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3 years ago

What set does 30 belong to

Mathematics

1 answer:

docker41 [41]3 years ago

7 0

Answer:

30 belongs to the set of real numbers, the set of rational numbers, the set of whole numbers, the set of natural numbers and the set of integers.

Step-by-step explanation:

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the expression 5n + 2 can be used to find the total cost in dollars of bowling where n is the number of games bowled and the 2 i

gogolik [260]

5(3)+2
15+2
17 monies

Just plug the amount of games in for n

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3 years ago

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Help me please, guys

devlian [24]

Answer:

7.7

Step-by-step explanation:

The area of the wall is 4 * 2 = 8.

The radius of each clock is 0.3 / 2 = 0.15.

The area of all 4 circles is 4 * (πr²) = 4 * 3.14 * 0.15² = 0.3.

8 - 0.3 = 7.7

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3 years ago

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Need help! will award brainiest

coldgirl [10]

Answer:

$\sf slope: \boxed{\bf \frac{1}{3} }$

Explanation:

$\sf slope :\dfrac{y_2-y_1}{x_2-x_1} =\dfrac{rise}{run} \ \ \ \ where \ (x_1, \ y_1) ,(x_2, \ y_2) \ are \ points$

Find slope:

$\rightarrow \sf \dfrac{x_2-x_1}{y_2-y_1}$

insert values

$\rightarrow \sf \dfrac{3-5}{-4-2}$

simplify

$\rightarrow \sf \dfrac{1}{3}$

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2 years ago

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three brothers and a sister shared a sum of money equally among themselves. if the brothers alone had shared the money, then the

Virty [35]

Answer:

1. t=240

2.t/3=(t/4)+20

Step-by-step explanation:

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3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

4 years ago