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3 years ago

Starting time 2:15 am elapsed time 45 minutes

1 answer:

4 0

One hour = 60 minutes
60-15=45 so I would just add your 45 minutes to your 2:15
2:15+:45= 3:00

so your answer will be 3:00 :)

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HELPPP THIS IS DUE SOON! I WILL MARK BRAINLIEST

expeople1 [14]

Answer:

7.3

Step-by-step explanation:

For this question, you must use the distance formula. The distance formula is based around Pythagorean Theorem, so you will see some similarities.

Distance = $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

X2 is 7 (X coordinate of school)

X1 is 5 (X coordinate of friend's house)

7 - 5 = 2

2^2 = 4

Y2 is 7 (Y coordinate of school)

Y1 is 0 (Y coordinate of friend's house)

7 - 0 = 7

7^2 = 49

4 + 49 = 53

$\sqrt{53}$ = 7.2801...

<em>Round to the nearest tenth...</em>

4 0

3 years ago

Read 2 more answers

How do you redraw a scale drawing with a new scale

rjkz [21]

Take measurements of what you want to draw, (e.g. a simple cube that is 100 cm x 100 cm x 100 cm.)
A full scale drawing would require you to draw the cube on a paper 100 cm x 100 cm x 100 cm.
If you want to 'scale', you would then redraw using new measurements. So, a 1/10 scale drawing would mean that you would redraw the cube on paper as 10 cm x 10 cm x 10 cm.
A 1/20 scale would be 5 cm x 5 cm x 5 cm. The same concept holds for all scale drawings.

3 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

The sum of two polynomials is –yz2 – 3z2 – 4y + 4. If one of the polynomials is y – 4yz2 – 3, what is the other polynomial? –2yz

marishachu [46]

The first polynomial is $y -4yz^2-3$ and let the second polynomial be P(y,z).

If the sum of two polynomials is $-yz^2-3z^2-4y+4$ , then $y-4yz^2-3+P(y,z)=-yz^2-3z^2-4y+4$ .

Take the first polynomial from the left side to the right side:

$P(y,z)=-yz^2-3z^2-4y+4 -(y-4yz^2-3),\\ P(y,z)=-yz^2-3z^2-4y+4 -y+4yz^2+3,\\ P(y,z)=3yz^2-3z^2-5y+7$ .

Answer: the second polynomial is $3yz^2-3z^2-5y+7$ .

3 0

3 years ago

Read 2 more answers

Susan went shopping to purchase new clothes. She

arlik [135]

Answer:

13.375

Step-by-step explanation:

you have to add 28 ± 17 = 45 ± 8.857 = 13.375

6 0

3 years ago