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neonofarm [45]
3 years ago
14

Is it possible for a system of linear equations to have no solutions ?

Mathematics
2 answers:
sveta [45]3 years ago
8 0
Yes it is possible,t<span>hey never intersect, so there is </span>no<span> point that lies on both lines, and </span>no solution<span> to the </span><span>system</span>
AveGali [126]3 years ago
4 0

Answer: true

Step-by-step explanation:

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What is the answer to this
ICE Princess25 [194]

2. -28

5. -60

8. -135

The answers are above. :)

7 0
3 years ago
If c varies directly as (r + 1) and c = 8, when r= 3, calculate the value of r when c= 20.
Aneli [31]

Answer:

r = 9

Step-by-step explanation:

Given that c varies directly as (r + 1) then the equation relating them is

c = k(r + 1) ← k is the constant of variation

To find k use the condition c = 8 when r = 3, then

8 = k(3 + 1) = 4k ( divide both sides by 4 )

2 = k

c = 2(r + 1) ← equation of variation

When c = 20, then

20 = 2(r + 1) ← divide both sides by 2

10 = r + 1 ( subtract 1 from both sides )

9 = r

5 0
3 years ago
Read 2 more answers
Which of these could be the graph of f(x)=5•(0.4)^
nikdorinn [45]
The answer is A buddy >:D
6 0
3 years ago
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Please help me solve this for 15 points! <br><br>-x2 + 4x - 54 = 0​
Mazyrski [523]

Answer:

x1 =2-5i*sqrt(2)

x2 =2+5i*sqrt(2)

Step-by-step explanation:

-x^2 +4x-54=0 (quadratic equation)

a=-1, b=4, c=-54

x1=(-b+sqrt(b^2-4ac))/2a

x1=(-4+sqrt(4^2 - 4*(-1)(-54))/2*(-1)

x1=(-4+sqrt(16-216))/(-2)

x1 =(-4+sqrt(-200))/(-2)

x1 =(-4+sqrt(200i^2))/(-2) i^2=-1

x1 =(-4+sqrt(100*2*i^2))/(-2)

x1 =(-4+10i*sqrt(2))/(-2)

x1 =2-5i*sqrt(2)

x2 =(-b-sqrt(b^2-4ac))/2a

x2 =(-4-10i*sqrt(2))/(-2)

x2 =2+5i*sqrt(2)

7 0
3 years ago
Read 2 more answers
Which are the solutions of x2 = –7x – 8?
vovangra [49]

Answer:

The solutions are:

x_1=\frac{-7+\sqrt{17}}{2}     x_2=\frac{-7-\sqrt{17}}{2}

Step-by-step explanation:

We have the following quadratic equation

x^2 = -7x - 8

We can rewrite the equation as follows

x^2+7x + 8=0

Now we use the quadratic formula to solve the equation

For an equation of the form ax ^ 2 + bx + c = 0 the quadratic formula is:

x=\frac{-b\±\sqrt{b^2-4ac}}{2a}

In this case:

a=1,\ b=7,\ c=8

Then:

x=\frac{-7\±\sqrt{7^2-4(1)(8)}}{2(1)}

x=\frac{-7\±\sqrt{49-32}}{2}

x=\frac{-7\±\sqrt{17}}{2}

x_1=\frac{-7+\sqrt{17}}{2}

x_2=\frac{-7-\sqrt{17}}{2}

7 0
3 years ago
Read 2 more answers
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