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Olegator [25]
3 years ago
8

I"ll make you brainiest if u answerthis

Mathematics
2 answers:
Andre45 [30]3 years ago
5 0
The train has 30 cars in total.
Flauer [41]3 years ago
4 0

Answer:

30 cars

Step-by-step explanation: 3 cars per 30 sec , their is 60 sec in a min

30x2=60

5x6=30

You multiply by 6 because you need to figure out how many cars pass in 1 min which is 6

Hopefully this makes sense.

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In two or more complete sentences, compare the number of x-intercepts in the graph of f(x)=x2 to the number of x-intercepts in t
musickatia [10]
Because f(x)=x^2 is the parent function of a quadratic, it has one x-intercept at (0,0), no shifts up or down. But because g(x)=(x+2)^2-7 is a quadratic being shifted 2 units to the left and most importantly 7 units down, the function will cross the x-intercept 2 times.

I hope this is the answer you were looking for!
4 0
2 years ago
Which relation is a function?
laila [671]

Answer:

Top Right

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
4(-4x-1)+x+3=44 what’s the value of x
Alja [10]
X= -3
Explanation below

7 0
3 years ago
I'm trying to rearange this equation so that t is the subject and i need help
gregori [183]
Work backwards.

s=ut+(at^2)/2

Subtract ut.

s-ut=(at^2)/2

Multiply by 2.

2(s-ut)=at^2

Divide by a.

2(s-ut)/a=t^2

Find the square root:

\sqrt{2(s-ut)/a}=t
6 0
4 years ago
Find the average value fave of the function f on the given interval. f() = 5 sec2(/6), 0, 3 2
Inessa05 [86]

Answer:

f_{ave} = {\frac{10 }{3\pi }

Step-by-step explanation:

To find - Find the average value fave of the function f on the given interval. f(x) = 5 sec²(x/6), [0, 3\pi/2]

Proof -

We know that,

Average value of f in the interval [a, b] is -

f_{ave} = \frac{1}{b - a}\int\limits^b_a {f(x)} \, dx

Now,

Here a = 0, b = \frac{3\pi }{2}, f(x) = 5sec^{2}(\frac{x}{6} )

Now,

f_{ave} = \frac{1}{\frac{3\pi }{2}  - 0}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx

      = {\frac{2 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx

      = {\frac{10 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {sec^2 ({\frac{x}{6} }) } \, dx

      = {\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) }|^{\frac{3\pi }{2} }_0 \, ]

      = {\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) - tan({0) } \, ]

      = {\frac{10 }{3\pi }}[ {1 - 0 }  ]

      = {\frac{10 }{3\pi }

⇒f_{ave} = {\frac{10 }{3\pi }

3 0
3 years ago
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