9514 1404 393
Answer:
$7.14
Step-by-step explanation:
Let p, d, q represent the numbers of pennies, dimes, and quarters in the collection, respectively.
p + d + q = 45 . . . . . . . . there are 45 coins in the collection
2p +5 = q . . . . . . . . . . . . 5 more than twice the number of pennies
p + 4 = d . . . . . . . . . . . . . 4 more than the number of pennies
Substituting the last two equations into the first gives ...
p +(p +4) +(2p +5) = 45
4p = 36 . . . . . . . . . . . . . subtract 9
p = 9 . . . . . . . . . . . divide by 4
d = 9 +4 = 13
q = 2(9) +5 = 23
The value of the collection is ...
23(0.25) +13(0.10) +9(0.01) = 5.75 +1.30 +0.09 = 7.14
The coin collection is worth $7.14.
Answer: i cant see the pic i would help but i cant see picture
Step-by-step explanation:
When you ask "what do you add to A to get B, you solve it by subtracting.
For example, what do you add to 5 to get 13?
Subtract 13 - 5 = 8
The answer is you add 8 to 5 to get 13.
Now you need to do the same with your polynomials.
What do you add to <span>x^2 - 2x + 6 to get 3x^2 + 7x?
To find the answer, subtract
(3x^2 + 7x</span>) - (x^2 - 2x + 6)
Answer: (x-2)^2+(y+3)^2 = 9Side notes
1) This circle has a center of (2,-3)
2) The radius of this circle is 3
3) The graph is shown in the attached image
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Work Shown:
x^2-4x+y^2+6y+4=0
x^2-4x+y^2+6y+4-4=0-4
x^2-4x+y^2+6y = -4
x^2-4x+4+y^2+6y = -4+4 ... see note 1 below
(x^2-4x+4)+y^2+6y = 0
(x-2)^2+y^2+6y = 0
(x-2)^2+y^2+6y+9 = 0+9 ... see note 2 below
(x-2)^2+(y^2+6y+9) = 9
(x-2)^2+(y+3)^2 = 9note 1: I'm adding 4 to both sides to complete the square for the x terms. You do this by first taking half of the x (not x^2) coefficient which in this case is -4. So take half of -4 to get -2. Then square this result to get 4
note 2: Like with note 1, I'm completing the square. What's different this time is that this is for the y terms now. The y coefficient is 6. Half of this is 3. Square 3 to get 9. So this is why we add 9 to both sides.
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So
the equation in standard form is (x-2)^2+(y+3)^2 = 9Note how
(x-2)^2+(y+3)^2 = 9
is equivalent to
(x-2)^2+(y-(-3))^2 = 3^2
So that second equation listed above is in the form (x-h)^2+(y-k)^2 = r^2
where
h = 2
k = -3
r = 3
making the center to be (h,k) = (2,-3) and the radius to be r = 3
The graph is attached.