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Helen [10]
3 years ago
9

The product of 11 and 5 times a number mutplied by 4

Mathematics
1 answer:
Vlada [557]3 years ago
3 0
11 + 5 x 4a


Hope this helps! Do you need an explanation?
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What is the average rate of change of the function f(x) = 5x ^ 2 + 20x between x = 0 and x = 0.5
Vitek1552 [10]

9514 1404 393

Answer:

  22.5

Step-by-step explanation:

The average rate of change on the interval [a, b] is the ratio ...

  (f(b) -f(x))/(b -a)

For your function and interval, it is ...

  (f(0.5) -f(0))/(0.5 -0)

Since f(0) = 0, this reduces to ...

  average rate of change = 2·f(0.5) = 2(5(0.5)² +20(0.5)) = 2(5/4 +10)

  average rate of change = 22.5

8 0
2 years ago
How do you solve 4836÷12
ANEK [815]
The answer is 403, just try simple long division.
8 0
2 years ago
Read 2 more answers
A random sample of 50 students from a high school with 900 students is surveyed. Each student is asked what science class he or
strojnjashka [21]

Answer: 50 and 900 is correct

Step-by-step explanation:

50 and or 100

4 0
2 years ago
A standard clock has a 1-cm hour hand and a 2-cm minute hand. At 12 pm, they are both pointing in the same direction and the dis
Levart [38]

Answer:

The distance between the hands is √(3)cm ≈ 1.73cm.

Step-by-step explanation:

In a standard clock, the angle between every number is 30°, therefore the angle between 12 and 2 will be 30° x 2 = 60°.

Looking at the diagram, to find c we can make use of our cosine formula

c² = a² + b² –2abCos(C°)

a = 2, b = 1 and C° = 60°

Therefore we have:

c² = 2² + 1² –2 x 2 x 1 x cos(60°) =

c² = 4 + 1 – 4 x 0.5 =

c² = 5 – 2 =

c² = 3

c = √(3) ≈ 1.73

Therefore, the distance between the hands is √(3)cm ≈ 1.73cm.

8 0
2 years ago
The velocity of an automobile starting from rest is given by the equation below, where v is measured in feet per second and t is
podryga [215]

Answer:

(a) 0.408m/sec^2

(b) 0.134m/sec^2

(c) 0.039m/sec^2

Step-by-step explanation:

We have given velocity as function of t v(t)=\frac{85t}{(8t+14)}

Acceleration is equation rate if change of velocity with respect to time

So a=\frac{dv}{dt}=\frac{(8t+14)85-85t\times 8}{(8t+14)^2}

(a) Acceleration at t = 5 sec

a=\frac{(8\times 5+14)85-85\times 5\times 8}{(8\times 5+14)^2}=0.408m/sec^2

(b) Acceleration at t = 10 sec

a=\frac{(8\times 10+14)85-85\times 10\times 8}{(8\times 10+14)^2}=0.134m/sec^2

(c) Acceleration at t = 20 sec

a=\frac{(8\times 20+14)85-85\times 20\times 8}{(8\times 20+14)^2}=0.039m/sec^2

5 0
3 years ago
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