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iogann1982 [59]
3 years ago
13

What would the unit label be for this

Mathematics
1 answer:
DerKrebs [107]3 years ago
6 0
The unit label is "degrees". 
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You want to buy a new bike, and your parents will contribute $32.50 toward the cost of the bike. If the most expensive bike is $
asambeis [7]

Answer:

They the answernis 17.50 fam


Step-by-step explanation:


4 0
3 years ago
from the set {-512,-8,8} ,use substitution to determine which value of x makes the equation true -64x=-1,152
Llana [10]

Answer:

18

Step-by-step explanation:

Just divide both sides by - 64

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3 years ago
In the numerical sentence below, A = 8, D = 1/2 , and F = 1/4. What are the values B, C, and E if the sentence below is equivale
a_sh-v [17]
B=3; C=16; and E=4096.

Using the information we have gives us

8^B \times C^{\frac{1}{2}} = E^{\frac{1}{4}}

We know that:
8^B = 2;
\\
\\C^{\frac{1}{2}}=4;
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\\E^{\frac{1}{4}}=8

We know that 2³ = 8, so B = 3.

To "undo" a square root, we square the number; 4²=16, so C = 16.

To "undo" a fourth root, we raise the answer to the fourth power; 8⁴ = 4096, so E = 4096.
3 0
3 years ago
Read 2 more answers
Factor: 2x + 2y<br> Show work please
bearhunter [10]

Answer:

2(x + y)

Step-by-step explanation:

You can pull out a 2 from the expression.

So divide 2x by 2 and 2y by 2.

2x/2 = x

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7 0
3 years ago
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}

7 0
3 years ago
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