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3 years ago

WILL GIVE 50 POINTS FOR CORRECT ANSWER ASAP AND WILL GIVE BRAINLIEST.

Mathematics

2 answers:

Rufina [12.5K]3 years ago

8 0

Answer:

(2, -1)

Step-by-step explanation:

because that's where the lines intersect

ladessa [460]3 years ago

8 0

(2,-1)

The solution is where the lines intersect

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Solve the system of equations by graphing.

k0ka [10]

Hope this helps lozzzzzzs

3 0

3 years ago

The sum of two numbers is 14 there diffrence is 6

Alexus [3.1K]

the two numbers are 10 and 4 because 10+4= 14 and 10-4=6

6 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

Please help me solve this 5^2x=7^x+1

denis23 [38]

Answer:

$x=\frac{ln7}{2ln5-ln7}$

Step-by-step explanation:

5^2x=7^x+1 ⇔ ln5^2x = ln7^x+1 ⇔ 2xln5 = (x+1)ln7 ⇔ (2ln5-ln7)x = ln7

⇔ x = ln7/(2ln5-ln7)

7 0

3 years ago

Read 2 more answers

For the following demand equation compute the elasticity of demand and determine whether the demand is elastic, unitary, or inel

adelina 88 [10]

Answer:

Demand is inelastic at p = 9 and therefore revenue will increase with

an increase in price.

Step-by-step explanation:

Given a demand function that gives <em>q</em> in terms of <em>p</em>, the elasticity of demand is

$E=|\frac{p}{q}\cdot \frac{dq}{dp} |$

If E < 1, we say demand is inelastic. In this case, raising prices increases revenue.
If E > 1, we say demand is elastic. In this case, raising prices decreases revenue.
If E = 1, we say demand is unitary.

We have the following demand equation $D(p)=-\frac{3}{4}p+29$ ; p = 9

Applying the above definition of elasticity of demand we get:

$E(p)=\frac{p}{q}\cdot \frac{dq}{dp}$

where

p = 9
q = $-\frac{3}{4}p+29$
$\frac{dq}{dp}=\frac{d}{dp}(-\frac{3}{4}p+29)$

$\frac{d}{dp}\left(-\frac{3}{4}p+29\right)=-\frac{d}{dp}\left(\frac{3}{4}p\right)+\frac{d}{dp}\left(29\right)\\\\\frac{d}{dp}\left(-\frac{3}{4}p+29\right)=-\frac{3}{4}$

Substituting the values

$E(9)=\frac{9}{-\frac{3}{4}(9)+29}\cdot -\frac{3}{4}\\\\E(9)=\frac{36}{89}\cdot -\frac{3}{4}\\\\E(9)=-\frac{27}{89}\approx -0.30337$

$|E(9)|=|\frac{27}{89}| < 1$

Demand is inelastic at p = 9 and therefore revenue will increase with an increase in price.

6 0

3 years ago