What is the domain of the function?

The area of the rug is 4 15/36<br> The rug is 2 3/4 meters in height <br> How wide is the rug

sineoko [7]

I assume the rug is rectangle.

The area of rectangle is width (W) multiply by hight (H)

$A=4\frac{15}{36}=4\frac{5\cdot3}{12\cdot3}=4\frac5{12}\\\\H=2\frac34\\\\\\A=W\cdot H\\\\4\frac{5}{12}=W\cdot2\frac34\\\\4\frac5{12}:2\frac34=W\\\\ \frac{4\cdot12+5}{12}:\frac{2\cdot4+3}{4}=W\\\\ \frac{53}{12}:\frac{11}4=W\\\\ \frac{53}{12}\cdot\frac4{11}=W\\\\ \frac{53\cdot4}{3\cdot4\cdot11}=W\\\\W=\frac{53}{33}\\\\W=1\frac{20}{33}$

The rug is $1\frac{20}{33}$ meters wide

4 0

4 years ago

BRAINLIEST!!!

Sonbull [250]

Answer:

$\sqrt{37}$

Step-by-step explanation:

The distance d between a point (m , n ) and a line in the form

Ax + By + C = 0 , is calculated as

d = $\frac{|Am+Bn+C|}{\sqrt{A^2+B^2} }$

Here (m, n ) = (6, 2 ) and

6x - y = - 3 ( add 3 to both sides )

6x - y + 3 = 0 → A = 6, B = - 1, C = 3

d = $\frac{|6(6)+-1(2)+3|}{\sqrt{6^2+(-1)^2} }$

= $\frac{|36-2+3|}{\sqrt{36+1} }$

= $\frac{|37|}{\sqrt{37} }$

= $\frac{37}{\sqrt{37} }$ × $\frac{\sqrt{37} }{\sqrt{37} }$

= $\frac{37\sqrt{37} }{37}$ ← cancel 37 on numerator/ denominator

= $\sqrt{37}$

7 0

3 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

Need to know this now pls !!! NEED THE PROCESS , TOO!!!<br><br> x squared - 18x = -117

nydimaria [60]

Answer:

x = 9 + 6i or x = 9 - 6i

Step-by-step explanation:

x² - 18x + 117 = 0

Complete the square and solve:

(x - 9)² - 81 + 117 = 0

(x - 9)² + 36 = 0

(x - 9)² = -36

x - 9 = ±sqrt(-36)

x - 9 = ±sqrt(-1).sqrt(36)

i = imaginary number = sqrt(-1)

x - 9 = ±6i

x = 9 ± 6i

There are no real solutions, only imaginary ones

9 + 6i and 9 - 6i

3 0

3 years ago

Read 2 more answers

What is the exponent of 50

german

2×5^{2} if you don't understand its 2 times 5 to the second power.

6 0

3 years ago