Answer:
Height of each triangle: 14cm
Area of each triangle: 168cm²
Step-by-step explanation:
So we know that the base of the area is 24. Lets find the height.
336 ÷ 24 = 14
If he makes two triangles out of the rectangle that just means he cuts it in half.
To find the area of one triangle lets do:
336 ÷ 2 = 168
Let's double check our answer:
(14 × 24) ÷ 2 = 168
Seems great!
Answer:
8.9
Step-by-step explanation:
Convert √
80 to a decimal.
8.9442719
Find the number in the tenth place 9 and look one place to the right for the rounding digit 4
. Round up if this number is greater than or equal to 5 and round down if it is less than 5
.
Answer: Choice D
(a-e)/f
=======================================
Explanation:
Points D and B are at locations (e,f) and (a,0) respectively.
Find the slope of line DB to get
m = (y2-y1)/(x2-x1)
m = (0-f)/(a-e)
m = -f/(a-e)
This is the slope of line DB. We want the perpendicular slope to this line. So we'll flip the fraction to get -(a-e)/f and then flip the sign from negative to positive. That leads to the final answer (a-e)/f.
Another example would be an original slope of -2/5 has a perpendicular slope of 5/2. Notice how the two slopes -2/5 and 5/2 multiply to -1. This is true of any pair of perpendicular lines where neither line is vertical.
First, we need to transform the equation into its standard form (x - h)²=4p(y - k).
Using completing the square method:
y = -14x² - 2x - 2
y = -14(x² + 2x/14) - 2
y = -14(x² + 2x/14 + (2/28)²) -2 + (2/28)²
y = -14(x + 1/14)² - 391/196
-1/14(y + 391/196) = (x + 1/14)²
This is a vertical parabola and its focus <span>(h, k + p) is (-1/14, -391/196 + 1/56) = (-1/14, -775/392).
Or (-0.071,-1.977).</span>
<h2>
<em>Answer:</em></h2><h2>
<em>4</em><em>2</em><em> </em><em>cm^</em><em>2</em></h2>
<em>please </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em> for</em><em> full</em><em> solution</em>
<em>hope </em><em>it</em><em> helps</em><em>.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em>
