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Romashka [77]
2 years ago
13

Hey there! BRAINLIEST! Lol, because I have no life whatsoever, but um if your on this question. Thank you. Please help me out! I

appreciate you! We need more people in this world like youu!

Mathematics
2 answers:
Lerok [7]2 years ago
6 0

Answer:

LCD=5 Final answer= 5/5 -> 5/1 -> to 5

Step-by-step explanation:

Anni [7]2 years ago
4 0

Answer:

LCD=5 Final answer= 5/5 -> 5/1 -> to 5

Step-by-step explanation:

First, find the LCD(least common denominator) which is 5. This is done listing the multiples of each number in this case the lowest number is 5.

3/5 will stay the same but for 1/10 you need to convert it to the fraction 2/5 Next add the fractions together to get 3/5 + 2/5 = 5/5 of 5/1 or just 5

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The product of 40 and distance to the finish line
AleksAgata [21]
Well, We have to Know What The Distance is to the Finish Line. product of is multiplication.
7 0
2 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
Does the given ordered pair, (-3,5/2) satisfies the given equation. 5x-2y=5 ?
anzhelika [568]

Answer:

No, it does not.

5 0
3 years ago
(Algebra II) Standard Form of a Quadratic Function - I'm having trouble with two questions on this quiz? Please don't just give
anastassius [24]
1.
the x value of the vertex in form
ax^2+bx+c=y
is
-b/2a
so

-2x^2+8x-18
x value of vertex is
-8/(2*-2)=-8/-4=2

plug it in to get y value
-2(2)^2+8(2)-18
-2(4)+16-18
-8-2
-10

vertex is at (2,-10)
or you could complete the square to get into y=a(x-h)^2+k, where the vertex is (h,k)
so as follows
y=(-2x^2+8x)-18
y=-2(x^2-4x)-18
y=-2(x^2-4x+4-4)-18
y=-2((x-2)^2-4)-18
y=-2(x-2)^2+8-18
y=-2(x-2)^2-10
vertex is (2,-10)






5.
vertex is the time where the speed is the highest
at about t=10, the speed is at its max
6 0
3 years ago
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suppose you have 76 feet of fencing to enclose a rectangular dog pen. The function A=38x-x^2, where x= width, gives you the area
yan [13]
If A=38x-x^2 then

dA/dx=38-2x

d2A/dx2=-2

Since the acceleration, d2A/dx2 is a constant negative, when velocity, dA/dx=0, it will be an absolute maximum for A(x)

dA/dx=0 only when 38=2x, x=19

A(19)=38(19)-19^2

A(19)=722-361

A(19)=361 ft^2

So the maximum possible area is 361 ft^2

(This will always be true as the maximum possible area enclosed by a given amount of material will always be a perfect square...)
3 0
3 years ago
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