for these kinds of problems the outside angle always equals the opposite 2. meaning angle x=64+70 which is 134 degrees. you can double check this because all triangles equal 180 degrees. meaning that the last angle in the triangle is 46. 46+70+64=180.
so your answer is 134.
hope this helps :)
4.5 boxes of nails are required for finishing 2 tables
Step-by-step explanation:
Given:
3.4 boxes of nails are required for finishing 1.5 tables
Required:
How much boxes of nails would he use for finishing 2 tables
Solution:
We can solve using Unitary Method:
Nails needed to finish 1.5 tables = 3.4 boxes
Nails needed to finish 1 table = 
Nails needed to finish 2 tables = 
So, 4.5 boxes of nails are required for finishing 2 tables.
Keywords: Word Problems
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Multiplying both sides of <span>−1/3x≤−6 by -3 results in "x is equal to or greater than 18."
Note that multiplying such an inequality requires reversing the direction of the inequality symbol.
I subst. 18 for x in </span><span>−1/3x≤−6 as a check, and found that the resulting inequality is true.</span>
Every calculation with multiple operations must abide to BODMAS (Brackets, Other, Divide, Multiple, Addition, Subtraction)
Thus we would multiple first, giving us 86138220. Then we would add giving us: 86145009
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.