1.75 + 0.25x ≤ 15
where x is equal the number of miles.

The key to solving an inequality is just temporarily making the ≤ symbol a =. Then solve for x like any other algebraic equation. Once you find the value of x, you will have solved the inequality. Make sure to add back the ≤ when you are done.

15 - 1.75 = 13.25
13.25/0.25 = x
x = 53
add back the ≤
x ≤ 53
This means Eddies can travel up to 53 miles with his $15.

4 0

3 years ago

Find a linear equation whose graph is the straight line with the given property.

jeka57 [31]

Answer:

y = 3/4x - 9/8

Then read explanation and see here with 2 coordinates would make

if 3/4 = -6 then 1 = -8

y = -6x + c

-6 / 3/4 + 9/8 / - 0.5 = -2.25

therefore c = -2.25

y = -6x - 2.25

y = -6x - 9/4 and again 9/4 can be found easily if round up fraction decimal to 225/100 and divide by 4

Step-by-step explanation:

C1) (0.5 , -0.75)

C2) ( 1. -3.75)

y = mx + c

-3.75 - - 0.75 / 1 - 0.5 = -3.75 + 0.75 / 0.5

= -3 / 0.5 = m

m = -6

y = -6x + c

then this works given just one coordinate to solve for 1 or 2 sets given points

y - y1 = mx+c

y - - 0.75 = 0.75 (x - x1)

y + 0.75 = 0.75 (x - 0.5)

y + 0.75 = 0.75x - 0.375

y = 0.75 - 0.75 = 0.75x( - 0.375 - 0.75)

y = 0.75x -1.125

y = 3/4x - 9/8 as common denominators of 1125/1000 is 9/8 as we divide by 125 into the fraction as we start at 250 and check its half just like when finding common denominators we 1/4 the lower number and see if we can find its half or its double etc with the other num er as first step for 3sf numbers.

5 0

2 years ago

Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

7 0

3 years ago