All categories

3 years ago

Write an equivalent expression for 10^6 / 10^4

Mathematics

2 answers:

creativ13 [48]3 years ago

6 0

The answer would 10^2 power

Yuri [45]3 years ago

3 0

Answer:

The answer would be 10^2 power.

You might be interested in

21 Last year Chelsea and Sonya took 236 pictures during Thanksgiving. This year Chelsea and Sonya each took the same amount of p

Leokris [45]

Let 'c' represent the number of pictures Chelsea took.

Let 's' represent the number of pictures Sonya took.

For last year's Thanksgiving, c + s = 236

For this year's Thanksgiving, let 'x' represent the number of photos taken in total. x = c + s, where c and s are two integers that are the same (c = s).

And we know that for both years, c + s + x = 500.

As we know that c + s is already 236 from last year, we can remove c + s from the equation in bold and replace it with 236 instead.

236 + x = 500.

Now we have to isolate the x term.

x = 500 - 236

x = 264.

We know that x = c + s, where c and s are the same, so we can just use one of the variables and double it (so you either get 2c or 2s - it doesn't matter which one you pick because they're both the same).

2c = 264

c = 132

c = s

s = 132.

Both took 132 pictures this year.

4 0

3 years ago

I'm really not sure how to go about solving this algebraic equation, please help! .513 = (2x)^2 / (.05-x)

LiRa [457]

Let's set it up like this:
$0.513=\frac{\left(2x\right)^2}{\left(0.05-x\right)}$
Multiply both sides by $0.05-x$
$0.513\left(0.05-x\right)=\frac{\left(2x\right)^2}{0.05-x}\left(0.05-x\right)$
$0.513\left(0.05-x\right)=4x^2$
We are then going to use the distributive property. Since we also know that the opposite of an number that is squared is the square root, we can also apply that. We would be left with something like this:
$x=-\frac{0.513+\sqrt{0.673569}}{8},\:x=\frac{\sqrt{0.673569}-0.513}{8}$
The variable $x$ can be both positive or negative.
We have found successfully our answer.

Let me know if you have any questions regarding this problem!
Thanks!
-TetraFish

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

(a) x=2, y=-1

(b) x=2, y=2

(c) $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) x=-2, y=-7

Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

Click the picture for the question please help!!!

sesenic [268]

Answer:

297.33 m

Step-by-step explanation:

3 0

3 years ago

Determine whether the statement below is always, sometimes, or never true. Justify your reasoning.

zloy xaker [14]

This is sometimes true I say this because if u substitute your answers that can be different or can be the same.
for ex. x=6 y=6; 6-3 and 6-3 will be the same but if x= 10 and y=15; 10-3 and 15-3 would give you different answers

5 0

3 years ago

Read 2 more answers