A car can travel 21 miles for each gallon of gas it uses. how many miles can the car travel using 51 gallons

1 answer:

3 0

21 mpg times 51 gallons gives a distance of 1,071 miles. Are you sure you're in highschool mathematics? This is basic multiplication, something most people learn in elementary school.

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Please help with this exercise.

algol [13]

for the lines k (x) = (- 9 / 5) x + (5 / 2) and h (x) = (2 / 5) x - (7 / 6), the solution for x will be 5 / 3 and y will be - 1 / 2.

We are given the two lines:

k (x) = (- 9 / 5) x + (5 / 2)

h (x) = (2 / 5) x - (7 / 6)

At the intersection of the two lines, we get that:

(- 9 / 5) x + (5 / 2) = (2 / 5) x - (7 / 6)

(2 / 5) x + (9 / 5) x = (5 / 2) + (7 / 6)

(11 / 5) x = 11 / 3

3 x = 5

x = 5 / 3

y = (2 / 5) (5 / 3) - (7 / 6)

y = 2 / 3 - 7 / 6

y = - 1 / 2

Therefore, we get that, for the lines k (x) = (- 9 / 5) x + (5 / 2) and h (x) = (2 / 5) x - (7 / 6), the solution for x will be 5 / 3 and y will be - 1 / 2.

Learn more about lines here:

brainly.com/question/24644930

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6 0

1 year ago

The graph shows the amount of money paid when purchasing bags of candy at the zoo:

Fofino [41]

Answer:

i think its $2

Step-by-step explanation:

because 1,2 could mean 1 bag = $2,

2,4 could mean 2 bags = $4, and so on.

4 0

2 years ago

Read 2 more answers

Solve the equation for x∈Z <br><br>-x² +8x -14 ≥ 0

kenny6666 [7]

$ax^2+bx+c=0\\\Delta=b^2-4ac\\if\ \Delta < 0\ then\ no\ solutions\\if\ \Delta=0\ then\ one\ solution:x=\frac{-b}{2a}\\if\ \Delta > 0\ then\ two\ solutions:x=\frac{-b\pm\sqrt\Delta}{2a}\\================================\\\\-x^2+8x-14\geq0\ \ \ \ |multiply\ both\ sides\ by\ (-1)\ \{change\ \geq\ on\ \leq\}\\\\x^2-8x+14\leq0\\\\a=1;\ b=-8;\ c=14\\\\\Delta=(-8)^2-4\cdot1\cdot14=64-56=8 > 0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2$

$x_1=\frac{-(-8)-2\sqrt2}{2\cdot1}=\frac{8-2\sqrt2}{2}=\frac{8}{2}-\frac{2\sqrt2}{2}=\boxed{4-\sqrt2}\\\\x_2=\frac{-(-8)+2\sqrt2}{2\cdot1}=\frac{8+2\sqrt2}{2}=\frac{8}{2}+\frac{2\sqrt2}{2}=\boxed{4+\sqrt2}\\\\============================$

$ax^2+bx+c=0\\\\if\ a > 0\ then\ the\ parabola\ o pen\ up\\if\ a < 0\ then\ the\ parabola\ o pen\ down\\========================\\\\a=1 > 0-therefore\ o pen\ up\ (look\ at\ the\ picture)\\\\===============================\\\\Answer:x\in\left$

$Solutions\ in\ \mathbb{Z}:\\\\\sqrt2\approx1.4\\\\threfore:4-\sqrt2\approx4-1.4=2.6\ and\ 4+\sqrt2\approx4+1.4=5.4\\\\look\ at\ the\ second\ picture:x=3\ or\ x=4\ or\ x=5\ (x\in\{3;\ 4;\ 5\})$