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3 years ago

PLEASE HELP ME!!! :)))

1 answer:

6 0

A = 1/2 h (b1 + b2)

Multiply both sides by 2.

2A = h(b1 + b2)

Divide both sides by h.

2A/h = b1 + b2

Subtract b2 from both sides.

2A/h - b2 = b1

Switch sides.

b1 = 2A/h - b2

$b_{1} = \frac{2A}{h} - b_{2}$

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Four more than a number is 27

son4ous [18]

Answer:

23 is the answer of the question

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3 years ago

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if you practice a musical instrument each day for 2/3 of an hour how many hours of practice would you get in each week?

RideAnS [48]

In order to find the answer of the given problem above, we need to make the solution below:
Given: 2/3 of an hour = practice each day
2/3 of an hour is equivalent to 40 minutes
And per week, there are 7 days. So we will multiply 40 minutes by 7, and the result is 280 minutes. Since there are 60 minutes in an hour, we divide 280 by 60 and we get 4.67 hours. Therefore, each week, you would get 4.67 hours of practice on playing a musical instrument.

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3 years ago

What is the remainder of x^5+2x^4+9x^3-6x^2+3x+3165 divided by x-5

riadik2000 [5.3K]

Answer:

8530

Step-by-step explanation:

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2 years ago

What is ( √2 - √3 )² =

allochka39001 [22]

just use FOIL (First, Outer, Inner, Last).

( √2 - √3 ) X ( √2 - √3 )

First: √2 X √2 = 2
Outer: √2 X -√3 = -√6
Inner: -√3 X √2 = -√6
Last: -√3 X -√3 = 3

and then add them together.

2 - √6 - √6 + 3
5 - 2(√6)

The answer is 5 - 2(√6) .

4 0

3 years ago

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Vavilen inflated a huge balloon in 19 minutes. Then he poked a hole in the balloon so that it slowly leaked air out until it was

kaheart [24]

Considering balloon to be an spherical object

Volume of sphere = $\frac{4}{3}\pi r^3$ , where r is the radius of sphere.

After 19 minutes ,If radius of balloon changes from r to R, the Volume changes from V to V'.

V'= $\frac{4}{3}\pi R^3$ ,

As given,rate at which the air leaked out of the balloon (in liters of air per minute) was half of the rate at which Vavilen inflated the balloon.

Also, $\frac{dR}{dt}=2 \times \frac{dr}{dt}$

Now, again the radius changes from R to r.So, Volume changes from V' to Back to V.

Time taken by the balloon to deflate = 38 minutes

But there is no relation between time and volume of Balloon.

So, we can't predict at which rate balloon is deflating or inflating.Apart from the statement given in Question: $\frac{dR}{dt}=2 \times \frac{dr}{dt}$

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3 years ago

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