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3 years ago

Find four consecutive integers whose sum is 234.

Mathematics

1 answer:

Agata [3.3K]3 years ago

5 0

Answer:

57, 58, 59, 60

Hope this Helps :)

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Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2.4 h, and Car B traveled the dis

S_A_V [24]

Given:
t A = 2.4 h
t B = 4 h
v A = 22 + v B

Solution:
Distance A and distance B is the same, distance could be defined using formula d = v × t
d A = d B
(v A × t A) = (v B × t B)

plug in the numbers
v A × 2.4 = v B × 4
(22 + vB) × 2.4 = 4 vB

remove the parenthesis using distributive property
(22 × 2.4) + (2.4 × vB) = 4vB
52.8 + 2.4vB = 4vB
add like terms
52.8 = 4vB - 2.4vB
52.8 = 1.6vB
52.8/1.6 = vB
vB = 33
the speed of car B is 33 mph

vA = 22 + vB
vA = 22 + 33
vA = 55
the speed of car A is 55 mph

6 0

3 years ago

What is the only real number which, when divided by itself, is 2020<br> times itself?

malfutka [58]

Answer:

Step-by-step explanation:

I'm not sure at all I'm sorry

5 0

3 years ago

Read 2 more answers

Six aides work 6-1/2 Hours at $4.00 an hour,what are the total wages

GaryK [48]

You multiply 6.5 *4 =26 then you multiply 26 *6= 156 so that's $156.00.

8 0

3 years ago

A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e

My name is Ann [436]

Answer:

$\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

Step-by-step explanation:

If the system contains $n$ atoms, we can arrange $r$ quanta of energy in

$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$

ways.

$\mathbf{a)}$

In this case,

$n = 2, r=1$ .

Therefore,

$\binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2$

which means that we can arrange 1 quanta of energy in 2 ways.

$\mathbf{b)}$

In this case,

$n = 20, r=10$ .

Therefore,

$\binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756$

which means that we can arrange 10 quanta of energy in 184 756 ways.

$\mathbf{c)}$

In this case,

$n = 2 \times 10^{23}, r = 10^{23}$ .

Therefore, we obtain that the number of ways is

$\binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

3 0

3 years ago

4 questions answer them by number them down try show you solved plzzzzz

AVprozaik [17]

31.51

r=−5

d=19

h<53

Step-by-step explanation:

7 0

3 years ago