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2 years ago

Identify the polynomial. x 3 y 3 monomial binomial trinomial

Mathematics

2 answers:

maxonik [38]2 years ago

6 0

I believe binomial since there are 2 monomials being multiplied.

eduard2 years ago

6 0

Answer:

x³y³ is monomial.

Step-by-step explanation:

Given :Polynomial x³y³ .

To find : Identify monomial binomial trinomial.

Solution : We have given that

Polynomial x³y³ .

Monomial : A polynomial with just one term.

Examples : 1) 3xy

2) 5y²x³ Both are monmial .

x³y³ is also a monomial because it has only one term.

Therefore, x³y³ is monomial.

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rachel is frosting the top of two cakes and wants to know which cake will require more frosting. one cake is round with a diamet

siniylev [52]

Answer:

The square cake is larger by 31 in^2

Step-by-step explanation:

A round cake has a circular area on top.

We need to find the radius

d= 2r

8 = 2r

r=4

The area of the circle is pi r^2

A = 3.14 * 4^2

A = 3.14 *16

= 50.24

The area of the square cake is

A =s^2

A = 9^2

A = 81

The square cake is larger

The difference is Asquare - A round

81 -50.24

30.76

Which is close to 31 in ^2

7 0

2 years ago

What is the multiplicative property of 0

shtirl [24]

Answer:

Step-by-step explanation:

What ever times 0 always is 0, take 100 × 0 it will make it Into 0. If you need help use a calculator, hope this helps

4 0

3 years ago

Find the sum of the first 3 terms of a geometric sequence with a=96 and r=1/4 the sum is____.

viva [34]

For Geometric sequence, the next term is multiplied by the common ratio.

Common ratio, r = 1/4

1st term, a = 96

2nd term = ar = 96*1/4 = 24

3rd term = ar² = 96*(1/4)*(1/4) = 6

Sum of the first three terms = 96 + 24 + 6 = 126

3 0

2 years ago

(12a^(2) - 3)/(2)*(2a + 1)^(-2)*((6)/((2a + 1)))^(-1)

Elena L [17]

Answer:

$\frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\$

Step-by-step explanation:

$\frac{(12a^2-3)}{2*(2a+1)^{-2}}*\frac{6}{(2a+1)^{-1} }\\ = \frac{(12a^2-3)}{2 * \frac{1}{(2a+1)^{2} }} * \frac{6}{\frac{1}{2a+1} }\\ =\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}\\=\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}$

$= \frac{(12a^2-3)(4a^{2} + 1 + 2 (2a)(1))}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{1} * \frac{3}{2a+1}\\= \frac{3(12a^2-3)(4a^{2} + 1 + 4a)} {2a+1}\\\\= \frac{3(12a^2(4a^{2} + 1 + 4a) - 3 (4a^{2}+1+4a)} {2a+1}\\= \frac{3(48a^{4}+12a^{2}+48a^{3} - 12a^{2} -12 - 4a)} {2a+1}\\= \frac{144a^{4}+36a^{2}+144a^{3} - 36a^{2} - 36 - 12a)} {2a+1}$