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3 years ago

7

Write an expression to represent the product of 6 and the square of a number plus 15.

1 answer:

Leona [35]3 years ago

5 0

Answer:

D (6)

Step-by-step explanation:

6x^2 + 15

the number next to the variable ( the unknown) is always the coefficient

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Compare the average rate of change for both functions over the interval [0,3].

Verizon [17]

So find g(0) and g(3)
G(0) = 0^2 -5(0) +3 = 3
And g(3) = 3^2 - 5(3) +3 = 9 - 15 + 3 = 3 so there is no change in g(x) for that interval.

H(0) = 8(0) + 10 = 10
And h(3) = 8(3) + 10 = 34
So the change in h(x) for that interval is 24.

5 0

2 years ago

Sorry i ment to put 14/15 - 1/3 then put as a fraction in simplest form

Sergio039 [100]

Answer:

Step-by-step explanation:

14/15 - 1/3

= 6/10

In simplest form

= 3/5

7 0

2 years ago

Determine if 9 is rational or irrational and give a reason for your answer.

Irina18 [472]

Answer:

9 is an rational number because it is a whole number

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Simplify expression X^2+x-2/x^3-x^2+2x-2

professor190 [17]

Answer:

$=\frac {x+2}{x^2+2}$ is the simplest form of given expression.

Step-by-step explanation:

The given question is $\frac{x^2+x-2}{x^3-x^2+2x-2}$

To solve the problem we have to group or split middle term and then factorise

$\frac{x^2+(2x-x)-2}{(x^3-x^2)+(2x-2)}$

Taking $x^2$ common from first two terms of denominator and 2 from next two terms

$= \frac{x^2+2x-x-2}{x^2(x-1)+2(x-1)}$

Now,taking x common from first two terms of numerator and -1 from next two terms and in denominator taking(x-1) common from both terms

$= \frac{ x(x+2)-1(x+2)}{(x-1)(x^2+2)}$

$=\frac{(x+2)(x-1)}{(x-1)(x^2+2)}$

Now cancel out x-1 from both numerator and denominator we get

$=\frac {x+2}{x^2+2}$ is the required simplest form.

7 0

2 years ago

Consider the sequence defined recursively by do = 0, an = an-1 + 3n – 1. a) Write out the first 5 terms of this sequence,

creativ13 [48]

Answer:

The first 5 terms of the sequence is 2,7,15,26,40.

Step-by-step explanation:

Given : Consider the sequence defined recursively by $a_0=0$ $a_n=a_{n-1}+3n-1$

To find : Write out the first 5 terms of this sequence ?

Solution :

$a_n=a_{n-1}+3n-1$ and $a_0=0$

The first five terms in the sequence is at n=1,2,3,4,5

For n=1,

$a_1=a_{1-1}+3(1)-1$

$a_1=a_{0}+3-1$

$a_1=0+2$

$a_1=2$

For n=2,

$a_2=a_{2-1}+3(2)-1$

$a_2=a_{1}+6-1$

$a_2=2+5$

$a_2=7$

For n=3,

$a_3=a_{3-1}+3(3)-1$

$a_3=a_{2}+9-1$

$a_3=7+8$

$a_3=15$

For n=4,

$a_4=a_{4-1}+3(4)-1$

$a_4=a_{3}+12-1$

$a_4=15+11$

$a_4=26$

For n=5,

$a_5=a_{5-1}+3(5)-1$

$a_5=a_{4}+15-1$

$a_5=26+14$

$a_5=40$

The first 5 terms of the sequence is 2,7,15,26,40.

5 0

3 years ago