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4 years ago

Which quadratic equation has no real roots?

Mathematics

2 answers:

Stella [2.4K]4 years ago

6 0

Answer:

The first choice and the last have no real roots.

Step-by-step explanation:

For an equation to have no real roots the value of the discriminant ( b^2 - 4ac) will be negative.

Working out the values:

1. D = (-2)^2 - 4* -1*-6 = -20 so it has no real roots.

2. D = (3)^2 - 4* -2* 4 = 41 so it does have real roots.

3. D = (1)^2 - 4* 2*-6 = 49 so real roots.

4. D = (-1)^2 - 4* 2*3 = -23 so no real roots

n200080 [17]4 years ago

3 0

Answer:

A and D

Step-by-step explanation:

By finding the roots of all the equations, we see that options A and D are the only ones that require an imaginary number in order to solve

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lara [203]

Answer:

4a+5b

Step-by-step explanation:

Given:

Apple=£a

Banana=£b

Quantity of Apple=4

Quantity of banana=5

Find the cost of 4 apples and 5 bananas

Total cost=PaQa+PbQb

Where,

Pa=price of Apple

Qa=quantity of Apple

Pb=price of banana

Qb=quantity of banana

Total cost=PaQa+PbQb

=4*a+5*b

=4a+5b

5 0

4 years ago

I don't understand. Please help me. I don't understand. Please help me.

sergeinik [125]

The answer to that are c and b or b and c I hope that this helps!

3 0

3 years ago

When randomly choosing bills from a cup of bills that contains a $1 bill a $2 bill a $5 bill a $10 bill a $20 bill and a $50 bil

Nesterboy [21]

Answer:

1/36

Step-by-step explanation:

Given the sample space

S= [$1, $2, $5, $10, $20, $50] = 6

Hence the probability of choosing a

$5 bill is

Pr($5) = 1/6

Also the probability of choosing a $20 bill is

Pr($20) = 1/6

Therefore the probability of choosing a $5 bill and a $20 bill

Pr($5,$20) = 1/6*1/6= 1/36

4 0

4 years ago

Read 2 more answers

What is -3 (4n+2) = -4n + -2(4n-6)

vampirchik [111]

Answer: There are no solutions.

Step-by-step explanation:

6 0

3 years ago

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

4 years ago