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11111nata11111 [884]
3 years ago
10

Solve this problem : (X-3)^2

Mathematics
2 answers:
krek1111 [17]3 years ago
7 0

Answer:

x^2 - 6x + 9

Step-by-step explanation:

(x - 3)^2

(x - 3)(x - 3)

x^2 - 3x - 3x + 9

x^2 - 6x + 9

Rufina [12.5K]3 years ago
5 0

Answer:x^2-6x+9

Step-by-step explanation:

(x-3)^2=(x-3)(x-3)=x^2-3x-3x+9

Collect like terms

X^2-6x+9

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2 cups of milk on 1/2 a quart of milk how many cups 1 quart of milk?
Wewaii [24]

Answer:

4 cups

Step-by-step explanation:

because if you have 2 cups for a half a quart then add another half which means add 2 more cups 2+2=4 cups

4 0
3 years ago
A recipe for giant soap bubbles uses 1/2 cup dishwashing liquid and 4 1/2 cups of water. How many cups of water will be used wit
GREYUIT [131]

Answer: D. 13 1/2.​

Step-by-step explanation:

1/2*3= 1 1/2 cups of dishwashing liquid.

4 1/2*3= 13 1/2 cups of water will be used with 1 1/2 cups of dishwashing liquid.

7 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
GIVING BRAINLY TO CORRECT answer
crimeas [40]

Answer:

False

Step-by-step explanation:

Range is defined as the greatest possible number minus the lowest possible number. The highest is 150 (in thousands) and the lowest is 25, meaning the range would be 125,000 and not 150,000.

Hope this helps!

8 0
2 years ago
Read 2 more answers
What number makes the expressions equivalent? Enter your answer in the box. 1/2(–1.4m + 0.4) =__m + 0.2A) -1.4B) 1.4C) -0.7D) 0.
Snowcat [4.5K]

Answer:

C. -0.7

Explanation:

Given the equation:

\frac{1}{2}(-1.4m+0.4)=\boxed{\square}_{}m+0.2​

First, distribute the bracket on the left-hand side:

\begin{gathered} \frac{1}{2}(-1.4m)+\frac{1}{2}(0.4) \\ =-0.7m+0.2 \end{gathered}

The number that makes the given expressions equivalent is -0.7.

The correct choice is C.

6 0
1 year ago
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