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2 years ago

How many significant numbers are in 15,000

Mathematics

2 answers:

padilas [110]2 years ago

8 0

There are 2 if no decimal point is provided

timofeeve [1]2 years ago

6 0

15,000 has 2 significant figures.
1050 has 3 significant figures. 
0.350 has 3 significant figures.

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The width of a rectangle measures (3s+t) centimeters,and it’s length measures (3s-9t) centimeters. Which expression represents t

andreyandreev [35.5K]

Answer:

12s - 16t

Step-by-step explanation:

2(3s + t) + 2(3s - 9t)

6s + 2t + 6s - 18t

12s + 2t - 18t

12s - 16t

3 0

3 years ago

How many lines of symmetry does the picture have? A. 3 B. 2 C. 1 D. 0

brilliants [131]

the answers is D cuz there is no line in the picture

3 0

1 year ago

If George received $ 78.00 for his work at Weaver's Store, and he was paid $ 6.00 per hour, how many hours did he work?

mihalych1998 [28]

13 hours
78.00 ÷ 6.00 equals 13 hours
6×13 equals 78

3 0

3 years ago

Read 2 more answers

An individual is at a restaurant with two other people and they are trying to determine the tip for their check total which came

zlopas [31]

Answer:

c. The tip is $10.00

Step-by-step explanation:

The total bill in the restaurant is = $ 68.42

They wanted to pay a gratuity of 15%

Therefore the correct tip can be calculated as :

= 68.42 of 15%

= $68.42 \times \frac{15}{100}$

= 68.42 x 0.15

= 10.263

≈ 10 (approx.)

Thus the individual along with the other two wishes to pay a tip $10.00 for the total value.

7 0

2 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago