D. 192in^2
This would be a simple area problem with a triangle. REMEMBER THIS EQUATION: a=bh*1/2 (b for base and h for height, these are multiplied together then that answer is halved out.)
So we just need to plug in our values into the equation, so the equation would look like a= (16)*(24)*1/2. 16 times 24 would then give you 384, you could either divide by 2 or multiply by 0.5 to get the next answer, as long as your HALVING the answer.
So we have our bh value so now we can multiply by 1/2 which will give us 384*1/2 which leaves us with 192.
I have also attached a photo of doing a longer(ish) way than this, that also proves that this equation works. Either one will provide you an answer.
Answer:
D. 4,704 ft²
Step-by-step explanation:
Giving a scale of 1 in. = 14 ft, to find the actual area of the field represented by the triangular drawing above, convert the length of the base and height to the actual lengths of the field using the scale.
Height of drawing = 8 in.
Actual height = 8*14 = 112 ft
Base of drawing = 6 in.
Actual base = 6*14 = 84 ft
Area of the field = area of ∆
Area = ½*base*height
Area = ½*84*112
Area = 4,704 ft²
Answer:
each notebook costs $2.70
each pack of pencils costs $1.50
Step-by-step explanation:
system of equations:
let p = pack of pencils
let n = notebook
3p + 5n = 18
4p + 4n = 16.8
I used the elimination method by multiplying the first equation by 4 and the second equation by -3
4(3p + 5n = 18) = 12p + 20n = 72
-3(4p + 4n = 16.8) = -12p -12n = -50.4
adding the new equations together you get: 8n = 21.6
n = 21.6/8
n = $2.70
solve for 'p':
3p + 5(2.7) = 18
3p + 13.5 = 18
3p = 4.5
p = $1.50
There is no solution for this equation
<span>The mid-point or the number that divides a series of values into two groups of equal numbers of values is referred to as the median. To get the median value, first sort the numbers from highest to lowest or lowest to highest and then pick the middle number. However, for even numbered sets of numbers, take the average of the two middle numbers and the resulting number is the median. </span>