Answer:
(a) AH < HC is No
(b) AH < AC is Yes
(c) △AHC ≅ △AHB is Yes
Step-by-step explanation:
Given
See attachment for triangle
Solving (a): AH < HC
Line AH divides the triangle into two equal right-angled triangles which are: ABH and ACH (both right-angled at H).
To get the lengths of AH and HC, we need to first determine the measure of angles HAC and ACH. The largest of those angles will determine the longest of AH and HC. Since the measure of the angles are unknown, then we can not say for sure that AH < HC because the possible relationship between both lines are: AH < HC, AH = HC and AH > HC
Hence: AH < HC is No
Solving (b): AH < AC
Length AC represents the hypotenuse of triangle ACH, hence it is the longest length of ACH.
This means that:
AH < AC is Yes
Solving (c): △AHC ≅ △AHB
This has been addresed in (a);
Hence:
△AHC ≅ △AHB is Yes
We have to add 3 2/3 + 2 2/3 using fraction strips. 3 2/3 and 2 2/3 are the mixed numbers. The mixed numbers consists of a whole number and a fraction. If we want to add those numbers we shoukld change them into improper fractions: 3 2/3 = ( 3 * 3 + 2 ) / 3 = 11 / 3; 2 2/3 = ( 2 * 3 + 2 ) / 2 = 8 / 3. Finally: 11 / 3 + 8 / 3 = ( 11 + 8 ) / 3 = 20 / 3 = 6 2/3. Answer: You reneme it turning them into improper fractions<span>. </span>
Answer:
Hi how are you doing today Jasmine
Let width equal X and length equal X+2. Use algebra 72 = X times X+2
Answer:
B) -x+8y=56
Step-by-step explanation:
y=1/8x+7
Multiply by 8 to clear the fraction
8y = 8(1/8x+7)
Distribute
8y = 8*1/8x +8*7
8y = x+56
Subtract x from each side
-x+8y = x-x+56
-x+8y = 56