All categories

3 years ago

A rectangular park had a perimeter of 18 miles. It is 7 miles wide. What is the area of the park?

1 answer:

7 0

Perimeter = 18 miles
Perimeter = 2a + 2b
2a + 2b = 18

a = ?
b = 7

18 = 2a + 2*7
18 = 2a + 14
2a = 18 - 14
2a = 4
a = 2

Area = a * b
Area = 2 * 7 = 14 miles2

You might be interested in

Which problem could be solved by evaluating the expression 7 x (-3)?

Tasya [4]

Answer:

The answer is -21

Step-by-step explanation:

To find the answer just multiply 7 by 3 and that's 21, so just add a negative sign to 21 and you get -21. I hope this helps :) please give me brainliest :)

6 0

3 years ago

Read 2 more answers

The value of 4 in 64513 is how many times the value of 4 in

il63 [147K]

It will be

(4,000) divided by (the value of 4 in the number you left out).

7 0

3 years ago

What is the equation of the Line Y=<br> X+

Korolek [52]

Answer:

uhm

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

olivia purchases gifts for her mom and grandma for mother's day. The first is a pair of sunglasses for $16.35 and the second is

PolarNik [594]

$23.40

Since we know how much Olivia spent total, and we know how much she spent on the sunglasses, we just subtract 16.35 from 39.74, and the final answer is 23.40

3 0

3 years ago

Read 2 more answers

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

4 years ago