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3 years ago

You deposit $50 in your saving account. One week later, you withdraw $20. Write each amount as an integer.

Mathematics

1 answer:

Tju [1.3M]3 years ago

3 0

Answer:

Depositing means that you added 50$ to your account. This means that you account increased by an amount of 50.

The integer describing this is +50.

Then, withdrawing means that you took money from the account. This means that your account decreased by an amount of 20.

This can be expressed as a -20 integer.

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Kareem ran three times last week. Here are the distances he rall

EleoNora [17]

Simply add the three decimals to find Kareem's total distance.

1.6 + 11.5 + 15 = 28.1

(0.6 + 0.5 = 1.1, 1 + 11 + 15 = 27, 27 + 1.1 = 28.1)

<h2>Answer:</h2>

Kareem ran 28.1 km on those three days.

I hope this helps :)

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2 years ago

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2 years ago

Val has a piece of wood 8 ft long. She is going to cut it into 9 equal-length pieces.

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Each piece will be 1/9 of the total length of the original board.

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3 years ago

Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A balloon, rising vert

Liula [17]

Answer:

(a) 2.79 seconds after its release the bag will strike the ground.

(b) At a velocity of 73.28 ft/second it will hit the ground.

Step-by-step explanation:

We are given that a balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 80 feet above the ground.

Assume the acceleration of the object is a(t) = −32 feet per second.

(a) For finding the time it will take the bag to strike the ground after its release, we will use the following formula;

$s=ut+\frac{1}{2} at^{2}$

Here, s = distance of the balloon above the ground = - 80 feet

u = intital velocity = 16 feet per second

a = acceleration of the object = -32 feet per second

t = required time

So, $s=ut+\frac{1}{2} at^{2}$

$-80=(16\times t)+(\frac{1}{2} \times -32 \times t^{2})$

$-80=16t-16 t^{2}$

$16 t^{2} -16t -80 =0$

$t^{2} -t -5 =0$

Now, we will use the quadratic D formula for finding the value of t, i.e;

$t = \frac{-b\pm \sqrt{D } }{2a}$

Here, a = 1, b = -1, and c = -5

Also, D = $b^{2} -4ac$ = $(-1)^{2} -(4 \times 1 \times -5)$ = 21

So, $t = \frac{-(-1)\pm \sqrt{21 } }{2(1)}$

$t = \frac{1\pm \sqrt{21 } }{2}$

We will neglect the negative value of t as time can't be negative, so;

$t = \frac{1+ \sqrt{21 } }{2}$ = 2.79 ≈ 3 seconds.

Hence, after 3 seconds of its release, the bag will strike the ground.

(b) For finding the velocity at which it hit the ground, we will use the formula;

$v=u+at$

Here, v = final velocity

So, $v=16+(-32 \times 2.79)$

v = 16 - 89.28 = -73.28 feet per second.

Hence, the bag will hit the ground at a velocity of -73.28 ft/second.

8 0

2 years ago