Pl algebra nation test
consider the following polynomial.

-2xy^5+3x^7-10x^3y^6+8x^6+4

which of the following statements are true? select all that apply

a)the polynomial has 5 terms.

b)the leading coefficient is -2.

c)the degree of the polynomial is 7.

d)the constant term is 4.

e)the polynomial is in decreasing degree order.

[A.] and [D.]

Step-by-step explanation:

Before we even solve to see which options are true and false, write the polynomial in descending order.

To do this, break the polynomial into terms.

1. -2xy^5

2. 3x^7

3. -10x^3y^6

4. 8x^6

5. 4

To write this in descending order, look at the exponents or degrees.

We have 5 terms [So A. is already a correct option choice] so look at the exponents or degrees for each term

1. -2xy^5  [degree: 6 (x has an exponent, 1)]

2. 3x^7 [degree: 7]

3. -10x^3y^6 [since there are 2 exponents, add them together (degree: 9)]

4. 8x^6 [degree: 6]

5. 4 [degree: 1 (even though it doesn't look like it, 4 actually has an exponent, 1, along with other numbers like 1, 2, 3, 4, 5 etc.)]

-10x^3y^6 has the largest degree, then 3x^7, then 8x^6 and -2xy^5, and lastly 4.

Now, rewrite the polynomial

-2xy^5 + 3x^7 -10x^3y^6 + 8x^6 + 4 = -10x^3y^6 + 3x^7 + 8x^6 -2xy^5 + 4

We already established A.) is a correct answer.

B.) is not correct since our leading term is -10x^3y^6 and -10 would be the leading coefficent.

C.) Is not correct since the degree of the polynomial is 9, since the highest degree is 9

D.) Is correct since 4 has no variables attached to it

E.) Is not correct since the polynomial was not in descending form, we had to do it for them.

Thus, A.) and C.) are the correct answers.

I hope this helps! Let me know if I did something wrong or you have any questions! Thanks! :)

1. f(x)=x²+10x+16

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

so, vertex = (-5, -9)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

2. f(x)=−(x−3)(x+1)

by multiplying the factors, the general form is f(x)= -x²+2x+3.

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

so, vertex = (1, 4)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

3. f(x)= −x²+4

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

so, vertex = (0, 4)

now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

4. f(x)=2x²+16x+30

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

so, vertex = (-4, -2)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

5. y=(x+2)²+4

the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a> 0 parabola is opened upward.

if a< 0 parabola is opened downward.

compare the given equation with general form of parabola.

-h=2 ⇒h=-2

k=4

so, vertex= (-2, 4)

as, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

hope this : )

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: