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2 years ago

Three times a number plus five is less than three fourths of the number

Mathematics

2 answers:

brilliants [131]2 years ago

8 0

3x+5<3/4x
4x+20/3 < x
20/3 < -3x
x < -2.2222...

tatiyna2 years ago

3 0

The answer would be:

3x + 5 > 3/4x

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At the start of a science experiment, the temperature is 10 degrees Celsius. If the temperature decreases by a constant rate of

notka56 [123]

Answer:

-5 If it is one of your answers it will be -5

Step-by-step explanation:

If you take 10 and have 3 decrease every minute and did it for 5 minutes it will go 10 7 4 1 -2 -5 That is 5 minutes of the temperature decreasing by 3 I hope it is right.

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2 years ago

A bus travels two different routes: the Green Route and the Blue Route.

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1 year ago

if the total value of the coins is 71$ and she has 9 rolls of dimes, how many rolls of nickels does she have?

Zanzabum

Answer:

18 rolls of nickles

Step-by-step explanation:

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Dont forget to thanks and rate!

4 0

3 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago

I answered wrong what’s the right answer?

olganol [36]

The answer is CBE :)))))

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2 years ago