Реши задание на изображении
Quick answers
The graph matches because the equation can be written as
\[ v=-\frac23c+\frac{40}{3}. \] Its \(v\)-intercept is \(13\frac13\), its \(c\)-intercept is \(20\), and its slope is \(-\frac23\).A. Oldest — Graph B (2)
B. Middle — Graph C (3)
C. Youngest — Graph A (1)A. \(\left(\frac35,0\right)\)
7. Why the graph represents \(4c+6v=80\)
Let \(c\) = number of cars and \(v\) = number of vans.
Solve for \(v\): \[ 4c+6v=80 \] \[ 6v=80-4c \] \[ v=\frac{80-4c}{6} =\frac{40}{3}-\frac23c \]
This is a straight line with:
\(v\)-intercept: when \(c=0\), \[ v=\frac{80}{6}=\frac{40}{3}=13\frac13 \] The graph crosses the \(v\)-axis between 12 and 14.
\(c\)-intercept: when \(v=0\), \[ 4c=80 \Rightarrow c=20 \] The graph crosses the horizontal axis at 20 cars.
Slope: \[ -\frac{4}{6}=-\frac23 \] In context, adding 3 cars gives \(3\cdot4=12\) seats, so you need 2 fewer vans, since \(2\cdot6=12\). The line goes down at that rate.
A point on the line such as \((5,10)\) works: \[ 4(5)+6(10)=20+60=80. \] Only nonnegative whole-number points make sense for actual cars and vans, but the line represents the equation.
8. Matching the siblings to the graphs
Write each sibling’s distance \(y\) after \(x\) hours.
Oldest: starts at 0 and runs 7 mph: \[ y=7x \] This is the steeper line through the origin → Graph B.
Middle: starts at 0 and walks 3.5 mph: \[ y=3.5x \] This is the less steep line through the origin → Graph C.
Youngest: starts 4 miles from the start and runs 5 mph: \[ y=4+5x \] This line starts at \(y=4\), not at the origin, and rises quickly → Graph A.
So:
\[ \boxed{\text{Oldest } \to \text{ Graph B},\quad \text{Middle } \to \text{ Graph C},\quad \text{Youngest } \to \text{ Graph A}} \]
9. Finding the \(x\)-intercept of \(y=3-5x\)
The \(x\)-intercept is where the graph crosses the \(x\)-axis, so \(y=0\): \[ 0=3-5x \] \[ 5x=3 \] \[ x=\frac35 \]
Thus the \(x\)-intercept is \[ \boxed{\left(\frac35,0\right)} \]
The answer is A.
Обсуждение
10 сообщений№7
Уравнение: \[ 4c+6v=80 \] Выразим \(v\): \[ 6v=80-4c \] \[ v=\frac{40}{3}-\frac{2}{3}c=13\frac13-\frac23c \]
Значит, график — прямая с:
- пересечением с осью \(v\): \(\left(0,\,13\frac13\right)\);
- пересечением с осью \(c\): \((20,0)\);
- угловым коэффициентом \(-\frac23\).
Проверим точки на прямой: \[ (8,8):\ 4\cdot8+6\cdot8=32+48=80 \] \[ (14,4):\ 4\cdot14+6\cdot4=56+24=80 \] \[ (20,0):\ 4\cdot20=80 \]
Эти точки лежат на изображённой прямой, поэтому график действительно соответствует уравнению \(4c+6v=80\).
№8
Пусть \(d\) — расстояние от старта, \(t\) — время в часах.
Старший:
\[ d=7t \] Прямая начинается в \((0,0)\) и проходит через \((1,7)\).Средний:
\[ d=3{,}5t \] Прямая начинается в \((0,0)\), но идёт менее круто.Младший:
\[ d=4+5t \] Прямая начинается не с \(0\), а с \(4\) миль: \((0,4)\).
По графику:
- график A — младший;
- график B — старший;
- график C — средний.
Ответ:
- A. Старший — 2. График B
- B. Средний — 3. График C
- C. Младший — 1. График A
№9
Нужно найти точку пересечения с осью \(x\), поэтому подставляем \(y=0\): \[ 0=3-5x \] \[ 5x=3 \] \[ x=\frac35 \]
Значит, \(x\)-intercept: \[ \left(\frac35,\,0\right) \]
Ответ: A. \(\left(\frac35,0\right)\).
Explain 8
For #8, match each sibling using two features of a distance–time graph:
- slope = speed;
- \(y\)-intercept = starting distance at time \(0\).
Let \(t\) be time in hours and \(d\) distance in miles.
Oldest: runs 7 mph
\[ d=7t \] Starts at \(0\), and the graph rises 7 miles every hour.
Middle: walks 3.5 mph
\[ d=3.5t \] Also starts at \(0\), but rises more slowly than the oldest’s graph.
Youngest: starts 4 miles from home and runs 5 mph
\[ d=4+5t \] This is the key one: at \(t=0\), \[ d=4, \] so the graph must start at 4 on the vertical axis, not at the origin.
Now look at the graphs:
- Graph A starts at \(4\), so it is the youngest.
- Graph B starts at \(0\) and is steeper, so it is the oldest.
- Graph C starts at \(0\) and is less steep, so it is the middle child.
Therefore:
\[ \boxed{\text{A. Oldest — Graph B (2)}} \]
\[ \boxed{\text{B. Middle — Graph C (3)}} \]
\[ \boxed{\text{C. Youngest — Graph A (1)}} \]
A quick check: the youngest’s graph should have slope \(5\), while the oldest’s has slope \(7\), so Graph B should be steeper than Graph A. Graph A’s head start of 4 miles is what makes it start above the horizontal axis.
Make it shorter
#8 — concise version
Distance–time graphs:
- Steeper slope = faster speed
- Starting height = starting distance
Equations:
- Oldest: \(d=7t\) — starts at 0, steepest → Graph B (2)
- Middle: \(d=3.5t\) — starts at 0, less steep → Graph C (3)
- Youngest: \(d=4+5t\) — starts at 4 miles, not 0 → Graph A (1)
\[ \boxed{\text{Oldest — B (2), Middle — C (3), Youngest — A (1)}} \]
Make it shorter
\[ \boxed{\text{Oldest — B (2) \quad Middle — C (3) \quad Youngest — A (1)}} \]
Oldest and middle start at 0; oldest is steeper. Youngest starts at 4 miles.
Explain nine
#9 — explained
An \(x\)-intercept is where the graph crosses the \(x\)-axis. At that point, \[ y=0. \]
Start with: \[ y=3-5x \]
Set \(y=0\): \[ 0=3-5x \]
Solve for \(x\): \[ 5x=3 \] \[ x=\frac35 \]
So the graph crosses the \(x\)-axis at: \[ \boxed{\left(\frac35,\,0\right)} \]
Answer: A.
