Answer:
x=4/15
Step-by-step explanation:
first you need to make common denominators for 1/5 and 1/3
the common denominator is 15
5*3=15, 3*5=15. 1*3=3, 1*5=5
new equation:3x-3/15=x+5/15
now subtract x from both sides: 3x-3/15=x+5/15
-x -x
2x-3/15=5/15
now you can add 3/15 to 5/15
2x=8/15
now for the last step, divide by 2 because there are 2 x's
2x/2=x 8/15 /2=4/15
x=4/15
Answer:
No they don’t have enough money
Step-by-step explanation:
No:
54+20=74=that’s not enough
211+189=400
Thats enough but they’re are two of them so it would only be enough for one of them
I HOPE THIS IS WHAT YOU’RE LOOKING FOR AND IF NOT THEN I’M SO SO SORRY
Answer:
a = -8
Step-by-step explanation:
19 = -3a - 5
Add 5 to both sides.
24 = -3a
Divide both sides by -3.
-8 = a
Switch sides.
a = -8
Answer:
The answer to this question can be defined as follows:
Step-by-step explanation:
Given:
Following are the graph attachment to this question:
The second function, that is 
is not even a function.
Remember that g(x) function is the inverted f(x) function. And when you see this pattern, a reflection on the Y-axis expects you.
Reflection in the axis.
In x-axis:
Increase the function performance by -1 to represent an exponential curve around the x-axis.
In y-axis:
Increase the input of the function by -1 to represent the exponential function around the y-axis.
Answer:
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
Step-by-step explanation:
If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -
x = ( 30 cos 20° )( time ),
y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2
To determine " ( 30 cos 20° )( time ) " you would do the following calculations -
( x = 30 * 0.93... = ( About ) 28.01t
This represents our horizontal distance, respectively the vertical distance should be the following -
y = 30 * 0.34 - 4.9t^2,
( y = ( About ) 10.26t - 4.9t^2 + 2
In other words, our solution should be,
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
<u><em>These are are parametric equations</em></u>
