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erastova [34]
3 years ago
15

Of the 80 house in a development, 50 have a two-car garage, 40 have an in-the-ground swimming pool, and 35 have both a two-car g

arage and an in-the-ground swimming pool. How many houses in the development have neither a two-car garage nor an in-the-ground swimming pool?A. 10B. 15C. 20D. 25E. 30
Mathematics
1 answer:
AVprozaik [17]3 years ago
6 0

Answer:

D. 25

Step-by-step explanation:

See the file attached.

Download pdf
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Solve for x, where x is a real number.
Alex787 [66]

Hello!

First of all, let's undo the square roots by squaring both sides.

6x+2=9x-8

We subtract 9x from both sides.

-3x+2=-8

We subtract 2 from both sides.

-3x= -10

We divide both sides by -3.

x=10/3

I hope this helps!

7 0
2 years ago
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{x|x€Z^x&lt;8}<br> Please someone help me what should be the elements in the set
joja [24]
What how is this a problem and how are you in middle school

5 0
3 years ago
If you have the fraction 8/25, and you are trying to turn it into a decimal, which of the following would be helpful?
AysviL [449]
In this situation you will need to do 8 divided by 25 which will give the answer 0.32.
8 0
3 years ago
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Why are asymptotes important in rational function graphs
leonid [27]

Answer:

when sketching the curves of functions.

Step-by-step explanation:

There is a wide range of graph that contain asymptotes and that includes rational functions, hyperbolic functions, tangent curves, and more. Asymptotes are important guides when sketching the curves of functions. This is why it’s important that we know the properties, general forms, and graphs of each of these asymptotes.

6 0
2 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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