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Sloan [31]
3 years ago
13

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

Mathematics
2 answers:
vodomira [7]3 years ago
7 0

Finding the discriminate of a quadratic formula determines the number and type of answers.

The formula is b^2 -4(ac)

a = 6. b = -7 and c = -4

-7^2 -4(6*-4) = 145

The answer is a positive number so this means there are 2 real solutions.

stepladder [879]3 years ago
4 0

Answer:

two real solutions

Step-by-step explanation:

We can use the discriminant to determine the type and number of solutions

b^2 -4ac  where ax^2 +bx +c

If b^2 -4ac > 0 we have 2 real solutions

b^2 -4ac = 0 there is 1 real solution

b^2 -4ac < 0 there are 2 complex solutions

6x^2 -7x -4 = 0

a=6  b =-7  c = -4

b^2 -4ac

(-7)^2 -4(6)(-4)

49 + 96

145>0

There are real solutions

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Differential Equations Problem
fgiga [73]

Answer:

y=\frac{t^2e^{2t}}{3}+ce^{2t}

Step-by-step explanation:

We have given differential equation \frac{dy}{dt}-2y=t^2e^{2t}

We know that linear differential equation is given by \frac{dy}{dt}+Py=Q

On comparing with standard equation P = -2 and Q= t^2e^{2t}

Now integrating factor IF=e^{-Pdt}

IF=e^{-2dt}=e^{-2t}

Now solution of differential equation is given by

y\times IF=\int\ IF\times Q\ dt

y\times e^{-2t}=\int\ e^{-2t}\times t^2e^{2t}\ dt

y\times e^{-2t}=\frac{t^2}{3}+c

y=\frac{t^2e^{2t}}{3}+ce^{2t}

7 0
3 years ago
Find the root of (M^3)+(M^2)-2=0. Show steps.
MrRissso [65]

Answer:

x = 1, -1 + i, and -1 - i are the 3 roots

Step-by-step explanation:

The Rational Root Theorem will help us find any real roots in this 3rd degree polynomial.  The c value in that polynomia is -2 and the d value is 1 (the leading coefficient).  

If c is -2, then the factors of -2 are:

1, -1, 2, -2

If d is 1, then the factors of 1 are:

1, -1

And c/d gives us all the possible roots:

1/1, -1/1, 2/1, -2/1 which of course simplify to

1, -1 2, -2

We will now use synthetic division to find the first real root.  I will use x = 1 first.

For synthetic division:

1|   1     1     0     -2

is how it is set up.  Bring down the first number, 1, multiply it by 1 and put that product up under the second "1":

1|    1     1     0     -2

           1

     1

Now add to get

1|     1     1     0     -2

            1

      1     2

Multiply 2 by 1 and put that product up under the 0 and add:

1|     1     1     0     -2

             1     2

      1     2    2

Again multiply the 2 by 1 and put that product up undeer the -2 and add:

1|     1     1     0     -2

            1      2      2

      1     2     2      0

That remainder of 0 tells us that x = 1 is a root of that polynomial and the depressed polynomial, made up from the bold numbers, is one degree less than what it started with:

x^2+2x+2=0

This does not factor without imaginary roots, so putting it into the quadratic formula looks like this:

x=\frac{-2+/-\sqrt{4-4(1)(2)} }{2}

which simplifies to

x=\frac{-2+/-\sqrt{-4} }{2}

The square root of -4 simplifies to 2i, so

x=\frac{-2+/-2i}{2}

which reduces to

x = -1 + i and

x = -1 - i

   

5 0
3 years ago
Assume that the random variable X is normally​ distributed, with mean mu equals 100 and standard deviation sigma equals 20. Comp
TEA [102]

Answer: 0.2119

Step-by-step explanation:

We assume that the random variable X is normally​ distributed.

Given : Population mean : \mu=100

Standard deviation : \sigma=20

Z-score : z=\dfrac{x-\mu}{\sigma}

Then, z-score corresponds to 116

z=\dfrac{116-100}{20}=0.8

By using the standard normal distribution table for z , we have

P(x>116)=P(z>0.8)=1-P(z\leq0.8)

=1-0.7881446\approx0.2119

Hence, the required probability = 0.2119

4 0
3 years ago
Solve and check <br> please please help i will mark u as a brill it is easy please
Lesechka [4]

Answer:

B=25

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What are all values of x for which the series shown converges? ​
adoni [48]

Answer:

One convergence criteria that is useful here is that, if aₙ is the n-th term of this sequence, then we must have:

Iaₙ₊₁I < IaₙI

This means that the absolute value of the terms must decrease as n increases.

Then we must have:

\frac{(x -2)^n}{n*3^n} > \frac{(x -2 )^{n+1}}{(n + 1)*3^{n+1}}

We can write this as:

\frac{(x -2)^n}{n*3^n} > \frac{(x -2 )^{n+1}}{(n + 1)*3^{n+1}} = \frac{(x -2)^n}{(n + 1)*3^n} * \frac{(x - 2)}{3}

If we assume that n is a really big number, then:

n + 1 ≈ 1

And we can write:

\frac{(x -2)^n}{n*3^n} > \frac{(x -2)^n}{(n)*3^n} * \frac{(x - 2)}{3}

Then we have the inequality

1 > (x - 2)/3

And remember that this must be in absolute value, then we will have that:

-1 < (x - 2)/3 < 1

-3 < x - 2 < 3

-3 + 2 < x < 3 + 2

-1 < x < 5

The first option looks like this, but it uses the symbols ≤≥, so it is not the same as this, then the correct option will be the second.

5 0
2 years ago
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