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Vanyuwa [196]
3 years ago
15

Please help me with this ​

Mathematics
1 answer:
sveticcg [70]3 years ago
4 0
C is the answer, looking at point N, after it is reflected across the y 1, you will see the point ends up at (4,1)
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It takes Priti 4 hours to drive from Ashdown to Bridgeton at an average speed of 50 mph. She then drives 30 miles from Bridgeton
Sveta_85 [38]

Answer:

Her average speed from Ashdown to Carton is approximately 48.4 mph

Step-by-step explanation:

The time and speed of the motion of Priti are as follows;

The time it takes Priti to drive from Ashdown to Bridgeton, t₁ = 4 hours

The average speed she drove from Ashdown to Bridgeton = 50 mph

The distance she drove from Bridgeton to Carton, d₂ = 30 miles

Her average speed from Bridgeton to Carton = 40 mph

The distance from Ashdown to Bridgeton, d₁ = 50 mph × 4 hours = 200 miles

The time it takes Priti to drive from Bridgeton to Carton, t₂ = 30 miles/(40 mph) = 0.75 hour

Average velocity, v_{ave} = (Total distance traveled, Δd)/(Total time, Δt)

Δd = d₁ + d₂, Δt = t₁ + t₂

Δd = 200 miles + 30 miles = 230 miles

Δt = 4 hours + 0.75 hour = 4.75 hours

∴ The average velocity, v_{ave} = (230 miles)/(4.75 hours)

∴ v_{ave} = (920/19) mph

Her average speed from Ashdown to Carton, rouded to 1 d,p, v_{ave} ≈ 48.4 mph

3 0
3 years ago
Manuela tiene una colección de 66 cromos que quiere repartir en montones iguales sin que sobre ninguno. ¿Cuántos cromos puede te
Karolina [17]

Answer:

umm

Step-by-step explanation:

WE speack english

4 0
3 years ago
Jenny plans to have two children but doesn't know if they will be boy-boy, girl-girl, or boy-girl. What is the probability that
lubasha [3.4K]
       B            G

B     <u>BB</u>         BG

G      GB        GG

.25

3 0
3 years ago
Read 2 more answers
The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha
Xelga [282]

Answer:

The average value of f over the interval [-2,5] is \frac{10}{7}.

Step-by-step explanation:

Let suppose that function f is continuous and integrable in the given intervals, by integral definition of average we have that:

\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6 (1)

\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20 (2)

By Fundamental Theorems of Calculus we expand both expressions:

\frac{F(3)-F(-2)}{3-(-2)} = -6

F(3) - F(-2) = -30 (1b)

\frac{F(5)-F(3)}{5-3} = 20

F(5) - F(3) = 40 (2b)

We obtain the average value of f over the interval [-2, 5] by algebraic handling:

F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)

F(5) - F(-2) = 10

\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}

\bar f = \frac{10}{7}

The average value of f over the interval [-2,5] is \frac{10}{7}.

4 0
3 years ago
Is this a function please help I’m failing
igomit [66]

Answer:

Yes

Step-by-step explanation:

The relation is a function. For a relation to be a function there must be a unique x value for each y value. So this means x's can not repeat, and in this relation, the x-values never repeat. Therefore this is a function.

7 0
3 years ago
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