For
ax+by=c
the slope is -a/b
20x+25y≥200
slope=-20/25=-4/5
negative slope

yint is where x=0
20(0)+25y≥200
25y≥200
y≥98
positive yint

x+y<10
slope=-1/1=-1
yint is where x=0
y<10
yint is at y=10

since it is equal, it is solid line
to tell if it is above then sub (0,0) and see if true
0≥200
false
shade on side that doesn't have (0,0), shade above line

x+y<10 doesn't have equal under so it is dashed
test (0,0)
0<10
true, it is shaded below

test point (4,5)

20(4)+25(5)≥200
80+125≥200
225≥200
true

4+5<10
9<10
true

so the ones that are true are

The line x + y < 10 has a negative slope and a positive y-intercept.
The line representing 20x + 25y ≥ 200 is solid and the graph is shaded above the line.
The overlapping region contains the point (4, 5).

3 0

2 years ago

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Jinghua hiked 4 1/2 miles through the woods in 2 1/4 hours. She hiked the return trip at the same average rate but by a differen

Anna007 [38]

Answer:

5 miles.

Step-by-step explanation:

Consider the question: Jinghua hiked 4 1/2 miles through the woods in 2 1/4 hours. She hiked the return trip at the same average rate but by a different route taking 2 1/2 hours. How many miles did Jinghua hike on the return trip ?

First of all, we will find Jinghua's speed using given information as:

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Convert mixed fractions into improper fractions:

$4\frac{1}{2}\Rightarrow \frac{9}{2}$

$2\frac{1}{4}\Rightarrow \frac{9}{4}$

$\text{Jinghua's speed}=\frac{\frac{9}{2}\text{ Miles}}{\frac{9}{4}\text{ Hours}}$

Using property $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}$ :

$\text{Jinghua's speed}=\frac{9*4\text{ Miles}}{9*2\text{ Hours}}$

$\text{Jinghua's speed}=\frac{2\text{ Miles}}{\text{ Hour}}$

We know that distance is equal to the product of speed and time.

$\text{Distance}=\text{Speed}\times\text{Time}$

Since we have been given that Jingua hiked the return trip at the same average rate, so distance covered by her on return trip would be speed (2 miles her hour) times given time (2 1/2 hours).

$\text{Distance covered by Jingua on return trip}=\frac{2\text{ Miles}}{\text{ Hour}}\times 2\frac{1}{2}\text{ Hours}$

$\text{Distance covered by Jingua on return trip}=2\text{ Miles}\times \frac{5}{2}$

$\text{Distance covered by Jingua on return trip}=5\text{ Miles}$

Therefore, Jingua hiked 5 miles on her return trip.

8 0

2 years ago