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3 years ago

Plz do this i will mark you brainlist and thats for 70 points!!!!

Mathematics

2 answers:

My name is Ann [436]3 years ago

8 0

Answer:

First find the circumference of both wheels

C = pi d

Large wheel = 3.14 * 60

= 188.4

Small wheel = 3.14 * 18

= 56.52

Divide them both with divide 600 by the circumference

Large wheel = 600/188.4 = 3.18 or 3 rotations

Small wheel = 600/56.52 = 10.61 or 11 rotations

b. 2 large wheels. This is because you can cover more distance with large wheels that small wheels. Large wheels are also worth the energy.

Hope this helps

finlep [7]3 years ago

8 0

Answer:

A: The large tire makes 3 rotations, and the small tire makes 11 rotations.

B: This part of the answer you answer because its asking for your opinion! ;)

Step-by-Step Explanation:

You must find the circumference of the big tire, and the small tire.

C= $\pi$ d

The large tire this means:

$C= 3.14(60)=188.4$

Dividing 600/188.4 we get 3 rounded to the nearest whole number.

For the small tire:

$C=3.14 (2*9)=3.14(18)=56.52\\Dividing 600/56.52, we get to the (nearest whole number), 11 small tire rotation.$

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Solve the equation. 8x2 – 4 = 28

Flura [38]

Hello,

Your answer would be 2, -2.

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2 , -2 would be the answer! :)

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3 years ago

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3 years ago

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3 years ago

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vagabundo [1.1K]

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3 years ago

Consider the function f(x)=9-x^2/x^2-4 For which intervals is f(x) positive? Check ALL that apply.

solong [7]

Answer:

a. f (x) < 0 for x ∈ (-∞ ,-3)

b. f (x) > 0 for x ∈ (-3,-2)

c. f (x) < 0 for x ∈ (-2,2)

d. f (x) > 0 for x ∈ (2,3)

e.f (x) < 0 for x ∈ (3,∞)

Step-by-step explanation:

Here, the given function is: $f(x)= \frac{9-x^2}{x^2-4}$

Now, to check for the sign of f(x) at x = k, put the value of x from the given interval.

We get:

a. (-infinity, -3)

put k = -4 from the given interval

We get $f(-4)= \frac{9-(-4)^2}{(-4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (-∞ ,-3)

b. (-3, -2)

put k = -2.5 from the given interval

We get $f(-2.5)= \frac{9-(-2.5)^2}{(-2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (-3,-2)

c. (-2, 2)

put k = 0 from the given interval

We get $f(0)= \frac{9-(0)^2}{(0)^2-4} = \frac{9}{-4} = -\frac{9}{4} < 0$

⇒ f (x) < 0 for x ∈ (-2,2)

d. (2, 3)

put k =2.5 from the given interval

We get $f(2.5)= \frac{9-(2.5)^2}{(2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (2,3)

e. (infinity, 3)

put k = 4 from the given interval

We get $f(4)= \frac{9-(4)^2}{(4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (3,∞)

7 0

4 years ago

Plz do this i will mark you brainlist and thats for 70 points!!!!​

Plz do this i will mark you brainlist and thats for 70 points!!!!