Answer:
1
Step-by-step explanation:
x+2/2x=3/2
2x×3=2x+4
6x=2x+4
4x=4
x=4/4
x=1
Answer: see below
<u>Step-by-step explanation:</u>
42) 11 < 3y + 2 < 20
<u> -2 </u> <u> -2 </u> <u> -2 </u>
9 < 3y < 18
<u> ÷3 </u> <u>÷3 </u> <u> ÷3 </u>
3 < y < 6
Graph: o----------o
3 6
44) 36 ≥ 1 - 5z > -21
<u> -1 </u> <u> -1 </u> <u> -1 </u>
35 ≥ -5z > -22
<u> ÷ -5 </u> ↓ <u> ÷ -5 </u> ↓ <u>÷ -5 </u>
7 ≤ z < 4.4
Graph: o------------ ·
4.4 7
46) 6b + 3 < 15 or 4b - 2 > 18
<u> -3 </u> <u> -3 </u> <u> +2 </u> <u>+2 </u>
6b < 12 4b > 20
<u> ÷6 </u> <u>÷6 </u> <u> ÷4 </u> <u>÷4 </u>
b < 2 or b > 5
Graph: ←--------o o---------→
2 5
48) 8d < -64 and 5d > 25
÷8 ÷8 ÷5 ÷5
d < -8 and d > 5
there is no number that is both less than - 8 and greater than 5
No Solution
Graph: (empty)
50) 15x > 30 and 18x < -36
<u> ÷15 </u> <u> ÷15 </u> <u>÷18 </u> <u>÷18 </u>
x > 2 and x < -2
there is no number that is both less than - 2 and greater than 2
No Solution
Graph: (empty)
Answer:
A. (y+z=6) -8
Step-by-step explanation:
You will use the process of adding together the like terms. Since in equation Q, we have 8y, we need -8y in equation P. Answer A is the only one that will give us -8. You have to distribute -8 across all the variables and numbers in the parenthesis.
A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number. I dont know about the other question
Answer:
The base is 4cm and the height is 5cm.
Step-by-step explanation:
This is a solve the system question. Call H the height of the triangle and B the base. The question tells us:
and
Sub the first equation into the second (as H is already isolated). You will end up with a quadratic equation - solve that any way you wish (e.g. quadratic formula). I've provided the factored form below which shows you the roots:
In this question, we take B=4. You can't have a negative side length so the other answer is eliminated. Now sub the value for B into either of the original equations. I'll use the first, again because H is already isolated:
