Can someone help me please ?

Mathematics

1 answer:

Katarina [22]3 years ago

3 0

Answer: you can always write a square root, cubic root, etc... As an exponent.

Step-by-step explanation:

A) y^1/2

B) (y+6)^1/5 - 3

Exercise 4

A) 49^1/2 = square root of 49

(Dont have the square symbol y my phone. Lol!)

Always remember... You can write:

A^x/y = y root of A^x

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Please help explain each problem thank you have a great day

boyakko [2]

Part A. 250+45a=t
That's because 'a' is the variable in which we don't know how many appointments she will have. 'a' times the $45 would be in the equation. Then you would have to add the $250 for the tests and medicines.So 250+45a=t
Part B.(250+45a)-(25%*t).
That's because the first set in parentheses is the total cost. You would have to subtract that buy the discount times the total cost which represents the the second set of the equation.:)
Part C.25%*6 appointments equals 1.5. You have to solve (250+45) and then multiply 25%*6.That product plus (250+45) would be your final answer.U won't tell you the answer because it would help.U would need explanation.This is how I would answer the question.Hope this helped

8 0

3 years ago

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

$y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{o}$ - Initial height of the ball, measured in feet.

$v_{o}$ - Initial speed of the ball, measured in feet per second.

$g$ - Gravitational constant, equal to $-32.174\,\frac{ft}{s^{2}}$ .

$t$ - Time, measured in seconds.

Given that $y_{o} = 0\,ft$ , $v_{o} = 95\,\frac{ft}{s}$ , $g = -32.174\,\frac{ft}{s^{2}}$ and $y = 120\,ft$ , the following second-order polynomial is found:

$120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

$-16.087\cdot t^{2} + 95\cdot t -120 =0$

The roots of this polynomial are, respectively:

$t_{1} \approx 4.075\,s$ and $t_{2} \approx 1.831\,s$ .

Both roots solutions are physically reasonable, since $t_{1}$ represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas $t_{2}$ represents the instant when the ball the same height after reaching maximum height.