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xxMikexx [17]
2 years ago
7

Find all values of k for which the equation 3x^2−4x+k=0 has no solutions.

Mathematics
2 answers:
dusya [7]2 years ago
5 0

The given equation has no solution when K is any real number  and k>12

We have given that

3x^2−4x+k=0

△=b^2−4ac=k^2−4(3)(12)=k^2−144.

<h3>What is the condition for a solution?</h3>

If Δ=0, it has 1 real solution,

Δ<0 it has no real solution,

Δ>0 it has 2 real solutions.

We get,

Δ=k^2−144 here Δ is not zero.

It is either >0 or <0

Δ<0 it has no real solution,

Therefore the given equation has no solution when K is any real number.

To learn more about the solution visit:

brainly.com/question/1397278

Leona [35]2 years ago
5 0

Answer:

k>1 1/3

Step-by-step explanation:

i have rsm too bro

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\bf \cfrac{10km}{55mins}\cdot \cfrac{0.621miles}{1km}\implies \cfrac{62.1miles}{55mins}\approx 1.12909909\approx \boxed{1.13\frac{miles}{mins}}
5 0
3 years ago
The train clock run fast and gain 4 minutes every 3 days. How many minutes and seconds will it have gained at the end of 2 days.
Katarina [22]

Answer:

2 minutes, 40 seconds

Step-by-step explanation:

set up a proportion and cross-multiply:

4/3 = x/2

3x = 8

x = 2 2/3 which is 2 minutes, 40 seconds

8 0
3 years ago
Solve the division problem. Round answer to the nearest hundredth. 9.252.063
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9.252.063 rounded to the nearest hundredth is 9.252.060

6 0
3 years ago
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Six friends attend a party. They form pairs for a game. How many different pairs are possible?
Leona [35]

Pairs, in this case, relates to a group of 2 or more. We have 6 friends. Let's call them A,B,C,D,E,F. This will allow us to make a [some sort of] combination tree:

1. ABC against DEF

2. ABD against CEF

3. ABE against CDF

4. ABF against CDE

5. ACD against BFE

6. ACE against BDF

7. ACF against BDE

8. ADE against BCF

9. ADF against BCE

10. AEF against BCD

I believe there are 12 combinations... I just can't think of the last 2 though.

5 0
3 years ago
Determine the rational zeros for the function f(x)=x^3+7x^2+7x-15
alexandr402 [8]

Answer:

1,-3,-5

Step-by-step explanation:

Given:

f(x)=x^3+7x^2+7x-15

Finding all the possible rational zeros of f(x)

p= ±1,±3,±5,±15 (factors of coefficient of last term)

q=±1(factors of coefficient of leading term)

p/q=±1,±3,±5,±15

Now finding the rational zeros using rational root theorem

f(p/q)

f(1)=1+7+7-15

   =0

f(-1)= -1 +7-7-15

     = -16

f(3)=27+7(9)+21-15

    =96

f(-3)= (-3)^3+7(-3)^2+7(-3)-15

     = 0

f(5)=5^3+7(5)^2+7(5)-15

    =320

f(-5)=(-5)^3+7(-5)^2+7(-5)-15

      =0    

f(15)=(15)^3+7(15)^2+7(15)-15

     =5040

f(-15)=(-15)^3+7(-15)^2+7(-15)-15

      =-1920

Hence the rational roots are 1,-3,-5 !

4 0
3 years ago
Read 2 more answers
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