HELP ASAP PLS

Which three values are greater than

solniwko [45]

Answer:

3/4 is greater than 1/2

45% = 9/20 is less than 1/2

1/3 is less than 1/2

0.62 is greater than 1/2

59% = 59/100 is greater than 1/2

0.315 is less than 1/2

So, A, D and E are greater than 1/2

Hope this helps!

6 0

3 years ago

What is the equation of the following graph in vertex form?

anygoal [31]

Answer:

I think the answer is 0,8

6 0

3 years ago

I'm stuck what is 15% of 32

Len [333]

In math, 'of' means 'multiplied by,' so 15% of 32 translates to '15% * 32.'
Now solve the equation:
x = 15% * 32
x = .15 * 32
x = 4.8

6 0

2 years ago

If f(x)=X-3/x and g(x) = 5x – 4, what is the domain of (fog)(x)?<br><br><br> NEED HELP FAST!!

dedylja [7]

Answer:

Step-by-step explanation:

This is function composition. We will take the g function and plug it into the f function wherever we see an x. That will look like this when it's done but before it's simplified:

$f(g(x))=\frac{5x-4-3}{5x-4}$ and that can be simplified down to

$f(g(x))=\frac{5x-7}{5x-4}$

The domain of this function will be all real numbers EXCEPT for the one(s) that cause the denominator to become undefined. Set the denominator equal to 0 and solve for x:

5x - 4 = 0 so

5x = 4 and

x = 4/5

In other words, D = {x | x ≠ 4/5}

6 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

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

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

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

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

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

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

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