All together: <span>3x + 0.05 + x + x + 0.10 = 7.50
5x + 0.15 = 7.50
5x = 7.35
x = 1.47
</span>1.47 +0.10=1.57<span>
so, the drink costs 1.57 </span>
October has 31 days.
31 days x 24 hours in a day x 60 minutes in one hour
= 44,640 minutes.
x = the first number
(x/2)² = 14
Hope this helps!
Can I get brainliest? thanks.
Okay, this is a simplified version of how to solve this complicated equation.
First add like terms but in this equation is not very efficient when solving.
So subtract the -2x squared from both sides and the equation should like mine
x⁴ + 4x³ + 12x + 45 = x ² -2x ² + 6x + 9
So keep the left side of the equation the same, but change the right like this because your adding the x squared.
-x² + 6x + 9
Now go back to the whole equation
x⁴ + 4x³ + 12x + 45 = -x² + 6x + 9
You can’t do anything on the left side of the equation do the same process with the other numbers like this:
x⁴ + 4x³ + 12x = -x² + 6x + 9 -45
You subtract the 45 from each side and it’s negative on the other side because you subtracted it.
Now add the 9 and -45 = -36
Now the equation should look like this:
x⁴ + 4x³ + 12x = -x² + 6x -36
Subtract -12x from each side and the 12x on the left side of the equation is canceled out and is gone
The equation should look like this:
x⁴ + 4x³ + = -x² + 6x -36 -12x
Then add the 6x and -12x because they are like terms
So this is the final simplified version of the complex problem
x⁴ + 4x³ + = -x² + -36 -6x
You can’t add any other like terms to find the value of x because x and x² can’t be added together
Answer:
The red arrow shows the resultant vector. We have a Side Angle Side triangle ABC so can use The Cosine Rule:
a2=b2+c2−2bccosA
This becomes:
R2=7002+402−(2×700×40×cos45)
R2=491,600−39,597.9
R=672.3xkm/hr
This is the groundspeed of the aircraft.
To find θ we can use The Sine Rule:
sinCc=sinAa
This becomes:
sinθ40=sin45672.3
sinθ=0.04207
θ=2.41∘
This is known as the drift angle and is the correction the pilot should apply to remain on course.
The heading is the direction the aircraft's nose is pointing which is 000∘.
The track is the actual direction over the ground which is 357.6∘
An alternative method to this would be to separate each vector into vertical and horizontal components and add.
The resultant can be found using Pythagoras.
