEASE HELP Graph 4x + y = 6x - 1 Show your work AND explain the method used to determine the graph
Anna007 [38]

Answer:

y=2x-1

Step-by-step explanation:

y= 6x-1-4x

y=2x-1

3 0
4 years ago
On a particular day, the wind added 5 miles per hour to Alfonso's rate when he was cycling with the wind and subtracted 5 miles
Nata [24]

now, this is pretty much the same as the one before it with Jaime, so I'll do this without much fuss.

recall d = rt.

a = Alfonso's rate

with the wind his speed is a + 5, against it is a - 5, 60 miles with it and 30 miles against it, all in the same time of t hours.

\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{with the wind}&60&a+5&t\\ \textit{against the wind}&30&a-4&t \end{array} \\\\\\ \begin{cases} 60=(a+5)t\implies \frac{60}{a+5}=\boxed{t}\\\\ 30=(a-4)t\\ -------------\\ 30=(a-4)\left( \boxed{\frac{60}{a+5}} \right) \end{cases} \\\\\\ 30(a+5)=(a-4)60\implies 30a+150=60a-240 \\\\\\ 390=30a\implies \cfrac{390}{30}=a\implies 13=a

8 0
3 years ago
The lines below are parallel. If the slope of the green line is -2, what is the slope of the red line?​
BlackZzzverrR [31]
Parallel lines have equal slopes, so if green line has slope = -2 then, red line's slope would be equal to -2
6 0
2 years ago
9. Calcule el valor de la fuerza que experimenta un ascensor cuando levanta 150 kg en los siguientes casos: a) Cuando asciende c
GenaCL600 [577]

Answer:

a) La fuerza neta que experimenta un ascensor cuando asciende con una aceleración de 4 metros por segundo al cuadrado es 600 newtons hacia arriba.

b) La fuerza neta que experimenta un ascensor cuando desciende con una aceleración de 6 metros por segundo al cuadrado es 900 newtons hacia abajo.

c) Por la fuerza de tensión sobre el cable del ascensor.

Step-by-step explanation:

a) La fuerza neta experimentada por el ascensor (F), en newtons, es:

F = m\cdot a (1)

Donde:

m - Masa, en kilogramos.

a - Aceleración, en metros por segundo cuadrado.

Si sabemos que m = 150\,kg y a = 4\,\frac{m}{s^{2}}, entonces la fuerza neta del ascensor es:

F = m\cdot a

F = 600\,N

La fuerza neta que experimenta un ascensor cuando asciende con una aceleración de 4 metros por segundo al cuadrado es 600 newtons hacia arriba.

b) Si sabemos que m = 150\,kg y a = -6\,\frac{m}{s^{2}}, entonces la fuerza neta del ascensor es:

F = m\cdot a

F = -900\,N

La fuerza neta que experimenta un ascensor cuando desciende con una aceleración de 6 metros por segundo al cuadrado es 900 newtons hacia abajo.

c) De manera simplificada, el ascensor experimenta dos fuerzas que definen la fuerza y aceleración netas: (i) La fuerza de tensión sobre el cable que traslada el ascensor y el peso total del ascensor, opuesta a la anterior y en función de la aceleración gravitacional. Puesto que la masa no varía en ningún caso, se concluye que el peso es constante, entonces la diferencia de valores se debe a la fuerza por tensión del cable. En el descenso, es fuerza es menor que en el ascenso.

8 0
3 years ago
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