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galben [10]
3 years ago
6

Find the missing number 1:2 = 3:_

Mathematics
2 answers:
Andrei [34K]3 years ago
7 0

Answer:

the answer is 6

Step-by-step explanation:

Angelina_Jolie [31]3 years ago
4 0
The answer is 6, hope this helped!
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A rocket is launched from the ground and travels in a straight path. The angle of inclination of the rocket's path is 1.2 radian
gladu [14]

Answer:

a)S=2.57

b)H=203.2yard

c)X=121.79yards

Step-by-step explanation:

From the question we are told that:

Angle \theta=1.2=68.755^o

a)

Generally the equation for Slope is mathematically given by

 S=tan \theta

 S=tan 68.755

 S=2.57

b)

Given the right Angle triangle with horizontal distance x=85yard

Generally the equation for Height traveled is mathematically given by

 H=tan\theta*x

 H=2.57*85

 H=218.45

 H=203.2yard

c)

Generally the equation for Horizontal Distance traveled at 313 height traveled is mathematically given by

X=\frac{313}{tan65.75}

X=\frac{313}{2.57}

X=121.79yards

3 0
3 years ago
Jasmine knows that the area of a rectangle is the product of its base and height. Help her write an expression that represents t
Andrews [41]

Answer:

8b and 80

Step-by-step explanation:

6 0
3 years ago
Did i do this correctly?
den301095 [7]
Yes u did do this correctly
5 0
3 years ago
Read 2 more answers
How do you find the inverse of this function?
ICE Princess25 [194]
Solve for x then replace x with f⁻¹(x) and y with x


y=(4^x-5)/3
times 3
3y=(4^x-5)
add 5
3y+5=4^x
take log₄ both sides
log₄(3y+5)=x
switch
f⁻¹(x)=log₄(3y+5)
5 0
3 years ago
Consider the region, R, bounded above by f(x)=x2−6x+9 and g(x)=−3x+27 and bounded below by the x-axis over the interval [3,9]. F
Salsk061 [2.6K]

Answer:

22.5

Step-by-step explanation:

The region R contains every point of the plane with coordinate x between 3 and 9, and with coordinate y positive such that y < f(x) and y < g(x).

We can note that both f and g are positive on [3,9] because g is a decreasing linear function and g(9) = 0, thus g is positive in every other point of the interval, and f(x) = (x-3)^2 is always positive excpept when x = 3, where it reaches the value 0.

The interception of the graphs takes place for a value x such that f(x) = g(x).

We compute x^2-6x+9 = -3x + 27, thus x^2-6x+9-(-3x + 27) = x^2-3x -18 = 0.

The roots of that quadratic function are

r_1, r_2 = \frac{3 ^+_- \sqrt { 9 +72}}{2} = \frac{3^+_-9}{2} , thus r1 = 6, r2 = -3. We dont care about -3 because it is outside the interval, but we know that f and g graphs intersects on x = 6. Thus, we obtain, due to Bolzano Theorem:

  • On the interval [3,6), the function f in smaller because it takes the value 0 on x=3, while g is always positive.
  • On the interval (6,9]. the function g is smaller because it takes the value 0 on x=9, while f is always positive

Hence, the upper bound is f on the interval [3,6) and g on the interval (6,9]. While the lower bound is the 0 function.

We need to calculate the following integral, using Barrow's rule

\int\limits^6_3 {x^2-6x+9} \, dx + \int\limits^9_6 {-3x+27} \, dx = (\frac{x^3}{3} - 3x^2 + 9x) |^6_3 + (\frac{-3x^2}{2} + 27x)|^9_6 = \\  (18 - 9) + (121.5-108) = 22.5

As a result, the area of the region R is 22.5

6 0
3 years ago
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