Solve :3(2x+1)^-1 + 5x=-1<br>

The sum of four and the product of three and a number x

Ulleksa [173]

4+3x is what it sound like. I hope this helps. :)

6 0

3 years ago

the product of 2 consecutive odd integers is equal to 27 plus six times the sum of the integers. what is the larger integer

olganol [36]

2 consecutive odd integers : x and x + 2

x(x + 2) = 27 + 6(x + x + 2)
x^2 + 2x = 27 + 6(2x + 2)
x^2 + 2x = 27 + 12x + 12
x^2 + 2x - 12x - 12 - 27 = 0
x^2 -10x - 39 = 0
(x + 3)(x - 13) = 0

x + 3 = 0
x = -3 (extraneous solution)

x - 13 = 0
x = 13

x + 2 = 15

so ur 2 integers are 13 and 15, with 15 being the largest

8 0

3 years ago

Need help on numbers 10 and 11 please!! sos

Harlamova29_29 [7]

Volume of prism B=800*6³/8³
=800*27/64
=25*27/2
=575/2
=287.5 cm³

5 0

3 years ago

Calculate the length of a triangle if another length is 15 feet and the hypotenuse is 39 feet.

Y_Kistochka [10]

We have to use Pythagorean theorem that says :

$a^2+b^2=c^2\\\Rightarrow a^2+15^2=39^2\\a^2=39^2-15^2\\a=\sqrt{1296}\\a=36$

The missing length of the triangle is : 36 feet

3 0

3 years ago

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

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

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

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

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

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

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

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