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2 years ago

A gallon of gasoline at the local gas station

Mathematics

2 answers:

Mandarinka [93]2 years ago

7 0

Answer: $21.52

Step-by-step explanation: 2.69 times 8 = $21.52

sveticcg [70]2 years ago

6 0

8 gallons of gasoline would cost $21.52

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4/x - 3/y = 1 ; 6/x + 15/y = 8 solve this equation

Soloha48 [4]

Answer:

x=2

y=3

Solution:

First we find common denominators. It is "xy". Then we multiply numerators by common denominator. We get followings:

(4y-3x)/xy=1; (6y+15x)/xy=8

Then

4y-3x=xy;

6y+15=8xy

Multiply first equasion by 5

20y-15x=5xy

Now we add two equasions to get one

20y-15x=5xy

6y+15x=8xy

We get

26y=13xy

Cut "y" and we will find "x"

26=13x

x=2

Put x value into the first equasion(4y-3x=xy) to find out "y"

4y-6=2y

2y=6

y=3

4 0

3 years ago

Mrs. King has a stack of papers to grade that contains 12 boy’s papers and 8 girl’s papers. What is the probability that the fir

Artist 52 [7]

Answer:3/5

Step-by-step explanation:

20 papers in total

12 boy papers

12/20

simplified to 3/5

7 0

3 years ago

Read 2 more answers

1.Sheena wants to make lime juice.

pochemuha

She needs 15 crates of limes

2x6=12
40x6=240 limes required to make 12 liters
240/16=15 crates needed

5 0

2 years ago

Write the number with the same value as 28 tens

pantera1 [17]

The answer would be 280 because 28x10 is 280

8 0

2 years ago

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago