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2 years ago

Plz need this asap ONLY ANSWER IF YOU KNOW IT will give brainliest

Mathematics

1 answer:

Natalija [7]2 years ago

8 0

Answer:

1 would be your y axis

Step-by-step explanation:

1. so the y axis is up and down

2. the x axis is horizontal side to side

3. the x axis is 0

4. the y axis is positive 1

5. So your anwser would be in the format (x,y)

6. It is (0,1)

7. It is positive

8. Plz name me brainliest I would really appreciate it if you need more help just ask me

Thank You

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I don’t know what to do with the 3 feet above the ground

Slav-nsk [51]

Answer:

3 seconds

Step-by-step explanation:

You want to set h(t)=3, then take the root such that t≠0. You would have:

3=3+48t−16t^2

Let's solve your equation step-by-step.

3=3+48t−16t2

Step 1: Simplify both sides of the equation.

3=−16t2+48t+3

Step 2: Subtract -16t^2+48t+3 from both sides.

3−(−16t2+48t+3)=−16t2+48t+3−(−16t2+48t+3)

16t2−48t=0

Step 3: Factor left side of equation.

16t(t−3)=0

Step 4: Set factors equal to 0.

16t=0 or t−3=0

t=0 or t=3

3 makes more sense than 0 because the ball can't have been in the air for 0 seconds.

So the ball was in the air for 3 seconds.

7 0

3 years ago

Five sisters each but matching scarves in two pairs of socks the scores are priced at $12 if the total bill for all five sisters

Illusion [34]

Answer:

the one pair of socks cost $2 dollars.

7 0

3 years ago

A projectile is launched from the ground at an angle of theta above the horizontal with an initial speed of v in ft/s. The rang

VARVARA [1.3K]

Answer:

$\theta=24.4^\circ$

Step-by-step explanation:

The formula you wrote has the fraction the other way around. You can find that the range of a projectile is in reality approximated by the equation $x=\frac{v^2}{g} sin(2\theta)=\frac{v^2}{32ft/s^2} sin(2\theta)$ , where we will use ft for distances.

From the given equation we have then $\frac{x32ft/s^2}{v^2} =sin(2\theta)$ , which means $\theta=\frac{Arcsin(\frac{x32ft/s^2}{v^2})}{2}$ since the arcsin is the inverse function of the sin.

Since we have x = 170 ft and v = 85 ft/s, we can substitute these values from the equation written, and we will have $\theta=\frac{Arcsin(\frac{(170ft)(32ft/s^2)}{(85ft/s)^2})}{2}$ , and from now on we have just to use a calculator, obtaining $\theta=\frac{Arcsin(0.75294117647)}{2}=\frac{48.84579804^\circ}{2}=24.4^\circ$