All categories

3 years ago

Answer??????????????????????????,?

Mathematics

1 answer:

ahrayia [7]3 years ago

5 0

This is not a multiple choice answer. You have to substitute the numbers given to the equation.

x=4

y=log(4)

y=log(2^2)

y=2log(2)
Use identity: logb^c = c x logb

x=0.25

y=log(0.25)

y=log(1/4)

y=log(1/2^2)

y=log(2^-2)

y=-2log(2)

x=0

Log is undefined

x=75
I’m not sure with this one. But if you put it in a calculator, you get 1.8751
If you find the derivative, it is 0, but I’m sure that it is not what you are looking for since you are only in algebra II.

Answers again: y=2log(2), y=-2log(2), undefined, and 1.8751

You might be interested in

Which relation is a function?

DIA [1.3K]

Answer: b

Step-by-step explanation:

You can determine which is a function by looking at the x values. If an x value repeats, it is not a function. In A, the x variable 2 repeats so it is not a function. In C, the x variable 5 repeats. In D, the x variable -5 repeats. The only answer choice where none of the x values are the same is B, so the answer is B. Hope this helps!

7 0

3 years ago

A math test is worth 100 points and has 38 problems. each problem is worth either 5 points or 2 points. how many problems of eac

blondinia [14]

Hey there,
There are 2 ways
1st way:
Since there are 8 5 point problems, 38 - 8 = 30, will give you the 2 point problems.

2nd way:
8 x 5 = 40
100 - 40 = 60
60 / 2 = 30
Thus there are 30, 2 point problems.

Hope this helps :))

~Top

4 0

3 years ago

Read 2 more answers

Please help I don’t think I remember this

nikdorinn [45]

Answer:

im not sure if this is right

Step-by-step explanation:

demension 1: 9 X 10 demension 2: 10 x 4

4 0

3 years ago

Read 2 more answers

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

4 years ago

Graph the system of equations.

Marat540 [252]

You can soustracte the first or the second one

and y cancel

x = 6

now you put 6 in x
you can take the first or the second one

6 - y = -2
6 - y = -2
y = -8

8 0

3 years ago