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2 years ago

Is 3/5 a rational number

Mathematics

2 answers:

Allushta [10]2 years ago

7 0

Yes all fractions are

alexdok [17]2 years ago

3 0

Yes, it is.

Rational numbers are in the form p/q where p and q are both real numbers, and q does not equal to 0.

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If f(x) = 3x-1 and g(x)= x+2 find (f-g)(x)

wariber [46]

Answer:

2x-3

Step-by-step explanation:

f(x) = 3x-1

g(x) = x+2

Find (f-g)(x), this means we need to subtract g from x.

(3x-1) - (x+2)

Subtract the like terms:

3x -x = 2x

-1 -2 = -3

Combine them to get 2x-3

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3 years ago

2x^4–5x^3+x^2+3x+2=?<br><br> X=5

DanielleElmas [232]

Answer:

$2x^4–5x^3+x^2+3x+2=$

$2 \times 5^4–5 \times 5^3+5^2+3 \times 5+2$

= 2 × 625 – 5 × 125 + 25 + 3 × 5 + 2

=1250 – 625 + 25 + 15 +2

= 1292 – 625

= 667

3 0

2 years ago

Two couples (adam/brenda and carl/darlene) have bought tickets for four adjacent seats to see a musical. in how many ways can th

DochEvi [55]

One second I know this :)

7 0

2 years ago

In one year, the Perseid meteor shower had a meteor appear every 1 1/5 minutes on average. That same year, the Leonid meteor sho

RoseWind [281]

Answer:

325,452 more meteors fell during the Perseid meteor shower

Step-by-step explanation:

An year has 365 days.

Each day has 24 hours.

Each hour has 60 minutes.

Minutes in an year:

365*24*60 = 525600

Persoid:

$1 + \frac{1}{5} = 1 + 0.2 = 1.2$

One each 1.2 minutes. How many during an year?

1 meteor - 1.2 minutes.

x meteors - 525600 minutes.

$1.2x = 525600$

$x = \frac{525600}{1.2}$

$x = 438000$

438,000 meteors fell during the Persoid shower.

Leonid:

$4 + \frac{2}{3} = 4.67$

One each 4.67 minutes. How many during an year.

1 meteor - 4.67 minutes.

x meteors - 525600 minutes.

$4.67x = 525600$

$x = \frac{525600}{4.67}$

$x = 112548$

112,548 meteors fell during the Leonid Shower.

How many more meteors fell during the Perseid meteor shower?

438,000 - 112,548 = 325,452

325,452 more meteors fell during the Perseid meteor shower

5 0

2 years ago

The equation, and you have to find what he did wrong. Please help me, i'm stuck- ive been stuck on it since yesterday

bezimeni [28]

9514 1404 393

Answer:

terms were improperly combined

Step-by-step explanation:

I find the easiest way to identify the error in a solution is to solve it myself and compare with what is shown as being done.

Here, we're starting with ...

$\dfrac{1}{5}g-\dfrac{1}{10}-g+1\dfrac{3}{10}g-\dfrac{1}{10}$

If we identify the g-terms and the constant terms, we can group them the way Tyson did:

$\dfrac{1}{5}g+(-1)g+1\dfrac{3}{10}g-\dfrac{1}{10}-\dfrac{1}{10}$

The next step is to combine like terms. We can factor out g, so the coefficients are easier to see and combine.

$\left(\dfrac{1}{5}+(-1)+1\dfrac{3}{10}\right)g-\dfrac{1}{10}-\dfrac{1}{10}\qquad\text{step 1a}\\\\\dfrac{1}{2}g-\dfrac{1}{5}\qquad\text{step 2}$

The error is 1 3/10 was combined with -2/10 as a constant instead of being combined with -4/5 as a coefficient of g.

3 0

2 years ago