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Tanzania [10]
3 years ago
10

Why does math homework exist?

Mathematics
2 answers:
mars1129 [50]3 years ago
6 0

Answer:

to annoy students

Step-by-step explanation:

lol just my opinion

gayaneshka [121]3 years ago
5 0

Answer:

Homework creates a bridge between school and home. Parents rarely get to spend much time with you while you're at school. Homework allows them to keep up with what you're doing in your classes on a daily basis. But you don't have homework purely for your parents' benefit. It's good for you, too, or maybe because teachers want to see us suffer

Step-by-step explanation:

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Alina [70]
It’s 3 loads and $195 because you multiply 52 by 2 to get the total amount of bricks and then you divide 104 by 45 and you have a remainder of 14 which would be the last load and then divide 65 or whatever the price of the load of bricks to get 195
7 0
3 years ago
The length of one leg of a right triangle is 24 feet and the length of the hypotenuse is 40 ft what is the length of the other l
Rama09 [41]
To solve for the other leg, you need to use Pythagorean theory.

The theory states that C^2 = A^2 + B^2

In your case, you have to solve for B.
C = 40 (hypotenuse)   A = 24 (leg 1)

The formula rewritten for B is b= \sqrt{c^{2}-a^{2} }

Now, you can solve.

b= \sqrt{c^{2}-a^{2} }

b= \sqrt{40^{2}-24^2 }

b = \sqrt{1600 - 576}

b = \sqrt{1024}

b = 32

Therefore, B = 32.

So the second leg is 32 feet.
7 0
3 years ago
Help me can you help with the second part
lapo4ka [179]

Answer:

12 packages with 2 toothbrushes and 3 soaps each

Step-by-step explanation:

5 0
3 years ago
Find the missing length indicated
Irina18 [472]

Answer:

144 hope it helps

Step-by-step explanation:

8 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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