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Rasek [7]
3 years ago
14

use the discrminant to determine all values of k which would result in the equation -3x^2-6x+k=0 having real,unequal roots.​ mus

t be right ty helppp asap
Mathematics
1 answer:
boyakko [2]3 years ago
7 0

Answer:

k>-3

Step-by-step explanation:

We have the equation:

-3x^2-6x+k=0

Where <em>a </em>= -3, <em>b</em> = -6, and <em>c</em> = <em>k.</em>

And we want to determine values of <em>k</em> such that the equation will have real, unequal roots.

In order for a quadratic equation to have real, unequal roots, the discriminant must be a real number greater than 0. Therefore:

b^2-4ac>0

Substitute:

(-6)^2-4(-3)(k)>0

Simplify:

36+12k>0

Solve for <em>k: </em>

<em />12k > -36<em />

<em />k>-3<em />

So, for all <em>k</em> greater than -3, our quadratic equation will have two real, unequal roots.

Notes:

If <em>k </em>is equal to -3, then we have two equal roots.

And if <em>k</em> is less than -3, then we have two complex roots.

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Mr. Jones lives in City A and drives at a constant speed to City B, which is 400 km 2 of the distance at that same away. On the
Sphinxa [80]

Answer:

He traveled at 72km/hr during the rain

Step-by-step explanation:

<em>Question is not well formatted. See comment</em>

Given

d=400km --- distance

t = 11hrs --- total time

Let his speed from city A till the rain starts on his return trip be s_1

Let his speed from city during the rain be s_2

So:

s_2 = s_1 -20

Required

Calculate s_2

From the question, we understand that; he drives the whole 400 km and 2/5 of 400 km at s_1

The distance covered during this period is:

d_1  = 400 + \frac{2}{5} * 400

d_1  = 400 + 160

d_1  = 560

And the time during this period is:

t_1 = \frac{2}{5} * 11

t_1 = 4.4

So, the distance during the rain is:

d_2 = 2 * 400 - d_1

d_2 = 2 * 400 - 560

d_2 = 800 - 560

d_2 = 240

And the time during the rain is:

t_2 = 11 - t_1

t_2 = 11 - 4.4

t_2 = 6.6

So, we have:

d_1  = 560 --- distance covered before the rain

d_2 = 240 --- distance covered when raining

s_2 = s_1 -20

t_1 = 4.4 ---- time spent before the rain

t_2 = 6.6 --- time spent in the rain

Speed is calculated as:

Speed = \frac{distance}{time}

Make distance the subject

distance = speed * time

So:

d_1 + d_2 = s_1 * t_1 + s_2 * t_2

Recall that:

s_2 = s_1 -20

Make s_1 the subject

s_1 = s_2 + 20

The expression d_1 + d_2 = s_1 * t_1 + s_2 * t_2 becomes:

560 + 240 = (s_2 + 20) * 4.4 + s_2 * 6.6

800 = 4.4s_2 + 88+ 6.6s_2

Collect like terms

6.6s_2 + 4.4s_2 = 880 - 88

11s_2 = 792

Solve for s_2

s_2= \frac{792}{11}

s_2=72km/h

4 0
3 years ago
Task 2
Softa [21]

Answer:

A) Carla: y = 1.5x + 30   <-- y is the total money, and x is the number of weeks

Darla: y = 3x   <-- y is the total money, and x is the number of weeks

B) No, they start with a different amount of money and also save different amounts each week. They will not always have the same amount of money.

C) See attachment

Step-by-step explanation:

We'll start with Carla.

Carla saves $1.5 every week. That means that if we multiply the number of weeks she has saved by 1.5 we'll get the amount of money she has. However, she starts at 30 so we have to add 30 to that at the beginning. So, Carla's equation is y = 1.5x + 30. y is the total money, and x is the number of weeks

Darla saves $3 every week. That means that if we multiply the number of weeks she has saved by 3 we'll get the amount of money she has. Unlike Carla, Darla starts at $0, so there is no need to add anything. So, Darla's equation is y = 3x. y is the total money, and x is the number of weeks

Since Carla and Darla start with a different amount of money and save different amounts each week, they won't always have the same amount of money. However, they will cross at some point. You can actually see this in the graph. Carla, there is red, and Darla is blue.

Now let's graph the 2 equations. First, we'll do Carla. her equation is y = 1.5x + 30. Since it is in the slope-intercept form we can just use y = mx + b to tell how we're going to graph this. In this formula, m is the slope, and b is the y-intercept. As you can see when you look back at Carla's equation, m is 1.5 and b is 30. First, we'll take the y-intercept, b, which is 30 as we just figured out, and plot it on the graph. All you have to do here is count 30 units up the y-axis (the one going up and down) and put a point there. Now we take the slope (1.5) and we start putting in points that the line will go through. 1.5 in fraction form is 3/2 meaning (starting from that point we marked earlier) we will go 3 units up, and 2 units to the right and put another point there. You can now take a ruler, line up the 2 points you marked and draw a big line through them that keeps going on forever on both sides (just put an arrow at both ends to mark that).

Now we'll do the same with Darla's equation. Her's is y = 3x. When we compare this with or slope-intercept formula, y = mx + b, we'll notice there is no b. That means it is 0 and that the y-intercept is also 0. So go ahead and mark a point in the middle of the graph where the 2 big lines cross. 3 in fraction form is just 3/1 meaning that we'll go 3 up and 1 to right (from the point you just marked) and put another point there. Now draw another big line through both of those points (WITH A RULER) and mark the ends with arrows.

4 0
2 years ago
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