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jarptica [38.1K]
3 years ago
10

Celeste has $40 she plans on using to pay for lunch with a friend. She always tips 17%. What is the maximum amount she can spend

on the food, so she can be sure to have enough left over to cover the tip? Round to the nearest cent.
Mathematics
1 answer:
irga5000 [103]3 years ago
5 0
Celeste can only spend 33.2 on food
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Tristen's cat had 3 baby kittens. Each kitten weighed 1 pounds. How many ounces did all of his kittens weigh together?​
Tema [17]
3 times 1. 3 pounds in all
5 0
3 years ago
An inequality is shown.
kykrilka [37]

Answer:

Step-by-step explanation:

subtracting 3 from both sides 3 x − 9 = 0 3 x − 9 + 9 = 0 + 9

Adding 9 to both sides 3 x = 9 3 x 3=9 3

Dividing both sides by 3

should give you 3

5 0
2 years ago
Richard has a rectangular plot of land that is 525 feet long and y feet wide. He decides to build a fence around the plot. If th
mart [117]

Answer:

Step-by-step explanation:

P= 2(L) + 2(w)

1504= 2(525) + 2(w)

1540= 1050 + 2(w)

1540 - 1050= 2(w)

490 = 2(w)

490/2 = w

245 = w

w is y

L is 525

5 0
3 years ago
Read 2 more answers
X-2y=-26 x-y=-2 solution of systems equation
svp [43]
If we subtract the 2 equations we eliminate x:-

x - x - 2y - (-y) = -26--2
-y = -24 
 y = 24
Plug this into the first equation:-
x - 2(24) = -26
x = -26 + 48  = 22


Answer is x = 22 and y = 24
4 0
4 years ago
NEED HELP ASAP!!
Solnce55 [7]

Answer:

Only Cory is correct

Step-by-step explanation:

The gravitational pull of the Earth on a person or object is given by Newton's law of gravitation as follows;

F =G\times \dfrac{M \cdot m}{r^{2}}

Where;

G = The universal gravitational constant

M = The mass of one object

m = The mass of the other object

r = The distance between the centers of the two objects

For the gravitational pull of the Earth on a person, when the person is standing on the Earth's surface, r = R = The radius of the Earth ≈ 6,371 km

Therefore, for an astronaut in the international Space Station, r = 6,800 km

The ratio of the gravitational pull on the surface of the Earth, F₁, and the gravitational pull on an astronaut at the international space station, F₂, is therefore given as follows;

\dfrac{F_1}{F_2} = \dfrac{ \dfrac{M \cdot m}{R^{2}}}{\dfrac{M \cdot m}{r^{2}}} = \dfrac{r^2}{R^2}  = \dfrac{(6,800 \ km)^2}{(6,371 \ km)^2} \approx  1.14

∴ F₁ ≈ 1.14 × F₂

F₂ ≈ 0.8778 × F₁

Therefore, the gravitational pull on the astronaut by virtue of the distance from the center of the Earth, F₂ is approximately 88% of the gravitational pull on a person of similar mass on Earth

However, the International Space Station is moving in its orbit around the Earth at an orbiting speed enough to prevent the Space Station from falling to the Earth such that the Space Station falls around the Earth because of the curved shape of the gravitational attraction, such that the astronaut are constantly falling (similar to falling from height) and appear not to experience gravity

Therefore, Cory is correct, the astronauts in the International Space Station, 6,800 km from the Earth's center, are not too far to experience gravity.

6 0
3 years ago
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