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3 years ago

Use the discriminant to determine how many solutions are possible for the following equation?(show work)

Mathematics

1 answer:

Gwar [14]3 years ago

3 0

Answer:

see the explanation

Step-by-step explanation:

<u><em>The correct question is</em></u>

Use the discriminant to determine how many solutions are possible for the following equation (show work).

5x^2-3x+4=0

we know that

The discriminant for a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$D=b^2-4ac$

If D=0 then the equation has only one real solution

If D>0 then the equation has two real solutions

If D<0 then the equation has no real solutions (two complex solutions)

in this problem we have

$5x^{2} -3x+4=0$

$a=5\\b=-3\\c=4$

substitute

$D=-3^2-4(5)(4)$

$D=-71$

The equation has no real solutions, The equation has two complex solutions

therefore

I know there are___No____

real solutions to the equation in problem 4 because ___the discriminant is negative___

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Type the expression that results from the following series of steps: start with x, divide 6, then subtract 1

Alexus [3.1K]

Answer:

x/6 - 1

Step-by-step explanation:

x divide 6 subtract 1

4 0

3 years ago

For which independent value do the equations generate the same dependent value? y1=6x-16 , y2=3x-10

Nikitich [7]

The question asks which x-value gives the same y-value for both equations, so if you need the ys to be equal, you can set both equations equal to each other, giving you 6x-16=3x-10. Now, solve for x. 6x-16=3x-10, subtract 3x and add 16, 3x=6, divide 3, x=2. Your answer is x=2.

7 0

3 years ago

10. Two lighthouses are located on a north-south line.

Kamila [148]

The question is an illustration of bearing (i.e. angles) and distance (i.e. lengths)

The distance between both lighthouses is 5783.96 m

I've added an attachment that represents the scenario.

From the attachment, we have:

$\mathbf{\angle A = 180^o - 120^o\ 43'}$

Convert to degrees

$\mathbf{\angle A = 180^o - (120^o +\frac{43}{60}^o)}$

$\mathbf{\angle A = 180^o - (120^o +0.717^o)}$

$\mathbf{\angle A = 180^o - (120.717^o)}$

$\mathbf{\angle A = 59.283^o}$

$\mathbf{\angle B = 39^o43'}$

Convert to degrees

$\mathbf{\angle B = 39^o + \frac{43}{60}^o}$

$\mathbf{\angle B = 39^o + 0.717^o}$

$\mathbf{\angle B = 39.717^o}$

So, the measure of angle S is:

$\mathbf{\angle S = 180 - \angle A - \angle B}$ ---- Sum of angles in a triangle

$\mathbf{\angle S = 180 - 59.283 - 39.717}$

$\mathbf{\angle S = 81}$

The required distance is distance AB

This is calculated using the following sine formula:

$\mathbf{\frac{AB}{\sin(S)} = \frac{AS}{\sin(B)} }$

Where:

$\mathbf{AS = 3742}$

So, we have:

$\mathbf{\frac{AB}{\sin(81)} = \frac{3742}{\sin(39.717)}}$

Make AB the subject

$\mathbf{AB= \frac{3742}{\sin(39.717)} \times \sin(81)}$

$\mathbf{AB= 5783.96}$

Hence, the distance between both lighthouses is 5783.96 m

Read more about bearing and distance at:

brainly.com/question/19017345

5 0

2 years ago

A science teacher has a supply of 50% sugar solution and a supply of 80% sugar solution. How much of each solution should the te

mash [69]

How much of each solution should the teacher mix together to get 105 ML of 60% sugar solution for an experiment?

1. Look at how 60% is closer to the solution of lower concentration (50%). You can deduce that you will be mixing a higher volume of the 50% solution.

2. All 4 answers add up to 105ml.

3. The intuitive answer is the first option:
70 ML of the 50% solution and 35 ML of the 80% solution

4. Let's check whether point 3 is true.
70ml/105ml X 0.5 + 35ml/105ml X 0.8 = (35 + 28)/105= 63/105= 60% / 105 ml = 105ml of 60% sugar solution

3 0

3 years ago

Law of sines?? side c=22 side a=16 side b= 21 solve for angle c??? help

den301095 [7]

Answer:

Angle C measures: $71.52^o$

Step-by-step explanation:

When you know the length of the three sides of the triangle and not a single angle, the law of sines is of no use. You need to use the law of cosines. In your case, since you want to find angle "C", you use the following;

$c^2=a^2+b^2-2\,a\,b\,cos(C)\\2\,a\,b\,cos(C)=a^2+b^2-c^2\\cos(C)=\frac{a^2+b^2-c^2}{2\,a\,b} \\cos(C)=\frac{16^2+21^2-22^2}{2\,(16)\,(21)}\\cos(C)=0.3169643\\C=arccos(0.3169643)\\C=71.52^o$

4 0

3 years ago