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Fiesta28 [93]
2 years ago
6

Rewrite 3^-4 without negative exponents

Mathematics
1 answer:
KengaRu [80]2 years ago
5 0

Answer:

\frac{1}{3^{4} }

Step-by-step explanation:

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A person who wants to get in shape goes to a local gym that advertises 31 training sessions for $1179. Find the cost of 101 trai
lesya [120]

Answer:

$3841.26

Step-by-step explanation:

The first step is to find the price of a single session. 1179/31 is about $38.03. Multiplying this by 101, you get about $3841.26. Hope this helps!

6 0
2 years ago
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Try box be<br> 1) Solve for m: 2m + 21 = 3<br> Help please…
Viktor [21]

first step is to subtract 21 from both sides

2m+21-21=3-21

2m=-18

2m/2 and -18/2

m=-9

So the final answer is m=-9 (negative nine equals m)

7 0
3 years ago
Danae is choosing between two jobs. One job pays an annual bonus of $1,500 plus $120 per day worked. The second job pays an annu
Zigmanuir [339]

Answer:

C


Step-by-step explanation:


<u>Job A:</u>

  • Bonus of $1500
  • $120 per day for d days is 120 multiplied by d. Which is 120d
  • Total Payment = 1500+120d

<u>Job B:</u>

  • Bonus of $2500
  • $110 per day for d days is 110 multiplied by d. Which is 110d
  • Total Payment = 2500+110d

Solving these 2 equations for d would tell us after how many days both job would pay the same. Answer choice C is correct.

4 0
3 years ago
Read 2 more answers
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
I want to know the answer to this . I don’t really get it . so how would I get the answer ?
Lapatulllka [165]

Answer:

0.444... (or 44.44...%)

Step-by-step explanation:

The experimental probability is the probability of an event actually happening. The experimental probability of a man having a beard here is 44.444...% (8/18).

8 0
3 years ago
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