Help Me!!!

1 answer:

7 0

I'll just help quickly with a couple of expressions that are equivalent.
3(m+5+9m)
If you solve it out, you get 3m+15+27m.
Simply the expression to get 30m+15.
You can regroup to make another expression 5(6m+3) which is also equal.

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Olivia ate two times as many nachos as Reagan. If Reagan ate r nachos, which equation can be used to find the number of nachos O

Rzqust [24]

Step-by-step explanation:

Olivia's nachos is v.

Raegan's nachos is r.

v = 2r

5 0

3 years ago

Read 2 more answers

Emma and Mila go to the movie theater and purchase refreshments for their friends. Emma spends a total of $117.00 on 9 bags of p

yuradex [85]

For this case, the first thing we must do is define variables.
We have then:
x: cost of each bag of popcorn
y: cost of each drink
writing the system of equations we have:
9x + 9y = 117
x + 3y = 23
Solving the system of equations we have:
x = 8 $
y = 5 $
Answer:
a system of equations that can be used to find the price of one bag of popcorn and the price of one drink is:
9x + 9y = 117
x + 3y = 23
The price of a bag of popcorn is $ 8.

7 0

4 years ago

I’ll put you as brainliest

kari74 [83]

The law of an object moving with constant acceleration is

$s(t)=s_0+v_0t+\dfrac{1}{2}at^2$

Where $s$ is space, $t$ is time, $s_0$ is the initial position, $v_0$ is the initial velocity and $a$ is the acceleration.

In this case, if we choose a reference grid with the vertical axis pointing upwards, the acceleration of gravity will point downwards (and thus be negative). The initial position is zero, because the rocket is on the ground, and the initial velocity is 100 (positive because pointing upwards).

So, its law is

$h(t)=100t-\dfrac{1}{2}gt^2$

(I changed $s$ for $h$ since the rocket is moving vertically, so its position is actually its height. Also, g is the acceleration due to gravity).

The rocket hits the ground if its height is zero, so if we set $h=0$ we have

$0=100t-\dfrac{1}{2}gt^2 \iff t(100-\dfrac{1}{2}gt)=0$

Solving for t, we have either t=0, or

$100-\dfrac{1}{2}gt=0 \iff 100=\dfrac{1}{2}gt \iff 200=gt \iff t=\dfrac{200}{g}$

The solution t=0 means that at the beginning the rocket is on the ground. So, we're interested in the other solution. Considering that g is about 32.2 feet/s^2, we have

$t=\dfrac{200}{g}\approx \dfrac{200}{32.2}\approx 6.21$