How do you Simplify -(1

Https://learning.k12.com/content/enforced/361756-COF_ID92747/Extended%20Answer%202.GIF?_&d2lSessionVal=6bj13oL4Dap6gjQyXIPmV

777dan777 [17]

Post a pic for me so i can answer

4 0

3 years ago

Mark wrote the highest score he made on his new video game as the product of 70 X 6000. Use the Associative and Commutative Prop

sesenic [268]

I think in commutative properties of multiplication:
70 x 6000 = 70 x 6000
420,000 = 420,000

I don't know associative properties, only commutative properties. Sorry :(

3 0

3 years ago

Read 2 more answers

There are 7/8 kilograms of salt in the kitchen. Mrs. Jackson used 2/15 of salt when preparing dinner . How much salt did she use

neonofarm [45]

so since Mrs Jackson used 2/15 of 7/8, the amount she used is simply their product.

$\bf \stackrel{\textit{how much is }\frac{2}{15}\textit{ of }\frac{7}{8}?}{\cfrac{7}{8}\cdot \cfrac{2}{15}}\implies \cfrac{7}{15}\cdot \cfrac{2}{8}\implies \cfrac{7}{15}\cdot \cfrac{1}{4}\implies \cfrac{7}{60}$

8 0

3 years ago

A student conducts several random surveys to find out the most preferred lunch option. Each survey was of 60 students. No studen

stira [4]

Answer:

A

C

is the answer

I think soo

5 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

How do you Simplify -(1 - x)+3