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romanna [79]
3 years ago
13

A person who weighs 120 pounds on Earth weighs 20 pounds on the moon. How much does a 93- pound person weigh on the moon?

Mathematics
2 answers:
olganol [36]3 years ago
4 0

Answer:-7

Step-by-step explanation:

KIM [24]3 years ago
4 0

Answer:

1.55 pounds

Step-by-step explanation:

120 divided by 20 equals 60

93 divided by 60 equals 1.55

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Is there another way to write 7.05.
german
Yes
seven and five tenths
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4 years ago
Solve using elimination 2x- 2y=-8 x+2y=-1
dimulka [17.4K]

Answer:( -3,1)

X=-3 y=1

Step-by-step explanation:

6 0
3 years ago
The radius of a 12 inch right circular cylinder is measured to be 4 inches, but with a possible error of ±0.2 inch. What is the
GuDViN [60]
Given:
r = radius = 4
h = height = 12
dr = error in radius = 0.2

What we want to find
dV = error in volume

Use the derivative to find the differential dV
V = pi*r^2*h
dV/dr = d/dr[ pi*r^2*h ]
dV/dr = 2pi*r*h
dV = 2pi*r*h*dr
dV = 2pi*4*12*0.2
dV = 19.2pi

So if the error in measuring the radius is +-0.2 inches, then the error in the cylinder volume is +-19.2pi cubic inches (we either measure the volume to be 19.2pi cubic inches too big, or 19.2pi cubic inches too small)

----------------------------------------------------------------------------------

Final Answer: Plus or minus 19.2pi cubic inches

Note: your teacher may want you to drop the "plus or minus" part
5 0
3 years ago
PLEASE HELP !!!!!!!!!!!!
V125BC [204]
The answer would be b - associate w it
3 0
3 years ago
X² + y² = 8
ohaa [14]

Answer:

(2,2)~ \text{and}~ (-2,-2)

Step by step explanation:

x^2 +y^2 =8~~~~~~...(i)\\\\x-y=0~~~~~~~~~...(ii)\\\\\text{From (ii):}\\\\x-y = 0 \implies x = y\\\\\text{Substitute x = y  in eq (i):}\\\\~~~~~~y^2 +y^2 =8\\\\\implies 2y^2 = 8\\ \\\implies y^2 = \dfrac 82\\\\\implies y^2 = 4\\\\\implies y = \pm\sqrt 4\\\\\implies y = \pm 2\\\\\text{Substitute y = x  in equation (i):}\\ \\~~~~~~x^2 +x^2 = 8\\ \\\implies 2x^2 = 8\\\\\implies x^2 = \dfrac 82\\\\\implies x^2 = 4\\\\\implies x = \pm\sqrt 4\\\\\implies x = \pm2\\\\

\text{Hence,}~ (x,y) = (\pm2, \pm 2)

3 0
2 years ago
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