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3 years ago

In the polynomial function F(x)=1/2x^2+8-5x^3-19x what is the leading the coefficient

Mathematics

2 answers:

Dima020 [189]3 years ago

8 0

The leading coefficient is in the term with the highest degree.
So its -5.

Dmitry_Shevchenko [17]3 years ago

6 0

Answer:

-5

Step-by-step explanation:

The leading term in a polynomial consist on the highest degree term. To get the highest degree term, we need to reorder the polynomial from left to right, starting with the highest degree term.

In this case:

$f(x)=\frac{1}{2} x^{2} +8-5x^{3} -19x$

reordering

$f(x)=-5x^{3} +\frac{1}{2}x^{2} -19x+8$

So, the leading coefficient is the one with the leading term:

$-5x^{3}$

So, it is <em>-5</em>

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What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

Please help me and explain how you got it-

BartSMP [9]

Answer:

i dont know

Step-by-step explanation:

8 0

3 years ago

PLEASE SOLVE THeSE QUESTIONS

g100num [7]

Heya!

Question #15:

To find the perimeter of the object, you can count the amount of squares that are on the outside of the object. After you country all around the object, the perimeters is 22 units (Option D)

Question #16:

Since we know the total perimeter, we can divide by the amount of sides a hexagon has because all of the sides are the same length. A hexagon has 6 sides. 42 / 6 = 7 inches (Option A)

Question #17:

To calculate the perimeter of the rectangle, you can add all the sides together. First, find common denominators.

6 1/2 = 6 2/4 and 3 1/4

Now, add all the sides together.

6 2/4 + 6 2/4 + 3 1/4 + 3 1/4 = 19 1/2 cm (Option B)

Question #18:

We can find the perimeter of the semi circle and square separately. Only take the perimeter of the square using 3 sides since the fourth sides is in the semi circle.

8 + 8 + 8 = 24 inches

Circumference of a semi circle formula: C = πd

C = (3.14)(8)

C = 25.12

Now, add both perimeters together.

24 + 25.12 = 49.12 inches (Option D)

Best of Luck!

3 0

2 years ago

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