Answer:
A. 15
Step-by-step explanation:
To solve this you need to compare the lengths given to you in the question statement.
Because the lines originate from a single point, they're like triangles. We can easily see a triangle AGF and a triangle ADE, right?
Both triangles are similar triangles, so we can see triangle ADE as a larger version of angle AGF.
They give you the dimension of A F and A E (through A F + F E) to establish a ratio... and they give you A G, asking for A D.
So, A F = 16, A E = 20 (16 + 4), A G = 12.
Since A D is to A G what A E is to A F, we can easily make the following cross-multiplication:
So, A D = (A G * A E)/A F
A D = (12 * 20) / 16 = 15
Answer:
<em>the</em><em> </em><em>perim</em><em>eter</em><em>=</em><em> </em><em>1</em><em>0</em><em>+</em><em>4</em><em>W</em><em>.</em>
<em>where</em><em> </em><em>W</em><em>=</em><em> </em><em>the</em><em> </em><em>wid</em><em>th</em><em>.</em>
Step-by-step explanation:
mathematically, perimeter (p)= 2L +2W
where L= length.and W=width
<em>from</em><em> </em><em>the</em><em> </em><em>exp</em><em>ression</em><em> </em><em>;</em><em>length is 5 more than the width</em>
<em>it</em><em> </em><em>is</em><em> </em><em>wri</em><em>tten</em><em> </em><em>as</em><em> </em>L = 5+W
<em>sub</em><em>stitute</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>formu</em><em>la</em><em> </em>
P= 2(5+W)+2W
P=10+2W+2W
P=10+4W as the perimeter of the rectangle.
Answer:
133
Step-by-step explanation:
Answer is 133
Answer:
-(-4)(-6)*3/5(10+15)
when there is a (-) in front of an expression in parentheses, change the sing of each term in the expression.
4*(-6)*3/5*(10+15)
Add the numbers (10+15)
4(-6)*3/5*25
multiplying an odd number of negative terms makes the product negative
-4*6*3/5*25
reduce the number with the greatest common factor 5
-4*6*3*5
calculate the product
solution
-360
Answer: A
Step-by-step explanation:
If you look only at the ranges provided for x A is the only one that fits the graph. The domain of the first function (
) is everything equal to or less than 2. The equal to is represented by the closed circle at point (2,5) which represents that this value is included for that function. The other function continues on with values greater than 2, but does not include the x value 2 as it has an open circle at point (2,10).
