Answer: 14x^2-93xy+60y^2 Hope that helps!
Step-by-step explanation:
1. Expand by distributing terms
(20x-12y)(x-4y)-(3x-4y)(2x+3y)
2. Use the Foil method:(a+b)(c+d)= ac+ad+bc+bd
20x^2-80xy-12yx+48y^2-(3x-4y)(2x+3y)
3. Use the Foil method : (a+b)(c+d)= ac+ad+bc+bd
20x^2-80xy-12yx+48y^2-(6x^2+9xy-8yx-12y^2)
4. Remove parentheses 20x^2-80xy-12yx+48y^2-6x^2-9xy+ 8yx+12y^2
5. Collect like terms (20x^2-6x^2)+(-80xy-12xy-9xy+8xy)+(48y^2+12y^2)
6. Simplify.
And your answer would be 14x^2-93xy+60y^2
Answer:
False
Step-by-step explanation:
y=x-2 y=3x+4
Substitute the point into both equations and see if it is true
2=4-2
2=2 true
and
2=3*4+4
2 = 12+4
2 =16
False
It is not a solution
Answer:
3x^2 -4x -4
Step-by-step explanation:
3x^2 – 8x – 2 + 4x – 2
Combine like terms
3x^2 -8x+4x -2 -2
3x^2 -4x -4
Answer:
x = 6.40 units
Step-by-step explanation:
4^2 + 5^2 = x^2
x = 6.40 units
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°.
