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2 years ago

What is 4 divided by 2

Mathematics

2 answers:

Vanyuwa [196]2 years ago

8 0

The answer would be 2

lakkis [162]2 years ago

8 0

Answer:

Step-by-step explanation:

4÷2=2

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Jeremy made a belt that was 6.4 decimeter long. How many centimeters long is the belt Jeremy made?

stira [4]

Answer:

Step-by-step explanation:

1 Decimeter is equal to 10 centimeters. To convert decimeters to centimeters, multiply the decimeter value by 10. In this case it is 64.

Hope I helped:)

5 0

2 years ago

im not sure wether to replace the minus signs with addition, so if you could help me that would be nice :) 1.2y+4.5-3.4y-6.3

Elenna [48]

Answer:

-2.2y - 1.8

Step-by-step explanation:

We are to simplify the expression:

1.2y + 4.5 - 3.4y - 6.3

Collect like terms:

1.2y - 3.4y + 4.5 - 6.3

Simplify:

-2.2y - 1.8

That is the answer.

6 0

2 years ago

How is estimation helpful when dividing multi-digit numbers.

Burka [1]

See, when you are traveling, you might get lost. if so, just remember "I need to go between _ and _ miles" estimating and rounding to the nearest number with 0 wiil help.

5 0

2 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

2 years ago

Mental Math A typical tip in a restaurant is 15% of the total bill. If the bill is $110, what would the

Katarina [22]

Answer:

$16.50

Step-by-step explanation:

15/100*110

15/10*11

3/2*11

33/3

16.50

7 0

2 years ago