To solve this we need to use Pythagorean theorem

Answer: x≈6.4
The first car consumed 40 gallons of gas and second car consumed 30 gallons of gas
<em><u>Solution:</u></em>
Let x = gallons consumed by car 1
Let y = gallons consumed by car 2
Fuel efficiency of car 1 = 15 miles per gallon
Distance covered in 1 gallon of gas = 15 miles
Fuel efficiency of car 2 = 25 miles per gallon
Distance covered in 1 gallon of gas = 25 miles
<em><u>Given a total gas consumption of 70 gallons</u></em>
Therefore,
gallons consumed by car 1 + gallons consumed by car 2 = 70
x + y = 70 ------ eqn 1
<em><u>The two cars went a combined total of 1350 miles</u></em>
Therefore,
gallons consumed by car 1 x distance covered in 1 gallon of gas of car 1 + gallons consumed by car 2 x distance covered in 1 gallon of gas of car 2 = 1350

15x + 25y = 1350 ----- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
From eqn 1,
x = 70 - y ------- eqn 3
<em><u>Substitute eqn 3 in eqn 2</u></em>
15(70 - y) + 25y = 1350
1050 - 15y + 25y = 1350
10y = 1350 - 1050
10y = 300
y = 30
<em><u>Substitute y = 30 in eqn 3</u></em>
x = 70 - 30
x = 40
Thus first car consumed 40 gallons of gas and second car consumed 30 gallons of gas
Answer:
(1/2)x+(1/4)=x
x=0.5
3x+3=2(x+7)
x=11
2(x-5)+7=x+8
x=11
0.8x+5=0.2(40-x)
x=3
5(x-1)-6x=-11x
x=0.5
6(x+2)-x=4(x+4)+7
x=11
(1/2)(x+8)+(3/2)x=10
x=3
Step-by-step explanation:
(1/2)x+(1/4)=x
x-0.5x=0.25
0.5x=0.25
x=0.5
3x+3=2(x+7)
3x+3=2x+14
3x-2x=14-3
x=11
2(x-5)+7=x+8
2x-10+7=x+8
2x-x=8+10-7
x=11
0.8x+5=0.2(40-x)
0.8x+5=8-0.2x
0.8x+0.2x=8-5
x=3
5(x-1)-6x=-11x
5x-5-6x=-11x
11x-x=5
10x=5
x=0.5
6(x+2)-x=4(x+4)+7
6x+12-x=4x+16+7
6x-4x-x=16+7-12
x=11
(1/2)(x+8)+(3/2)x=10
0.5x+4+1.5x=10
2x=6
x=3
Answer:
24 in^2
Step-by-step explanation:
(1/2)bh
(1/2)(6)(8)
24
Answer:
The Answer is 1
Step-by-step explanation:
It's 1 because x is -3 and its getting squared so its -3 times -3 which is 9. And then Y is -2 cubed so -2x-2x-2 which equals -8.
9-8=1
Hope this helps!