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MrRissso [65]
2 years ago
5

Jimmy is preparing a slide show to teach the grade three students the importance of cleanliness. which element should he use in

order to hold their attention?
Answers: A) Plain Background, B) Minimal images, C) Animation, D)layout, E) Speaker notes
Computers and Technology
2 answers:
melomori [17]2 years ago
7 0

Its Animation, if he uses animation the slide show will be more interesting and creative.

allochka39001 [22]2 years ago
3 0
Animation or minimal images sry stuck between both
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Naomi is giving a presentation on historical figures. She has all the pictures of the people she wants to talk about. After each
kykrilka [37]
The answer would be A. The most efficient way is putting all the pictures on the presentation first and then just put a new text box after each picture and then just repeat the process for all the pictures. 
4 0
2 years ago
Which of the following facts determines how often a nonroot bridge or switch sends an 802.1D STP Hello BPDU message?
irinina [24]

Answer:

b. The Hello timer as configured on the root switch.

Explanation:

There are differrent timers in a switch. The root switch is the only forwarding switch in a network, while non root switches blocks traffic to  prevent looping of BPDUs in the network. Since the root switch is the only forwarding switch, all timing configuration comes from or is based on the configuration in the root.

The hello timer is no exception as the nonroot switch only sends 802.1D DTP hello BPDU messages forwarded to it by the root switch and its frequency depends on the root switch hello timer.

8 0
2 years ago
The area of a square is stored in a double variable named area. write an expression whose value is length of the diagonal of the
cestrela7 [59]

Answer:

The correct answer to the following question will be "Math.sqrt(area*2)".

Explanation:

  • The Math.sqrt( ) method in JavaScript is used to find the squares root of the given figure provided to the feature as a variable.
  • Syntax Math.sqrt(value) Variables: this function takes a single variable value that represents the amount whose square root is to be determined.
  • The area of the figure is the number of squares needed to cover it entirely, like the tiles on the ground.

Area of the square = side times of the side.

Because each side of the square will be the same, the width of the square will only be one side.

Therefore, it would be the right answer.

5 0
2 years ago
1)When the liquid is spun rapidly, the denser particles are forced to the bottom and the lighter particles stay at the top. This
Delvig [45]

Answer:

Centrifugation.

Explanation:

When the liquid is spun rapidly, the denser particles are forced to the bottom and the lighter particles stay at the top. This principle is used in centrifugation.

Centrifugation can be defined as the process of separating particles from a liquid solution according to density, shape, size, viscosity through the use of a centrifugal force. In order to separate these particles, the particles are poured into a liquid and placed in a centrifuge tube. A centrifuge is an electronic device used for the separation of particles in liquid through the application of centrifugal force. Once the centrifuge tube is mounted on the rotor of the centrifuge, it is spun rapidly at a specific speed thereby separating the solution; denser particles are forced to the bottom (by moving outward in the radial direction) and the lighter particles stay at the top as a result of their low density.

6 0
2 years ago
Modify the TimeSpan class from Chapter 8 to include a compareTo method that compares time spans by their length. A time span tha
My name is Ann [436]

Answer:

Check the explanation

Explanation:

Here is the modified code for

TimeSpan.java and TimeSpanClient.java.

// TimeSpan.java

//implemented Comparable interface, which has the compareTo method

//and can be used to sort a list or array of TimeSpan objects without

//using a Comparator

public class TimeSpan implements Comparable<TimeSpan> {

   private int totalMinutes;

   // Constructs a time span with the given interval.

   // pre: hours >= 0 && minutes >= 0

   public TimeSpan(int hours, int minutes) {

        totalMinutes = 0;

        add(hours, minutes);

   }

   // Adds the given interval to this time span.

   // pre: hours >= 0 && minutes >= 0

   public void add(int hours, int minutes) {

        totalMinutes += 60 * hours + minutes;

   }

   // Returns a String for this time span such as "6h15m".

   public String toString() {

        return (totalMinutes / 60) + "h" + (totalMinutes % 60) + "m";

   }

   // method to compare this time span with other

   // returns a negative value if this time span is shorter than other

   // returns 0 if both have same duration

   // returns a positive value if this time span is longer than other

   public int compareTo(TimeSpan other) {

        if (this.totalMinutes < other.totalMinutes) {

            return -1; // this < other

        } else if (this.totalMinutes > other.totalMinutes) {

            return 1; // this > other

        } else {

            return 0; // this = other

        }

   }

}

// TimeSpanClient.java

public class TimeSpanClient {

   public static void main(String[] args) {

        int h1 = 13, m1 = 30;

        TimeSpan t1 = new TimeSpan(h1, m1);

        System.out.println("New object t1: " + t1);

        h1 = 3;

        m1 = 40;

        System.out.println("Adding " + h1 + " hours, " + m1 + " minutes to t1");

        t1.add(h1, m1);

        System.out.println("New t1 state: " + t1);

        // creating another TimeSpan object, testing compareTo method using the

        // two objects

        TimeSpan t2 = new TimeSpan(10, 20);

        System.out.println("New object t2: " + t2);

        System.out.println("t1.compareTo(t2): " + t1.compareTo(t2));

        System.out.println("t2.compareTo(t1): " + t2.compareTo(t1));

        System.out.println("t1.compareTo(t1): " + t1.compareTo(t1));

   }

}

/*OUTPUT*/

New object t1: 13h30m

Adding 3 hours, 40 minutes to t1

New t1 state: 17h10m

New object t2: 10h20m

t1.compareTo(t2): 1

t2.compareTo(t1): -1

t1.compareTo(t1): 0

4 0
2 years ago
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