Which subsets of numbers does -8 belong to?

Mathematics

1 answer:

Softa [21]3 years ago

8 0

Answer:

real, rational, integer

Step-by-step explanation:

You might be interested in

I need to find out the answer to the following questong.HELP!!!!!!!! 3×9+10×36/6

____ [38]

The answer would be 87.

6 0

3 years ago

Read 2 more answers

5x/-6 + 9/ -6 = 3x/-3 + 6/ -3 step by step

Ipatiy [6.2K]

These are the steps. Just follow the numbers on the left hand side. Hopes this helps. The answer is -3.

3 0

3 years ago

A rectangular cake has a length of 22 1/2 inches and a width of 10 inches. How many whole, square pieces of cake with side lengt

faltersainse [42]

Well, we need to find the area of the cake. 22.5 times 10 is 225. That is the area then we need to find the area of each piece of cake. 2.5 times 2.5 is 6.26.
so divide the area of the cake by the area of each piece to find how many pieces can be cut.
225/6.25=36 36 pieces of cake can be cut from the cake

5 0

3 years ago

If the interest rate on a savings account is 0.018%, approximately how much money do you need to keep in this account for 1 year

Diano4ka-milaya [45]

We are not told how often the interest is compounded, so assuming it is compounded yearly, you need to keep $9.99 in the account to pay the fee.

Explanation: 
Compound interest follows the formula A=p(1+r)^t,
where:
A is the total amount in the account,
p is the amount of principal,
r is the interest rate as a decimal number,
and t is the number of years.

For our problem: 
A = 9.99,
p is unknown,
r = 0.018% = 0.00018,
and t=1.

This gives us: 
9.99=p(1+0.00018)^1;
9.99=p(1.00018).

Divide both sides by 1.00018: 
9.99=p.

3 0

3 years ago

Read 2 more answers

Integrate e^x(sin(x) cos(x))

Karo-lina-s [1.5K]

$I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx$

$\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}$

Looking for $I_2$ :

$\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I$

Replacing:

$2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}$

6 0

3 years ago