Let X = the distance you need to find.
Using the Pythagorean Theorem we can solve for x
x = SQRT( 500^2 - 300^2)
x = SQRT (250,000 - 90,000)
x = SQRT(160,000)
x = 400
It is 400 yards
Answer:
3 * (x + 7)
Step-by-step explanation:
Mark number as x
Then the sum of a number and seven is x + 7
Three times x + 7 can be written as
3 * (x + 7)
Parentheses mean that we first add x and 7 and then multiply their sum by 3
Alternatively you can write it as 3x + 21, since
3 * (x + 7) = 3 * x + 3 * 7 = 3x + 21
<u>Angie buys</u>
1 Software package + 3 months of game play.
Each software package costs = $20.
Let us assume cost of one month of game play = $x.
Therefore, total cost to Angie for 1 Software package + 3 months of game play = 20*1 + 3x = 20 +3x.
<u> Kenny buys </u>
1 software package + 2 months of game play.
Therefore, total cost to Kenny for 1 software package + 2 months of game play = 20*1 + 2*x = 20+2x.
Their total cost = $115.
Adding their costs and set it equal to 115, we get
<h3>20 +3x + 20+2x = 115.</h3>
Now, we need to solve it for x.
40 + 5x = 115.
5x = 115 - 40.
5x = 75.
Dividing both sides by 5, we get
x= 15.
<h3>Therefore, $15 is the total cost of one month of game play.</h3>
<u>Answer:</u>
$6.00
<u>Step-by-step explanation:</u>
To solve this problem, we have to construct two separate equations for each family, and then use any of the methods for solving simultaneous equations.
Let's consider <em>h </em>to represent the cost of 1 hot dog, and <em>w </em>to mean the cost of 1 water bottle.
• For the first family:
We can rearrange the equation to make <em>w</em> the subject:
⇒ 
⇒ 
• For the second family:
Since we have previously obtained an expression for <em>w</em> in terms of <em>h</em>, we can substitute that expression for <em>w</em> in the above equation, and then solve for <em>h</em>:
⇒ 
⇒ 
⇒ 
[Multiplying both sides of the equation by 3]
⇒ 
⇒ 
⇒ 
∴ The price of one hot dog is $6.00.
<span>they appear to be using the (2x) as the variable;
so,
</span><span>e^(t) + t^2 ;
now fill in the interval [0,2x]
e^(2x) + (2x)^2 -e^(0)
D{t} [e(2x) +4x^2 - 1]
2e^(2x) + 8x</span>
