Given that €1 = £0.72 a) how much is €410 in £? b) what is the £ to € exchange rate

What does -8+(-12) / (-4) equal too?

zalisa [80]

Answer:

5

Step-by-step explanation:

1. Add -8 to -12. This equals -20.

2. Now you have -20/(-4)

3. Solve this by dividing -20 and -4.

4. Two negatives make a positive, so your answer would be 5.

Hope this helps

5 0

2 years ago

Read 2 more answers

Compute the dimensions of gravitational constant G where F=Gm1m2/r^2

VARVARA [1.3K]

Answer:

G=L^3/(T^2M)

Step-by-step explanation:

G=(Fr^2)/Mm
G=(ML^3)/(T^2M^2)
G=L^3/(T^2M)

8 0

2 years ago

A carton measures 3 feet by 2 feet by 2 feet. A machine can fill the carton with packing material in 3 seconds. How long would i

True [87]

1) we calculate the volume of the first carton:
volume=length x width x height
volume=3 ft * 2 ft * 2 ft=12 ft³

Therefore:
A machine can fill 12 ft³ in 3 seconds.

2) we calculate the volume of the second carton.
volume=length x width x heigth
volume=4 ft * 5 ft * 6 ft=120 ft³

3) we calculate the time that the machine needs for fill the second carton with packing material by the rule of three.

12 ft³---------------------3 seconds
120 ft³-------------------- x

x=(120 ft³ * 3 seconds) / 12 ft³=30 seconds.

answer: 30 seconds.

5 0

3 years ago

Match each system of equations to its graph. y = 2x + 1 y = x + 2 y = 3x y = x + 3 y = 2x − 2 y = x − 2 y = 2x + 3 y = x + 5 y =

GalinKa [24]

The matchup are:

(1st picture): y=2x+3 and y=x+5
(2nd picture): y=4x+2 and y =3x+2
(3rd picture): y=2x+1 and y =x+2
(4th picture): 3x and x+3

<h3>What does the graph of an equation shows?</h3>

The graph of the linear equation is known to be one that often brings or set the points that can be found on the coordinate plane and it is one that shows all the solutions to the equation.

Note that when all variables stands for real numbers, a person can be able to use graph to show the equation and this is often done through plotting the points to show a pattern and then link up the points to have all the points.

From the above, the pictures that have all the points as shown on the graph are:

(1st picture): y=2x+3 and y=x+5
(2nd picture): y=4x+2 and y =3x+2
(3rd picture): y=2x+1 and y =x+2
(4th picture): 3x and x+3

Learn more about equations from

brainly.com/question/2972832

#SPJ1

4 0

1 year ago

The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s

oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is $A(6) = 28$ .

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable $t$ .

Notice that the interval of interest $0 \le t \le 6$ is closed on both ends. In other words, this interval includes both endpoints: $t = 0$ and $t= 6$ . Over this interval, the value of $A(t)$ might be maximized when $t$ is at the following:

One of the two endpoints of this interval, where $t = 0$ or $t = 6$ .
A local maximum of $A(t)$ , where $A^\prime(t) = 0$ (first derivative of $A(t)\!$ is zero) and $A^{\prime\prime}(t)$ (second derivative of $\! A(t)$ is smaller than zero.)

Start by calculating the value of $A(t)$ at the two endpoints:

$A(0) = 10$ .
$A(6) = 28$ .

Apply the power rule to find the first and second derivatives of $A(t)$ :

$\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}$ .

$\displaystyle A^{\prime\prime}(t) = 6\, t - 15$ .

Notice that both $t = 1$ and $t = 4$ are first derivatives of $A^{\prime}(t)$ over the interval $0 \le t \le 6$ .

However, among these two zeros, only $t = 1\!$ ensures that the second derivative $A^{\prime\prime}(t)$ is smaller than zero (that is: $A^{\prime\prime}(1) < 0$ .) If the second derivative $A^{\prime\prime}(t)\!$ is non-negative, that zero of $A^{\prime}(t)$ would either be an inflection point (if $A^{\prime\prime}(t) = 0$ ) or a local minimum (if $A^{\prime\prime}(t) > 0$ .)

Therefore $\! t = 1$ would be the only local maximum over the interval $0 \le t \le 6\!$ .

Calculate the value of $A(t)$ at this local maximum:

$A(1) = 15.5$ .

Compare these three possible maximum values of $A(t)$ over the interval $0 \le t \le 6$ . Apparently, $t = 6$ would maximize the value of $A(t)\!$ . That is: $A(6) = 28$ gives the maximum value of $\! A(t)$ over the interval $0 \le t \le 6\!$ .

However, note that the maximum over this interval exists because $t = 6\!$ is indeed part of the $0 \le t \le 6$ interval. For example, the same $A(t)$ would have no maximum over the interval $0 \le t < 6$ (which does not include $t = 6$ .)

4 0

3 years ago