So you distrubute and get
-2p-8+2-3+5p
add like terms
3p-9
ANSWER: 3p-9
If triangle KMN is congruent with triangle LMN, then:
1) KM must be congruent with LM→KM=LM→
x+2y=3x-y→x+2y-x-2y=3x-y-x-2y→0=2x-3y→2x-3y=0 (1)
2) KN must be congruent with LN→KN=LN→
5x+3y=8x-7→5x+3y-5x-3y+7=8x-7-5x-3y+7→7=3x-3y→3x-3y=7 (1)
We have a system of two equations and two unkowns:
(1) 2x-3y=0
(2) 3x-3y=7
Subtracting equation (1) from equation (2)
(3x-3y)-(2x-3y)=7-0
3x-3y-2x+3y=7
x=7
Replacing x=7 in equation (1)
(1) 2x-3y=0→2(7)-3y=0→14-3y=0→
14-3y+3y=0+3y→14=3y→14/3=3y/3→14/3=y→y=14/3
The length of the hypotenuse is:
KN=5x+3y=5(7)+3(14/3)=35+14→KN=49
or
LN=8x-7=8(7)-7=56-7→LN=49
Answer: Option C. 49
The answer should be 0.40(1,125)=450
<span>If tan theta= 15/8, then theta is expected that theta is found in the first and third quadrant. the y-component is 15 and the x-component is 8. hence the hypotenuse is 17 from the pythagoren theorem. cos theta thus is equal to 8/17, sec theta is equal to 17/8. csc theta is equal to 17/15. answer are A and D</span>
Answer: D: -2
explanation:
1. divide fractions inside brackets by multiplying by the reciprocal
1/4 • 4/7
= 4/28
2. multiply 4/28 by 7/5
= 28/140 which can be reduced to 1/5
3. multiply by -10
= 1/5 • -10
= -2
