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2 years ago

Solve for V. 3y=39 Simplify your answer as much as possible.

Mathematics

1 answer:

Crank2 years ago

3 0

Answer:

3y=39 :3

y=13

Step-by-step explanation:

prove me wrong

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A softball is tossed into the air upward from a first floor balcony. The distance of the ball 14 +32t -16t^2 above the ground at

snow_tiger [21]

You can either find the roots of the quadratic formula using the quadratic formula two find the halfway point or you can use the formula t=-b/2a. Both of the two methods gives you the axis of symmetry (the t value for the vertex).
I will use the second method since I don't like using the quadratic formula.
t=-32/(2x-16)
t=1 second
After you find the t value for the axis of symmetry, you plug it into the formula of the graph to find the height.
h=-16+32+14
h=30 feet

3 0

2 years ago

Use the elimination method to solve the system of equations.<br><br>3x-5y= -16<br><br>2x+5y=31

Sholpan [36]

5x = 15
x = 3

3(3) - 5y = -16
9 - 5y = -16
5y = -16 - 9
5y = - 25
y = - 5

(3, -5)

8 0

2 years ago

Complete the table no links

olya-2409 [2.1K]

Answer:

24.50, 10

Step-by-step explanation:

4 0

1 year ago

I really need help!If someone answers this,I will be happy to mark the brainest answer!EXTRA:you can answer the top question for

andreev551 [17]

9) They traveled 31.9 hours over the 3-day trip.

10.9 + 8.6 + 12.4 = 31.9

10) She does not have enough money. She has $12, but everything she wants to purchase would cost $12.39.

Large drink = 1.79

Turkey sandwich = 4.85

1.79 + 4.85 = 6.64

6.64 + 5.75 = 12.39

5 0

2 years ago

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

8 0

2 years ago