Answer:
Step-by-step explanation:
f(x) = x is an odd function; the power of x is x^1.
f(x) = -x is also an odd function.
Answer:
![\boxed{ \bold{ \huge{ \boxed{ \sf 240 \: {inches}^{2} }}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7B%20%5Chuge%7B%20%5Cboxed%7B%20%20%5Csf%20240%20%5C%3A%20%20%7Binches%7D%5E%7B2%7D%20%7D%7D%7D%7D)
Step-by-step explanation:
Given,
Length of a rectangle = 20 inches
Perimeter of a rectangle = 64 inches
Area of a rectangle = ?
Let width of a rectangle be ' w ' .
<u>Fi</u><u>rst</u><u>,</u><u> </u><u>finding </u><u>the</u><u> </u><u>width</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>rectangle</u>
![\boxed{ \sf{perimeter = 2(l + w)}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Csf%7Bperimeter%20%3D%202%28l%20%2B%20w%29%7D%7D)
plug the values
⇒![\sf{64 = 2(20 + w)}](https://tex.z-dn.net/?f=%20%5Csf%7B64%20%3D%202%2820%20%2B%20w%29%7D)
Distribute 2 through the parentheses
⇒![\sf{64 = 40 + 2w}](https://tex.z-dn.net/?f=%20%5Csf%7B64%20%3D%2040%20%2B%202w%7D)
Swap the sides of the equation
⇒![\sf{40 + 2w = 64}](https://tex.z-dn.net/?f=%20%5Csf%7B40%20%2B%202w%20%3D%2064%7D)
Move 2w to right hand side and change it's sign
⇒![\sf{2w = 64 - 40}](https://tex.z-dn.net/?f=%20%5Csf%7B2w%20%3D%2064%20-%2040%7D)
Subtract 40 from 64
⇒![\sf{2w = 24}](https://tex.z-dn.net/?f=%20%5Csf%7B2w%20%3D%2024%7D)
Divide both sides of the equation by 2
⇒![\sf{ \frac{2w}{2} = \frac{24}{2} }](https://tex.z-dn.net/?f=%20%5Csf%7B%20%5Cfrac%7B2w%7D%7B2%7D%20%20%3D%20%20%5Cfrac%7B24%7D%7B2%7D%20%7D)
Calculate
⇒![\sf{w = 12 \: inches}](https://tex.z-dn.net/?f=%20%5Csf%7Bw%20%3D%2012%20%5C%3A%20inches%7D)
Width of a rectangle ( w ) = 12 inches
<u>Now</u><u>,</u><u> </u><u>finding</u><u> </u><u>the</u><u> </u><u>area</u><u> </u><u>of </u><u>a</u><u> </u><u>rectangle</u><u> </u><u>having</u><u> </u><u>length</u><u> </u><u>of</u><u> </u><u>2</u><u>0</u><u> </u><u>inches</u><u> </u><u>and </u><u>width </u><u>of</u><u> </u><u>1</u><u>2</u><u> </u><u>inches</u>
![\boxed{ \sf{area \: of \: rectangle = length \: \times \: \: width}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Csf%7Barea%20%5C%3A%20of%20%5C%3A%20rectangle%20%3D%20length%20%5C%3A%20%20%5Ctimes%20%20%5C%3A%20%5C%3A%20width%7D%7D)
plug the values
⇒![\sf{area \: of \: rectangle =20 \times 12 }](https://tex.z-dn.net/?f=%20%5Csf%7Barea%20%5C%3A%20of%20%5C%3A%20rectangle%20%3D20%20%5Ctimes%20%2012%20%7D)
Multiply the numbers : 20 and 12
⇒![\sf{area \: of \: rectangle = 240 \: {inches}^{2} }](https://tex.z-dn.net/?f=%20%5Csf%7Barea%20%5C%3A%20of%20%5C%3A%20rectangle%20%3D%20240%20%5C%3A%20%20%7Binches%7D%5E%7B2%7D%20%7D)
Hence, Area of a rectangle = 240 inches²
Hope I helped !
Best regards!
Answer:4.5 for triangle + 56= 60.5 squuare units
Step-by-step explanation:
Answer:
$128
Step-by-step explanation:
1. We need to find how much 1 yard of material cost.
If needed, use a calculator*
120/75=1.6
2. Now we need to know how much 15 yards of material cost.
1.6 times 15=24
3. The buyer ordered an additional 90 yards. If we know how much 75 yards cost, and 15 yards, all you have to do is...
120+24=144
90 yards=$144
4. The additional cost is $144
(Note: the total cost would be $264 (120+144=264))
Ok like what kind of math problems