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MakcuM [25]
3 years ago
10

Do not solve the system below. Its solution is (x, y)

Mathematics
1 answer:
Natalija [7]3 years ago
8 0
Answer:

(x,y) = (1,1)
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Solve the system:<br> 4x-3y=2<br> 16x+9y=15
Nookie1986 [14]

x=3/4, y=1/3. (x,y)=(3/4,1/3)

5 0
3 years ago
Show work for this fraction equations 30 1/2 -13 3/4
loris [4]
30\frac12-13\frac34

Let's put these as improper fractions.
30 = 60/2. add the 1/2 to get 61/2
13 = 52/4. add the 3/4 to get 55/4

\frac{61}2-\frac{55}4

To add/subtract fractions, they must have the same denominator. (bottom no.)
If we multiply the top and bottom of a fraction by something, it stays equal.
We can conclude that 61/2 = 122/4. (bottom multilpies by 2, so does the top)

\frac{122}4-\frac{55}4

Now that we have a common denominator, we can subtract.
The "fourths" just sort of acts like a unit. Subtract 55 from 122 to get \frac{67}4.

We can convert this to an improper fraction simply by dividing 67 by 4.
This leaves us with 16, and a remainder of 3. Our answer is \boxed{16\frac34}
7 0
4 years ago
12x - 6y = -24 in "function form" is which one of the following? y = -2x + 4
marusya05 [52]

Answer:

Step-by-step explanation:

Answer to Consider the area between the graphs x+6y=12 and x+4=y^2. ... This Can Be Computed As A Single Interval Where Alpha= Beta= And H(y)=

3 0
3 years ago
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Ms. Barnes washed more.
5 0
3 years ago
write the first five terms of the sequence defined by the recursive formula a_n=(5a_n-1)-1, with a_1=0
nadezda [96]

Answer:

a_1 = 0

a_2 = -1

a_3 = -6

a_4 = -31

a_5 = -156

Step-by-step explanation:

a_n=(5a_n-1)-1

a_1 = 0

to find a2  substitute a1

a_2 = (5* a_1)-1

a_2 = 5* 0 -1

a_2 = 0-1

a_2 = -1

to find a3  substitute a2

a_3 = (5* a_2)-1

a_3 = (5* -1 ) -1

a_3 = -5 -1

a_3 = -6

to find a4  substitute a3

a_4 = (5* a_3)-1

a_4 = (5* -6 ) -1

a_4 = -30 -1

a_4 = -31

to find a5  substitute a4

a_5 = (5* a_4)-1

a_5 = (5* -31 ) -1

a_5 = -155 -1

a_5 = -156

3 0
4 years ago
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