now, we know that B and F are midpoints, namely BF is a midsegment of the triangle, and therefore BF || AE, so, the base AE is twice BF.
![\bf 2BF=AE\implies 2(23)=5x-4\implies 46=5x-4\implies 50=5x \\\\\\ \cfrac{50}{5}=x\implies \boxed{10=x}](https://tex.z-dn.net/?f=%5Cbf%202BF%3DAE%5Cimplies%202%2823%29%3D5x-4%5Cimplies%2046%3D5x-4%5Cimplies%2050%3D5x%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B50%7D%7B5%7D%3Dx%5Cimplies%20%5Cboxed%7B10%3Dx%7D)
Step-by-step explanation:
hello,
the shape is a right trapezium.
you use the following formula:
area of a trapezium = ½ × sum of parallel sides× height.
or simply,
a = ½ ( a + b ) h
where a and b are the parallel sides
and h is the height.
thus,
a = ½ × ( 4 + 6 )× 4
a = ½ ( 10 ) × 4
a = 5 × 4
a = 20
have a great day! au revoir !
Digit 3 on the 10 side has a value of 3 tens or 30. We want to find the digit that has exactly 10 times that value. well, 10 × 30 = 300. So which digit has the value of 300? The other 3 is in the 100's place and it's value is 3 one hundreds or 300. So that is your digit... the 3 in the hundreds place.
Answer:
Quotient of a number and 3 = x/3
Ten less the quotient of a number and 3 will equal 6
So...
10 - x/3 = 6
+ x/3 +x/3
10 = x/3 +6
-6 - 6
3 × 4=(x/3) × 3
12 = x