Problem 1
The two angles 6x and 30 are vertical angles, so they are congruent or the same in measure.
6x = 30
x = 30/6
x = 5 is the answer
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Problem 2
The two angles form a straight angle which is 180 degrees. We consider them a linear pair (since they are adjacent and supplementary).
So,
5x+(3x+12) = 180
8x+12 = 180
8x = 180-12
8x = 168
x = 168/8
x = 21 is the answer
Answer:
325 pencils, 650 markers, 975 pens
Step-by-step explanation:
in picture.
Answer:
W= a number greater then -3= -2,-1,0,1,2,3,4,5,6,7,8,9,10...
Step-by-step explanation:
If you use this equation then you say that the ground is h=0 and solve as a quadratic.
The quadratic formula is (-b±<span>√(b^2-4ac))/2a when an equation is in the form ax^2 + bx + c
So the equation you have been given would be -16t^2-15t-151 = 0
This equation has no real roots which leads me to believe it is incorrect.
This is probably where your difficulty is coming from, it's a mistake.
The equation is formed from S=ut+(1/2)at^2+(So) where (So) is the initial height and S is the height that you want to find.
In this case you want S = 0.
If the initial height is +151 and the initial velocity and acceleration are downwards (negative) and the initial velocity (u) is -15 and the initial acceleration is -32 then you get the equation S=-15t-16t^2+151
Solving this using the quadratic formula gives you t = 2.64 or t = -3.58
Obviously -3.58s can't be the answer so you're left with 2.64 seconds.
Hope this makes sense.
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Answer:
Y≤-3/4x+2
Step-by-step explanation:
less than because the bottom of the line is shaded
equal to because the line is solid
slope you go down three and go right four which is negative
y intercept is where x=0
