In the first step when the student did -2x-3 they didn’t make the sign positive. It should be 4=-2+6, not 4=-2-6
Salinas will use 32 gallons for 480 miles
<h3>How to calculate the number of gallons required ? </h3>
Salinas used 15.6 gallons of gas for 234 miles
Miles can be described as the distance travelled by an object or an individual
The first step is to calculate the number of miles travelled using 1 gallon
15.6= 234
1= x
cross multiply both sides
15.6x= 234
x= 234/15.6
x= 15
1 gallon will take her 15 miles
Therefore, the number of gallons required for 480 miles can be calculated as follows
1= 15
x= 480
cross multiply both sides
15x= 480
x= 480/15
x= 32
Hence Salinas will require 32 gallons for 480 miles of the journey
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Answer:
There is no equals sign in that expression so it is not actually an equation and has no solution.
If you are looking to expand the two polynomials, then you get:
(x² - x + 5)(x - 3)
= x(x² - x + 5) - 3(x² - x + 5)
= x³ - x² + 5x - 3x² + 3x - 15
= x³ - 4x² + 8x - 15
Answer: The area is 1 1/4. The perimeter is 6.
Step-by-step explanation: 2.45 is about 2.5. 5/8 is about 1/2. To find the area we need to multiply length times width. 2.5 is 2 and 1/2 which is 5/2. 5/2 times 1/2 is 5/4. To simplify it divide 5 by 4 which is 1 with remainder 1. 5/4 is 1 1/4. Therefore the area is 1 1/4 meters squared. To find the perimeter we have to add length + length + width + width. . 2.5 is the same thing as 2 1/2. 5/8 is about 1/2 if we use benchmarks. 2 1/2 + 2 1/2 + + 1/2 + 1/2 is 6. The perimeter is 6.
Answer: B 0.5 kWh per day
Step-by-step explanation:
Since the graph represents the function where electricity usage in kilowatts per hour of a clock radio varies directly with the number of days it is plugged into the wall current
The constant of variations will be the gradient ( slope ) of the graph.
Gradient = (Y2 - Y1)/(X2 - X1)
Where X1 = 2, X2 = 6, Y1 = 0.5, Y2 = 1.5
Substitute all the values into the formula
Gradient = (1.5 - 0.5)/( 6 - 4 )
Gradient = 1/2
Gradient = 0.5.
Therefore, a reasonable estimate of the constant of variation is 0.5 kWh per day
