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3 years ago

This should be quick answer pls help

Mathematics

1 answer:

stiks02 [169]3 years ago

5 0

Answer:

3+18=21

Step-by-step explanation:

because 3x6 is 18, and 18+3 is 21.

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Perform the indicated operation. 9z^3/16xy . 4x/27z^3

ziro4ka [17]

Answer:

$\frac{1}{12y}$

Step-by-step explanation:

This is a multiplication problem.

We want to multiply $\frac{9z^3}{16xy}\cdot \frac{4x}{27z^3}$

We factor to get:

$\frac{9z^3}{4\times 4xy}\cdot \frac{4x}{9\times 3z^3}$

We now cancel out the common factors to get:

$\frac{1}{4\times y}\cdot \frac{1}{1\times 3}$

We now multiply the numerators and the denominators separately to get.

This simplifies to $\frac{1}{12y}$

Therefore the simplified expression is $\frac{1}{12y}$

8 0

3 years ago

Melissa earns 7.25% commission on everything she sells at the electronics store where she works. She also earns a base salary of

Gnoma [55]

.725 multiplied by 4500=362.5
750+3262.5=4012.5

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3 years ago

A survey is made in a neighborhood of 80 voters. 65 are democrats and 15 are republicans. of the democrats, 35 are women, while

Aleksandr [31]

Total voters=v=80
total no. of democrats=d=65
total no. of republicans=r=15
probability of democrats=p1=65/80=0.8125
probability of republicans=p2=15/80=0.1875
probability of a democrat or a republican=p1+p2=0.8125+0.1875=1

7 0

3 years ago

Read 2 more answers

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 $\pi/4$ .

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3 years ago

Rectangular Prism Rectangular Pyramid Cylinder Cone ?????

Liono4ka [1.6K]

If it’s rotating around something, it’s most likely a cylinder I believe

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3 years ago