Answer:
$848
Step-by-step explanation:
Calculation for the cost of this familiy's clothing in 1991
First step is to calculate the amount that is increase per year
Increase per year= ($1000-$620)/(1995-1985)
Increase per year= 380/10
Increase per year= $38
Now let y be :38*x + $620 and let x be 6 years (1991-1985)
Second step is to calculate the cost of the clothing in 1991
y = 38*6 + 620
y=228+620
y = $848
Therefore the cost of this familiy's clothing in 1991 will be $848
Answer:
its 1
Step-by-step explanation:
<span>3x+30+6=3x+36</span>
<span>⇒3x+36=3x+36</span>
<span>⇒3x−3x=36−36</span>
<span>⇒0⋅x=0
</span>infinite solutions of <span>x </span><span> for all </span><span> x∈<span>R
</span></span>
Let x be the price for each night and y be the price for each meal. Then, express the statements into mathematical equations:
Choice A: 3x + y = 250
Choice B: 3x + 6y = 330
Subtract both equations, the difference would be -5y = -80
-5y = -80
y = -80/-5
y = $16 per meal
3x + y = 250
3x + 16 = 250
3x = 250 - 16
3x = 234
x = 234/3
x = $78 per night
The answer is D.
