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liq [111]
2 years ago
10

Use Descartes’s rule of signs to determine the possible number of positive and negative real zeros of f(x)=8x^4-6x^3-4x^2-7x+3

Mathematics
1 answer:
pav-90 [236]2 years ago
5 0

let's recall Descartes rule of signs, to get the Real Positive ones we start by checking the sign changes for f(x), and to get the Real Negative ones we check the sign changes for f( -x ).

Check the picture below.

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Is the order pair (-5,9) a solution to the equation y = 5x -4
s2008m [1.1K]

Answer:

No

Step-by-step explanation:

9=5(-5)-4 is not correct, 5(-5)-4 is actually -29 so no, the coordinates are not a solution to the equation.

8 0
3 years ago
Read 2 more answers
HELP IVE BEEN STRUGGLING FOR THE PAST 15 MIN PLEASEEEEE
joja [24]

<u>Answer:</u>

c=  -1/2

<u>Step-by-step explanation:</u>

Let's solve your equation step-by-step.

4 /3  = −6c − 5/ 3

Step 1: Simplify both sides of the equation.

4 /3  =−6c +  −5 /3

Step 2: Flip the equation.

−6c +  −5 /3  =  4/3

Step 3: Add 5/3 to both sides.

−6c +  −5 /3  +  5 /3  = 4/3  +  5 /3  −6c

=3

Step 4: Divide both sides by -6.

−6c  −6   =  3 − 6 c =  −1 /2

4 0
3 years ago
Read 2 more answers
How many terms are in the arithmetic sequence 1313, 1616, 1919, ……, 7070, 7373?
Ede4ka [16]
The formula in getting the arithmetic sequence:
an = a1<span> + (n – 1)d.
</span>
We can substitute the values in order for us to know how many terms the sequence has.

an = 7373
a1 = 1313
d = 303

an = a1<span> + (n – 1)d
</span>7373 = 1313 + (n-1) 303
7373 = 1313 + 303n - 303
-303n = 1313 - 303 - 7373
303n = -1313 + 303 + 7373
303n = 6363
n = 21

So, there are 21 terms in the sequence given.

3 0
3 years ago
suppose it takes 48 chickens fingers to feed Mr . Youngs 4th grade class of 20 students how many chicken fingers would be needed
Mrrafil [7]
30 is 1.5 times greater than 20 so 48*1.5 = 72
it would take 72 chicken fingers to feed 30 students
6 0
3 years ago
Graph the ellipse with equation x^2 /25 + y^2 /4 = 1.
nexus9112 [7]
Center: 0,0
vertex: 5,0
vertex2 (-5,0)
4 0
3 years ago
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