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77julia77 [94]
4 years ago
14

Eric earned 60 points on his last tes t.If there were 75 points.what his precent grade.

Mathematics
2 answers:
Slav-nsk [51]4 years ago
4 0
In this problem, you would just divide your test results by the maximum test points you can get. 
60 divided by 75 is 0.8, which is 80%, which is probably a grade I will never see in my life :P
Hope I helped!
Cloud [144]4 years ago
4 0
So he got 60/75. let's divide it by 15!
which is then 4/5. and if you multiply that by 2 you have a base of 10!---> 8/10 now that's 80%
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7 0
3 years ago
3. At the start of summer, Ben has $250 . He takes a summer
earnstyle [38]

Answer:

Week 4

Step-by-step explanation:

Ben has $250 in the beginning.  He saves $150 per week.

y = 150x + 250

Tim has $1,650 in the beginning.  He spends $200 per week.

y = 1650 - 200x

We are trying to find which x-value produces the same y-value for both equations.  You can do this by setting both equations equal to each other.

150x + 250 = 1650 - 200x

(150x + 250) + 200x = (1650 - 200x) + 200x

350x + 250 = 1650

(350x + 250) - 250 = (1650) - 250

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3 0
3 years ago
5x + y = -5<br> I need to graph this pls help
Lisa [10]
Hope this helps
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4 0
3 years ago
Si un buzo se sumerge 4 metros luego sube 2 metros y finalmente desciende 5 metros más, ¿a qué profundidad se encuentra al final
wolverine [178]

Así, concluimos que la posición final del buzo será 7 metros bajo la superficie del agua.

<h3 /><h3>¿a qué profundidad se encuentra al final de su recorrido?</h3>

Primero, vamos a definir la superficie del agua como el 0 metros.

Sí sabemos que primero el buzo se sumerge 4 metros, entonces en este punto la posición del buzo es:

P = 0m - 4m = -4m

Luego el buzo sube 2 metros, entonces la nueva posición será:

P = -4m + 2m = -2m

Finalmente, el buzo desciende otros 5 metros, entonces la posición final del buzo será:

P = -2m - 5m = -7m

Así, concluimos que la posición final del buzo será 7 metros bajo la superficie del agua.

Sí quieres aprender mas sobre posiciones:

brainly.com/question/21853903

#SPJ1

7 0
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