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svp [43]
4 years ago
13

What is the equation of this circle in general form?

Mathematics
1 answer:
jekas [21]4 years ago
5 0

Answer:

The answer is B

Step-by-step explanation:

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How many one-half cubes with dimensions of 1/2x1x1 fit in a unit cube? 1
yanalaym [24]

Answer:

Option 2

Step-by-step explanation:

A cube with 1/2 * 1* 1 will have a volume of 0.5 cubic cm

While a unit cube has a volume of 1 cubic cm

So,

2 cubes of these dimensions can fit into a unit cube.

5 0
3 years ago
Read 2 more answers
What is 0.675 in its simplest form?
Musya8 [376]
27 over 40 (as a fraction)

WORK:
.675 is 675 over 1000
5 goes into both of those so
675/5=135 1000/5=200
135/5=27
200/5=40
3 0
3 years ago
The solutions of a quadratic equation are ​-4,9 Write a related quadratic function in factored form.
Yuri [45]

Answer:

y = (x + 4)(x - 9)

Step-by-step explanation:

x = {-4, 9}

y = (x + 4)(x - 9)

3 0
3 years ago
En un triángulo rectángulo A es un ángulo agudo y Sen A = 4/5 ¿Cuál será el valor de Tan A?
Nonamiya [84]

Answer:

\displaystyle \tan A=\frac{4}{3}

Step-by-step explanation:

<u>Funciones Trigonométricas</u>

La identidad principal en trigonometría es:

sen^2A+cos^2A=1

Si sabemos que A es un ángulo agudo (que mide menos de 90°), su seno y coseno son positivos.

Dado que Sen A = 4/5, calculamos el coseno:

cos^2A=1-sen^2A

Sustituyendo:

\displaystyle cos^2A=1-\left(\frac{4}{5}\right)^2

\displaystyle cos^2A=1-\frac{16}{25}

\displaystyle cos^2A=\frac{25-16}{25}

\displaystyle cos^2A=\frac{9}{25}

Tomando raíz cuadrada:

\displaystyle cos\ A=\sqrt{\frac{9}{25}}=\frac{3}{5}

La tangente se define como:

\displaystyle \tan A=\frac{sen\ A}{cos\ A}

Substituyendo:

\displaystyle \tan A=\frac{\frac{4}{5}}{\frac{3}{5}}

\displaystyle \tan A=\frac{4}{3}

6 0
3 years ago
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'
vodka [1.7K]

Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

5 0
3 years ago
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