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Svet_ta [14]
3 years ago
14

Equivalent to 10 minus 8

Mathematics
1 answer:
Reptile [31]3 years ago
5 0
1+1


Hope this helps!
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Find the discriminant of the following equation.<br><br> 4x2 + 12x + 9 = 0
oksano4ka [1.4K]
For
ax^2+bx+c=0
discriminatn is b^2-4ac

a=4
b=12
c=9
discriinat is
12^2-4(4)(9)=
144-144
0

the discrminat is 0

8 0
3 years ago
A car is traveling up a steep ramp to a parking garage.the ramp is 445 feet long and rises a vertical distance of 80 feet. What
Alecsey [184]
With the information provided in the problem, we can create a right triangle with the ramp as its hypotenuse, and  the vertical rise as its opposite side from its angle of elevation.
Let E be the angle of elevation from the car to the end of the ramp. We now know that the hypotenuse of our triangle measures 445 feet, and the opposite side measures 80 feet, so we need a trig function that relates our angle of elevation with the hypotenuse and the opposite side. That function is sine:
sine(E)= \frac{80}{445}
E=arcsine( \frac{80}{445} )
E=10.36

We can conclude that the angle of elevation from the car to the end of the ramp is 10.36°.

3 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 need help for bi dont get it
sergij07 [2.7K]

Answer:

EWWWWWWWWWWWWW UR BI???!!

Step-by-step explanation:

7 0
3 years ago
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Use the image to answer the question. Use the drop-down menus to complete the statements. It isfor Acacia to pick a purple piece
I am Lyosha [343]

Answer:

1. less likely

2. more likely

Step-by-step explanation:

test took

3 0
3 years ago
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