EASY WORK!!!plz look at the photo.

The reactants of two reactions are shown.

katovenus [111]

Answer:

The answer to your question is letter B.

Step-by-step explanation:

The correct answer is letter B because both Copper and Iron as reactants have an oxidation number of zero and when they have reacted their oxidation numbers would be different from those.

A redox reaction is a reaction that involves a transfer of electrons from one atom to another and the oxidation numbers of the reactants are different from the oxidation numbers from the products.

4 0

3 years ago

What is 50% of 20% = ?

algol [13]

10% because .5x.2=.1

4 0

3 years ago

Read 2 more answers

Find the slope and y intercept of the graph of the graph of each equation.

AleksandrR [38]

Solve each equation for y and then the slope is m and y-intercept is b as in:
$y = mx+b$

So, if you do that, you'll get (remember, this is <em>after</em> solving for y):
1. m = -6, b = -2
2. m = 5, b = -1
3. m = -5/3, b = 5
4. m = 0, b = -1/4
5. m = -1, b = -3
6. m = 3, b = -4
7. m = 1/2, b = -5/2
8. m = 7/2, b = 1

Hope this helps.

7 0

3 years ago

Read 2 more answers

At a 33-foot depth underwater, the pressure is 29.55 pounds per square inch (psi). At a depth of 66 feet, the pressure reaches 4

andrezito [222]

I am thinking it is times 2 everytime

4 0

3 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression:

$x_v=\frac{-b}{2a}$

Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$

See the graph produced in the attached image.

4 0

2 years ago