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vitfil [10]
3 years ago
12

Graph the system of equations. then determine wheather the system has no solution, one solution, or infinitely many solutions. I

f the systems has one solution, name it.
y= -x + 5
y= x - 3

A. one solution; (1,4)
B. infinitely many
C. no solution
D. one solution; (4, 1)
Mathematics
2 answers:
dimaraw [331]3 years ago
8 0

Answer:

See below in bold.

Step-by-step explanation:

If we add the 2 equations  we eliminate x  and we get 2y =2.

So y = 1.

Substituting y = 1 in the second equation 1 = x - 3.

So x = 4.

A. One solution: (4, 1).

If we drew a graph we would have 2 lines which intersect at the point (4, 1).

kondor19780726 [428]3 years ago
6 0

Answer:

Step-by-step explanation:

If we add the 2 equations  we eliminate x  and we get 2y =2.So y =1.

Substituting y = 1 in the second equation 1 = x - 3.So x = 4.

A.

One solution: (4, 1).If we drew a graph we would have 2 lines which intersect at the point (4, 1).

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A line passes through $A(1,1)$ and $B(100,1000)$. How many other points with integer coordinates are on the line and strictly be
jonny [76]

Answer:

98

Step-by-step explanation:

Integers are whole numbers or opposite of whole numbers.

The slope is 1.

How?

Change of y is 100-1=99.

Change of x is 100-1=99.

The slope is 99/99=1 or 1/1.

So if we start at (1,1) and we rise 1 and run 1 right, we get (2,2).

If we do that again from (2,2) we get (3,3).

Following the pattern of going up 1 and right from each new location discovered we get all the of these points:

Start (1,1)

(2,2)

(3,3)

(4,4)

(5,5)

(6,6)

....

(96,96)

(97,97)

(98,98)

(99,99)

end (100,100)

So we just need to count all the numbers from 2 to 99.

99-2+1

97+1

98

7 0
2 years ago
Evaluate the following expression. cos135°
Mrac [35]
Negative 1 / Square Root of 2
4 0
3 years ago
How do you solve 2=11+3(3x+3)
nadezda [96]

Answer:

Just do it Nike

Step-by-step explanation:

4 0
3 years ago
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lana66690 [7]
Angle EBD and Angle DBE are adjacent
6 0
3 years ago
Expressed as a product of its prime factors in index form, a number N is
Elena-2011 [213]

<u>ANSWER</u>



5N^2=3^{2} \times 5^{5} \times x^{6}



<u>EXPLANATION</u>


N=3\times5^2 \times x^3.


5N^2=5(3\times5^2 \times x^3)^2


Recall this property of exponents;


(a^m)^2=a^{m} \times a^m



So our product becomes;


5N^2=5(3\times5^2 \times x^3) \times (3\times5^2 \times x^3)



5N^2=5\times 3\times 3 \times 5^2 \times 5^2 \times x^3 \times x^3


5N^2=3\times 3\times 5 \times 5^2 \times 5^2 \times x^3 \times x^3



Recall this law of exponents:


a^m \times a^n =a ^{m+n}


5N^2=3^{1+1} \times 5^{1+2+2} \times x^{3+3}


5N^2=3^{2} \times 5^{5} \times x^{6}




7 0
3 years ago
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