Answer:
11x-15
Step-by-step explanation:
4x-8+7x-7
11x-15
Answer:
Yes
Step-by-step explanation:
The fraction 4/5 is equivalent to 24/30 (To get this I multiplied both the numerator and denominator by 6). The fraction 5/6 is equivalent to 25/30 (To get this I multiplied both the numerator and denominator by 5).... Since 25 is greater than 24, 25/30 (or originally 5/6) is the larger fraction
Step 1
Find the area of one equilateral triangle
Applying the law of sines
in this problem
a=b=7 cm
C=60 degrees
so
cm²
Step 2
To calculate the area of the hexagon multiply the area of one equilateral triangle by 
![Area Hexagon=6*[49*\frac{\sqrt{3}}{4}] \\ Area Hexagon=73.5*\sqrt{3}](https://tex.z-dn.net/?f=%20Area%20Hexagon%3D6%2A%5B49%2A%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5D%20%20%5C%5C%20Area%20Hexagon%3D73.5%2A%5Csqrt%7B3%7D%20%20)
cm²
therefore
the answer is the option
73.5 sqrt 3cm²
Answer:
Review how to find the percent of an amount to find the percent of an amount or of a whole we write
Step-by-step explanation:
Answer: 0.5
Step-by-step explanation: In this problem, we're asked to solve the following equation for <em>p. </em>Let's first switch 14 and 14p around so we have 14p + 14 = 21.
<em />
To solve this equation for <em>p</em>, we must first isolate the term containing <em>p</em> which in this case is 14p.
Since 14 is being added to 14p, we need to subtract 14 from both sides of the equation.
14p + 14 = 21
-14 -14
On the left side of the equation, the positive 14 and negative 14 cancel each other out and we have 14p. On the right side of the equation, we hav 21 - 14 which gives us 7.
Now we have the equation 14p = 7.
Since <em>p</em> is being multiplied by 14, to get <em>p</em> by itself, we divide both sides of the equation by 14.
On the left side of the equation the 14's cancel and we are left with <em>p</em>. On the right side of the equation, 7 divided by 14 is 0.5 which is our answer.
Therefore, p = 0.5 which is the solution for our equation.
Remember, you can always check your solution by substituting a number in for a variable to make sure the equation is true.
