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3 years ago

Ill give 30 points plz asnwer ( there are multiple answers btw) tyyy

Mathematics

2 answers:

iogann1982 [59]3 years ago

3 0

Answer:

7,13,

Step-by-step explanation:

that's just what I think

alexandr1967 [171]3 years ago

3 0

Answer:i believe the answer is infinite.

Step-by-step explanation:

As we are free to decide what A is then all that has to be done is add the remaining answer to achieve the difference. Example i pick .25 as A which leaves 0.75 to equal 1. Then i use 0.75 and add 1.25 to equal 2. Then again 1.25 plus 2.75 to get 4. A is .25 and D is 2.75 equals 3.

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What is the greatest common factor of 12 and 90?

qwelly [4]

The greatest common factor (gcf) is 6.

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3 years ago

Solve 5n – 7p + 3n = 25p for n.

saul85 [17]

5n – 7p + 3n = 25p

Combine like terms

8n - 7p = 25p

Add (7p) to both sides

8n - 7p + 7p = 25p + 7p

Simplifying

8n = 32p

Divide both sides by 8

8n / 8 = 32p / 8

Simplifying

n = 4p

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3 years ago

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A Triangle on a coordinated plane is translated

motikmotik

Answer:

Translated how?

Step-by-step explanation:

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2 years ago

A cylindrical can of cocoa has the dimensions shown at the right.

IRINA_888 [86]

Answer:

What are the dementions? 9 and 2?

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Step-by-step explanation:

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2 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