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Alex
2 years ago
9

Convert 3.2 repenting decimal into a fraction

Mathematics
1 answer:
Olenka [21]2 years ago
7 0
x =3.\bar{2}\\10x = 32.\bar{2}\\\\ 10x - x = 32.\bar{2} - 3.\bar{2}\\9x = 29\\x = \frac{29}{9}\\\\\\ \boxed{\bf{\frac{32}{9} = 3.\bar{2}}}
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Let f be a continuous function on the closed interval [ − 2 , 5 ]. If f((-2)=-7 and f(5)=1, then the Intermediate Value Theorem
azamat

Answer:

The function has at least 1 zero within the interval [-2,5].

Step-by-step explanation:

The intermediate value theorem states that, for a function continuous in a certain interval [a,b], then the function takes any value between f(a) and f(b) at some point within that interval.

This theorem has an important consequence:

If a function f(x) is continuous in an interval [a,b], and the sign of the function changes at the extreme points of the interval:

f(a)>0\\f(b) (or viceversa)

Then the function f(x) has at least one zero within the interval [a,b].

We can apply the theorem to this case. In fact, here we have a function f(x) continuous within the interval

[-2,5]

And we also know that the function changes sign at the extreme points of the interval:

f(-2)=-70

Therefore, the function has at least 1 zero within the interval [-2,5], so there is at least one point x' within this interval such that

f(x')=0

6 0
3 years ago
The kittens, Annie and Josie, are pushing a ball, Annie with a force magnitude of 80 N in a direction of 133 degrees, and Josie
kvv77 [185]

Answer: Then the magnitude of the force is 37.86N, and the direction is 54.35°

Step-by-step explanation:

We can write the forces as vectors.

We know that Annie pushes with a magnitude of 80N in a direction of 133° (Remember that the angles are always measured from the x-axis)

The components of this force, (Ax, Ay), are then:

Ax = 80N*cos(133°)

Ay = 80N*sin(133°)

And we know that Josie pushes with a magnitude of 95N in direction of 290°

The components of this force, (Jx, Jy), are:

Jx = 95N*cos(290°)

Jy = 95N*sin(290°)

When we add these forces, the total force acting on the ball is:

F = (80N*cos(133°) ,  80N*sin(133°)) + (95N*cos(290°), 95N*sin(290°))

F = (80N*cos(133°) + 95N*cos(290°), 80N*sin(133°) + 95N*sin(290°))

Now, the third kitten wants to do a force K, in a direction θ, such that the net force acting on the ball is zero.

Then we must have that, each component of the force of the third cat (K*cos(θ)  on the x-axis and k*sin(θ) on the y-axis), is such that:

K*cos(θ) + 80N*cos(133°) + 95N*cos(290°) = 0

k*sin(θ)  + 80N*sin(133°) + 95N*sin(290°) = 0

Now we need to solve that system for k and θ

if we simplify the equations we get:

k*cos(θ) - 22.07N = 0

k*sin(θ)  -30.76N  = 0

Now we can rewrite them as:

k*cos(θ) = 22.07N

k*sin(θ)   = 30.76N  

Now we can take the quotiet between both equations to get:

(k*sin(θ))/(k*cos(θ)) = 30.76N/22.07N

Tan(θ) = 1.394

θ = Atan(1.394) = 54.35°

Now that we know the angle, we can find the value of the magnitude k, by using one of the two equations of the system:

k*cos(54.35°) = 22.07N

k = 22.07N/cos(54.35°) = 37.86N

Then the magnitude of this force is 37.86N, and the direction is 54.35°

7 0
2 years ago
Consider the curve defined by 2x2+3y2−4xy=36 2 x 2 + 3 y 2 − 4 x y = 36 .
8_murik_8 [283]

Answer:

Step-by-step explanation:

Given a curve defined by the function 2x²+3y²−4xy=36

The total differential of this function with respect to a variable x makes the function an implicit function because it contains two variables.

Differentiating both sides of the equation with respect to x we have:

4x+6ydy/dx-(4xd(y)/dx+{d(4x)/dx(y))} = 0

4x + 6ydy/dx -(4xdy/dx +4y) = 0

4x + 6ydy/dx - 4xdy/dx -4y = 0

Collecting like terms

4x-4y+6ydy/dx - 4xdy/dx = 0

4x-4y+(6y-4x)dy/dx = 0

4x-4y = -(6y-4x)dy/dx

4y-4x = (6y-4x)dy/dx

dy/dx = (4y-4x)/6y-4x

dy/dx = 2(2y-2x)/2(3y-2x)

dy/dx = 2y-2x/3y-2x proved!

5 0
3 years ago
Find the number of permutations in the word “embarrass”.
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4 I think i'm sorry if I get it wrong.  Hope I helped ≥ω≤
7 0
3 years ago
Please help due tomorrow Fri
ch4aika [34]
5 feet is the answer
3 0
3 years ago
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