Answer: 8.5
Step-by-step explanation: 8.5 because 34 divided by 4 is 8.5. To find the unit rate, make 4 hours turn to 1 hour. To do that, divided 4 by 4. Do the same with the other number, which is 34. 34 divided 4 is equal to 8.5
The angle addition postulates states that if an angle UVW has a point S lying in its interior, then the sum of angle UVS and angle SVW must equal angle UVW, or ?UVS + ?SVW= ?UVW.
Answer:
(x+5) (x-2)
SO -5 and 2 option A
Step-by-step explanation:
factor it. and set equal to zero.
Answer:
There are 24 nickels
Step-by-step explanation:
Let x represent the number of nickels
Let y represent the number of quarters
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Value Value
Type Number of of
of of each all
Coin Coin Coin Coin
—————————————————————
Nickels | x | $0.05 | $0.05x
Quarters | y | $0.25 | $0.25y
—————————————————————
Totals 28 ——— $2.20
•••••••••••••••••••••••••••••••••••••••••••••••••
The first equation comes from the “Number of coins” column.
(Number of nickels) + (Number of quarters) = (total number of coins)
Equation: x + y = 28
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The second equation comes from the “value of all coins” column.
(Value of all nickels) + (Value of all quarters) = (Total value of all coins)
0.05x + 0.25y = 2.20
Remove the decimals by multiplying each term by 100:
5x + 25y = 220
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So we have the system of equations:
{x + y = 28
{5x + 25y = 202
Solve by substitution. Solve the first equation for y:
x + y = 28
y = 28 - x
Substitute (28 - x) for y in 5x + 25y = 220
5x + 25 (28 - x) = 220
5x + 700 - 25x = 220
-20x + 700 = 220
-20x = -480
x = 24
The number of nickels is 24.
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Substitute in y = 28 - x
y = 28 - (24)
y = 4
The number of quarters is 4.
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Checking:
24 nickels is $1.20 and 4 quarters is $1.00
That’s 28 coins.
Indeed $1.20 + $1.00 = $2.20
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Answer:
DIVIDING COMPLEX NUMBERS
<em>Dividing complex numbers is a little more complicated than addition, subtraction, and multiplication of complex numbers because it is difficult to divide a number by an imaginary number. For dividing complex numbers, we need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary part of the denominator so that we end up with a real number in the </em><em>DENOMINATOR</em>
