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2 years ago

Yo help NO LINKS PLS

Mathematics

2 answers:

s344n2d4d5 [400]2 years ago

7 0

Answer:

19 square millimeter

Step-by-step explanation:

gavmur [86]2 years ago

3 0

Answer:

Step-by-step explanation:

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Help Asap Please!! mark as Brainliest

nekit [7.7K]

1. $\frac{1}{2}$ (10m - 14) + 9 = 12 1. Given

2. 5m - 7 + 9 = 12 2. Distributive Property

3. 5m + 2 = 12 3. Simplify (added like terms)

4. 5m = 10 4. Subtraction Property of Equality

5. m = 2 5. Division Property of Equality

3 0

2 years ago

Read 2 more answers

Solve the inequality: PLS SHOW YOUR WORK PLEASE<br><br> -45 < 5(p-2)

andrezito [222]

Step-by-step explanation:

Given

-45 < 5(p-2)

Dividing both sides by 5

9 < p - 2

Adding two on both sides

9 + 2 < p - 2 + 2

11 < p

p > 11

Hope it will help :)❤

3 0

2 years ago

7) 2x + y = 7<br> What is the intercept

posledela

Answer:

2x + y = 7

x=7/2

y=7

Step-by-step explanation:

to find y itercept =0

to find x intercept=0

5 0

2 years ago

An orchard sells 4 pounds of apples for one dollar how much would 5 pounds of apples cost

Mrac [35]

Answer:

1.25

Step-by-step explanation:4 pounds equal to 1 dollar so that means that each pound equals to 0.25 cents. SO, now you multiply 0.25 x 5= 1.25

5 0

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

2 years ago