2 years ago

What is the maximum number of x-intercepts that a polynomial of degree 7 can have

Mathematics

1 answer:

Daniel [21]2 years ago

5 0

Polynomial of degree seven can have the maximum 7 real solutions,
so maximum number of x-intercepts is also 7.

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Which of these points are on a line that has a slope of -3?

Ira Lisetskai [31]

Simple....

using the formula $\frac{ y_{1}- y_{2} }{ x_{1}- x_{2} }$ to find slope....

1.) (9,2)(4,3)

$\frac{2-3}{9-4}= -\frac{1}{5}$ (No)

2.) (9,2)(3,4)

$\frac{2-4}{9-3} = -\frac{2}{6} =- \frac{1}{3}$ (No)

3.) (2,9)(3,4)

$\frac{9-4}{2-3} = \frac{5}{-1} =-5$ (No)

4.) (2,9)(4,3)

$\frac{9-3}{2-4} = \frac{6}{-2} =-3$ (YES!)

Thus, your answer.

3 0

3 years ago

I also need help with this one

icang [17]

It is c.

explanation:

20+130+h =180

when you solve you get h=50

130+g = 180

g= 50

50+60+f = 180

f=70

7 0

2 years ago

Read 2 more answers

Soccer field dimensions according to FIFA must be no greater than 120 meters in length. What is the length of the field in kilom

Yuri [45]

Answer: 0.12 Kilometer

Step-by-step explanation:

120 divided by 1000

6 0

2 years ago

2x 2 (9x 2 −16)

xz_007 [3.2K]

Is the equation2x2(9x+2-16) if so you would do -16+2 first which is -14

Then multiply 2x2 equals 4

4(9x-14)

Multiply the 4 to 9x and -14

You get 36x-56

7 0

2 years ago

What is the minimum value of 2x + 2y in the feasible region?<br> (0,7) (3,7) (6,3) (6,0)

topjm [15]

Graph the feasible region for the following constraints: x + y < 5. 2x + y > 4 ... y = 10/3. x = 30/3 - 10/3 = 20/3. Intersects at (20/3, 10/3). -x + 2y = 0. x - 2y = 0.
hence i would go for
<span>(6,3)</span>

8 0

3 years ago

What is the maximum number of​ x-intercepts that a polynomial of degree 7 can​ have

What is the maximum number of x-intercepts that a polynomial of degree 7 can have