Match each item with a statement below. designed to play and possibly record music, photos, movies, and video on a television or extra-large monitor screen the act of printing to a printer on a network connection uses a print head that moves across the paper creating one line with each pass data sent to a printer without any formatting or processing a type of file similar to a PDF, can be viewed, edited, printed, faxed, emailed, or posted on websites long-handled brush made of nylon fibers that are charged with static electricity a type of electrophotographic printer that can range from a small, personal desktop model to a large, network printer capable of handling and printing large volumes continuously creates a printed page by using some mechanism that touches or hits the paper a bunch of bits in rows and columns used to align and/or clean inkjet nozzles, can solve problems when colors appear streaked or out of alignment a. Home Theater PC b. remote printing c. Inkjet printer d. Raw data e. XPS file f. Extension magnet brush g. Laser printer h. Impact printer i. Bitmap j. Calibration

Answer:

arc(AD)=74\°

Step-by-step explanation:

Givens

• AC and BD are diameters.
• Angle BCA measures 53°.

Notice that line segment BE is congruent to line segment CE, because they are the radius of the circumference. That means the triangle that contains these sides is isosceles.

\angle BCA = 53\° = \angle CBE, by definition of triangles isosceles.

Then,

\angle BEC = 180-53-53=74\°, by sum of internal angles.

And,

\angle BEC = \angle DEA, by vertical angles definition.

\angle DEA = 74\°

But, we know that the measured of a subtended arc is equal to its angle.

Therefore, arc(AD)=74\°

The measure of arc AD is 74°.

Given that,

In circle E, and are diameters. Angle BCA measures 53°. Circle E is shown. Line segments AC and BD are diameters.

Lines are drawn to connect points B and C and points A and D.

Angle BCA is 53 degrees.

We have to determine,

What is the measure of arc AD?

According to the question,

The line segment BE is congruent to line segment CE because they are the radius of the circumference. That means the triangle that contains these sides is isosceles.

Lines are drawn to connect points B and C and points A and D.

Angle BCA is 53 degrees.

By the definition of the isosceles triangle,

∠BCA = 53° =∠CBE

The sum of internal angles is,

∠BEC = 180° - 53° - 53° = 74°

And By vertical angle definition,

∠BEC = ∠DEA

So, ∠DEA = 74°

The measurement of a subtended arc is equal to its angle is,

Therefore, arc(AD) = 74°

Hence, The measure of arc AD is 74°.

Answer:

The grooves behind the chain have been cut over decades by motion of the chain as hikers grabbed it, and the potholes were cut by water swirling rocks around during the rare floods over much longer times.

Explanation:

In the question above, we can see that the physical action of some factors found in this environment, such as the movement of the currents and the movement of stones in the water, which managed to cause some changes in the environment, building a totally new scenario, which would not occur if these factors did not exist.

By reading the question and imagining the scenario described, we can conclude that the grooves behind the chain have been cut over decades by motion of the chain as hikers grabbed it, and the potholes were cut by water swirling rocks around during the rare floods over much longer times.

Explanation:

Join G⭕new in how many years will 24000 amount to 27783 at 5% per annum compounded annually?

Answer: The ratio of \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}  in the cell is 4.07\times 10^{-4}

Explanation:

The half reactions for the cell is:  

Oxidation half reaction (anode):  Sn(s)\rightarrow Sn^{2+}+2e^-

Reduction half reaction (cathode):  Sn^{2+}+2e^-\rightarrow Sn(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.10 V

\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}  = ?

Putting values in above equation, we get:

0.10=0-\frac{0.0592}{2}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}

\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}=4.016\times 10^{-4}

Hence, the ratio of \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}  in the cell is 4.016\times 10^{-4}

The ratio of the Sn2+ concentrations in the two half-cells is 0.00042.

The cell reaction is; Sn(s) ------> Sn^2+(aq) + 2e

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Ecell = 0.10 V

n = 2

Q = ?

E°cell = 0.00 V

0.10 V = 0.00 V - 0.0592/2 log Q

-(0.10 V × 2/0.0592) = log Q

-3.378 = log Q

Q = 0.00042