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ira [324]
3 years ago
13

If string str1 = "forest" and string str2 = "school", then what is the value of str1.compareto(str2);?

Mathematics
1 answer:
REY [17]3 years ago
8 0
<span>This is an java command program. String 1 is stands for str1= '' forest '' accepted within quote marks. String 2 is stands for str1= '' school'' accepted within quote marks the command compareto will cause the output will display as shown below : str1.compareto(str2); when run this command we will get the output like this ; forest school ================================================== the model program is like this; { String str1 = " forest " ; String str2 = "school "; System.out.println(); System.out.println("String 1: " + str1); System.out.println("String 2: " + str2);</span>
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I need help ASAP. What would go in the middle??
Trava [24]

Answer:

C

Step-by-step explanation:

<h3>7.4 x 10 <em>-3</em>= 0.0074</h3><h3>1.3 x 10 <em>-2</em>= 0.013</h3>

In between them would be 0.0092.

5 0
3 years ago
Zachary bought 8 hot dogs at the fair. He paid a total cost of $12.00 for the food. Find the unit cost of one hot dog.
masya89 [10]

Answer:.66

Step-by-step explanation:

You divide 8 and 12

3 0
4 years ago
Read 2 more answers
When a number is substituted for the same variable in two expressions, how many times must those two expressions have different
zysi [14]

Answer:

Hope this helps!!!

Step-by-step explanation:

An equation is a mathematical statement that two expressions are equal. The solution of an equation

is the value that when substituted for the variable makes the equation a true statement.

Our goal in solving an equation is to isolate the variable on one side of the equation and a number on

the other side so the equation reads:

Variable = Number

To achieve our goal, we use two principles of equality, the addition principle and the multiplication

principle.

• Use the addition principle to move terms from one side of the equation to the other side. To

move a term, add it's opposite to both sides of the equation.

• Use the multiplication principle to solve for the variable. If the variable is multiplied by a

number, divide both sides of the equation by that number. If the variable is divided by a

number, multiply both sides of the equation by that number.

To solve equations, use the procedure outlined below.

Steps for Solving Equations

Step 1: Clear fractions and decimals by multiplying each term of the equation by the LCD (least

common denominator).

Step 2: Remove the parentheses by distributing.

Step 3: Combine any like terms found on the same side.

Step 4: Use the addition principle to move the variable term to one side of the equation and the

number to the other side.

Step 5: Multiply or divide to solve for the variable.

Step 6: Check the result in the original equation.

4 0
3 years ago
Read 2 more answers
1000+20029293848493+2832783920
jonny [76]
An easy way is since there are 2 big numbers and 1 small, add the small to the larger
let's add it to the third number
1,000+2,832,783,920=2,832,784,920
now add 20,029,293,848,493 to 2,832,784,920 (use calculator)
basically you add up the1's  place with the 1's place, 10's place with the 10's place and so on
1's=3+0=3
10's=9+2=11 or 1 and move 1 up to next place
100's=4+9+1=14 or 4 and move 1 up to next place
1,000's= 8+4+1=13 or 3 and move 1 up to next place
10,000's= 4+8+1=13 or 3 and move 1 up to next place
100,000's= 8+7+1=16 or 6 and move 1 up to next place
1,000,000's= 3+2+1=6
10,000,000= 9+3=12 or 2 and move 1 up to next place
100,000,000=2+8+1=11 or 1 and move 1 up to next place
1,000,000,000= 9+2+1=12 or 2 and move 1 up to next place
therer are no more number on the smaller number so just add the bigger ones at the top with the added 1 so
20,032,126,633,413
6 0
3 years ago
Read 2 more answers
If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi
Finger [1]

<u>ANSWER: </u>

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

<u>SOLUTION: </u>

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = \mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2} --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval  

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16

34 \times 1+\frac{1}{2} \times a \times 1^{2}=18

34+\frac{a}{2}=18

\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0
3 years ago
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