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3 years ago

Anybody know the answer?

Mathematics

2 answers:

madam [21]3 years ago

5 0

Answer:

Step-by-step explanation:

You can read f(x) to mean y. y is above the x axis between -2 and - 6. To put it more formally,

-6 <x <-2

Why don't you use ≤ ?

It's because where is f(x) positive? 0 is not a positive value,, so you cannot use the endpoints.

Sliva [168]3 years ago

4 0

I don’t know
Sorry I’m typing this, brainly says I have to answer 1 question in order to ask more and I’m taking a test right now, just ignore this I can’t risk failing, sorry.

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Step-by-step explanation:

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4 years ago

Which values of c will cause the quadratic equation –x2 + 3x + c = 0 to have no real number solutions? Check all that apply.

rodikova [14]

Answer:

-5

negative nine-halves

Step-by-step explanation:

we know that

In the quadratic equation $ax^{2} +bx+c=0$

If $b^{2}-4ac < 0$

then

The system has no real numbers solutions

we have

$-x^{2} +3x+c=0$

$a=-1,b=3$

substitute

$3^{2}-4(-1)c < 0$

$9+4c < 0$

$c < -\frac{9}{4}$

Verify each case

case 1) -5

For c=-5

substitute

$-5 < -\frac{9}{4}$

$-20 < 9$ ------> is true

therefore

The value of c=-5 will cause the quadratic equation to have no real number solutions

case 2) negative nine-halves

For c=-9/2

substitute

$-9/2 < -\frac{9}{4}$

$-36 < -18$ ------> is true

therefore

The value of c=-9/2 will cause the quadratic equation to have no real number solutions

case 3) negative one-quarter

For c=-1/4

substitute

$-1/4 < -\frac{9}{4}$

$-4 < -36$ ------> is not true

therefore

The value of c=-1/4 will not cause the quadratic equation to have no real number solutions

case 4) 1

For c=1

substitute

$1 < -\frac{9}{4}$ ------> is not true

therefore

The value of c=1 will not cause the quadratic equation to have no real number solutions

case 5) 9 Over 4

For c=9/4

substitute

$9/4< -\frac{9}{4}$ ------> is not true

therefore

The value of c=9/4 will not cause the quadratic equation to have no real number solutions

5 0

3 years ago

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