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2 years ago

Which is the polynomial function of lowest degree with leading coefficient of 4 and roots 7 and 3?

Mathematics

1 answer:

laiz [17]2 years ago

3 0

Answer:

Step-by-step explanation:

If the roots are 7 and 3, then x = 7 and x = 3 are solutions. In factor form, x = 7 is (x - 7); in factor form, x = 3 is (x - 3). We FOIL these together to get

$x^2-3x-7x+21$ which simplifies to

$x^2-10x+21$ and we distribute in the 4:

$4(x^2-10x+21)$ to get

$4x^2-40x+84$ which is another way to write the last choice you're given.

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What if for step 1 you picked (6,3)?

Inessa05 [86]

Answer: Then you picked (6,3) and there is nothing you can do about it

7 0

2 years ago

5. – 3(1 + 6r) = 14 - 7

MAVERICK [17]

Answer:

r= -1.8

Step-by-step explanation:

Distribute

-3-18r=14-7

Combine like terms

-3-18r=7

add 3 to both sides

-18r=10

divide by -18

r=-1.8

3 0

2 years ago

Min has a shelf filled with books. The books have different thicknesses. This line plot shows the thickness of each book. He pil

Vladimir [108]

Answer:

$2\frac{1}{4} \ inch$

Step-by-step explanation:

We have the four thickest books from the line plot given by

Two books of $\frac{1}{2}$ inch thick and the two books of $\frac{5}{8}$ inch thickness.

So when the two $\frac{1}{2}$ inch thick book is being piled on the top of a $\frac{5}{8}$ inch thick books.

Therefore, the total thickness of the pile of books is given by :

$=2 \times \frac{1}{2}+2 \times \frac{5}{8}$

$=\frac{9}{4}$

$=2\frac{1}{4} \ inch$

Therefore, the thickness of the pile is $2\frac{1}{4} \ inch$ .

7 0

2 years ago

Find the vertex and zeros of the following function f(x)=x^2-6x-7

elena-s [515]

Try this solution:

Step-by-step explanation:

1) for zeros of the function: x²-6x-7=0; ⇒ x₁=7; x₂= -1.

2) for the vertex of the function: x₀= -b/2a, where a;b;c - numbers from ax²+bx+c=0 (from x²-6x-7=0).

x₀=6/2=3;

y₀= (4ac-b²)/4a; ⇒ y₀= -16.

It means, the coordinates of the vertex are (3;-16).

6 0

2 years ago

PLEASE help me solve this problem. I really need help.

mote1985 [20]

When:
a = b = 0 we have point (0,0).
a = 0 and b ≠ 0 vertical line segment from (0, -b) to (0, b)
a ≠ 0 and b = 0 horizontal line segment from (-a, 0) to (a, 0)
a ≠ 0 and b ≠ 0 a ≠ b ellipse
a ≠ 0 and b ≠ 0 a = b circle

3 0

2 years ago