Answer:
how many square centimeters is the box??
Step-by-step explanation:
2a) there is a right angle on T so 180-90-37=53
2b) do pythogrean theorem 22sq+ 12sq= 628 sq rt of 628 is 25.06 which is 25
5a) 5*8=10x so X=4
5b) 47*2=94 -57 = 37
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°.
Answer:
x^2+7x=30
x^2+7x-30=0
Here a=1,b=7 and c=-30
Now,
Discriminant(D)=b^2-4ac
=7^2-4×1×(-30)
=49+120=169
By applying Quadratic formula
x=-b+- root over b^2-4ac÷2a
=-7+- root over D ÷ 2×1
=-7 +- root over 169 ÷2
=-7 +- 13 ÷2
Now,
Either x=-7+13÷2=6÷2=3
Or x=-7 -13 ÷2=-20÷2=-10
Hence, ans is (C) x=-10 and x=3
