Jake spent a total of 70 cents.
b = black-and-white = 8 cents
c = color = 15 cents
70 = 8b + 15c
he made a total of 7 copies
b + c = 7
system of equation:
70 = 8b + 15c
b + c = 7
--------------------------
b + c = 7
b + c (-c) = 7 (-c)
b = 7 - c
plug in 7 - c for b
70 = 8(7 - c) + 15c
Distribute the 8 to both 7 and - c (distributive property)
70 = 56 - 8c + 15c
Simplify like terms
70 = 56 - 8c + 15c
70 = 56 + 7c
Isolate the c, do the opposite of PEMDAS: Subtract 56 from both sides
70 (-56) = 56 (-56) + 7c
14 = 7c
divide 7 from both sides to isolate the c
14 = 7c
14/7 = 7c/7
c = 14/7
c = 2
c = 2
---------------
Now that you know what c equals (c = 2), plug in 2 for c in one of the equations.
b + c = 7
c = 2
<em>b + (2) = 7
</em><em />Find b by isolating it. subtract 2 from both sides
b + 2 = 7
b + 2 (-2) = 7 (-2)
b = 7 - 2
b = 5
Jake made 5 black-and-white copies, and 2 color copies
hope this helps
In order to solve this, you have to set up a systems of linear equations.
Let's say that children = c and adults = a
30a + 12c = 19,080
a + c = 960
I'm going to show you how to solve this system of linear equations by substitution, the easiest way to solve in my opinion.
a + c = 960
- c - c
---------------------- ⇒ Step 1: Solve for either a or c in either equation.
a = 960 - c
20(960 - c)+ 12c = 19,080
19,200 - 20c + 12c = 19,080
19,200 - 8c = 19,080
- 19,200 - 19,200
---------------------------------- ⇒ Step 2: Substitute in the value you got for a or c
8c = -120 into the opposite equation.
------ ---------
8 8
c = -15
30a + 12(-15) = 19,080
30a - 180 = 19,080
+ 180 + 180
-------------------------------
30a = 19,260
------- -----------
30 30
a = 642
__________________________________________________________
I just realized that there can't be a negative amount of children, so I'm sorry if these results are all wrong.
Two figures are similar if one is the scaled version of the other.
This is always the case for circles, because their geometry is fixed, and you can't modify it in anyway, otherwise it wouldn't be a circle anymore.
To be more precise, you only need two steps to prove that every two circles are similar:
- Translate one of the two circles so that they have the same center
- Scale the inner circle (for example) unit it has the same radius of the outer one. You can obviously shrink the outer one as well
Now the two circles have the same center and the same radius, and thus they are the same. We just proved that any two circles can be reduced to be the same circle using only translations and scaling, which generate similar shapes.
Recapping, we have:
- Start with circle X and radius r
- Translate it so that it has the same center as circle Y. This new circle, say X', is similar to the first one, because you only translated it.
- Scale the radius of circle X' until it becomes
. This new circle, say X'', is similar to X' because you only scaled it
So, we passed from X to X' to X'', and they are all similar to each other, and in the end we have X''=Y, which ends the proof.
Answer:
Step-by-step explanation:
Part A
In applying the synthetic division, the polynomial would be fully expressed as
2x³ + 0x² + 5x - 6
The coefficients are 2, 0, 5 and - 6
The working becomes
2| 2 0 5 -6
4 8 16
2 4 13 10
The student that correctly used synthetic division to find f(2) is Claudia
Part B
The value of f(2) is
2(2)^3+5(2) - 6
= (2 × 8) + 10 - 6
= 16 + 10 - 6 = 20
Part C
x - 2 is not a factor of 2x³ + 5x - 6 because there is a remainder of 20.
Part D
The methods that can be used to find out if x - 2 is a factor of 2x³ + 5x - 6 are synthetic division, long division, remainder theorem
