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Vesnalui [34]
2 years ago
10

3(x-2) < 2(x+9) 1. x<12 2.x>12 3. x<24 4.x>24

Mathematics
1 answer:
Nataly_w [17]2 years ago
7 0

Answer:

3rd option x<24

Step-by-step explanation:

3×(x-2)<2×(x+9)

3 times everything in the bracket and 2 times everything in the bracket.

3x(-2x)-6<2x(-2x)+18 take away 2x from both sides

( Opposite operators EG, Add becomes subtract)

x-6(+6)<18(+6) add 6 on both sides

x<24

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Please help meeeeee with this
aleksandrvk [35]
Alright, let's do all of these (though this is a bit long).
1.
The constant is 1.8. All other values are coefficients to variables, which as the name implies will change.
2.
1 hour is 60 minutes, 1 minute is 60 seconds.
So, 4.2 *60 *60 = 15120 seconds.
3.
<span>−5x−4(x−6)=−3-5x-4(x-6)=-3
Let's move all x to one side, and all other numbers to another.
-5x-4(x-6)=-3-5x-4(x-6)=-3
x can be any value you want, if you actually solve this you'll only end up with -3 = -3, which is correct, of course.
Let me show you:
</span><span>−5x−4(x−6)=−3-5x-4(x-6)=-3
+5x +4(x-6)        +5x +4(x-6)
-3 = -3
The value of x is irrelevant, then. X can be any real number.
4.
I'm going to assume it was an error in printing with this? If not please correct me.
m=a+2b(or b2)
subtract 2b from each
a=m-2b
(This question seems kind of odd. We should probably address this in the comments.)
5.
</span><span>5(x−2)<−3x+6
Move all x to one side, numbers to other.
5x-10<-3x+6
+3x     +3x
    +10      +10
8x<16
/8
<span>x < 2
</span>6.
y-3=3(x-5)
alright, to find zeros set one variable to zero and solve
x first
-3=3x-15
+15    +15
3x=12
/3
x=4
x-int is (4,0)
now y
</span>y-3=3(0-5)
y-3=-15
+3     +3
y=-12
so y-int is (0,-12)
i've got to sleep now so i'll do the rest tomorrow. Sorry for the incomplete answer.

8 0
3 years ago
Evaluate the variable expression when a=-4, b=2, c=-3, and d =4. b-3a/bc^2-d​
gtnhenbr [62]

Answer:

Therefore, the variable expression when a=-4, b=2, c=-3, and d =4 is

\dfrac{b-3a}{bc^{2}-d}=1

Step-by-step explanation:

Evaluate:

\dfrac{b-3a}{bc^{2}-d}

When a=-4, b=2, c=-3, and d =4

Solution:

Substitute, a=-4, b=2, c=-3, and d =4 in above expression we get

\dfrac{b-3a}{bc^{2}-d}=\dfrac{2-3(-4)}{2(-3)^{2}-4}\\\\=\dfrac{2+12}{18-4}\\\\

\dfrac{b-3a}{bc^{2}-d}=\dfrac{14}{14}=1

Therefore, the variable expression when a=-4, b=2, c=-3, and d =4 is

\dfrac{b-3a}{bc^{2}-d}=1

6 0
3 years ago
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