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3 years ago

How many 3/5 make a whole ?

Mathematics

2 answers:

kow [346]3 years ago

7 0

Answer:

5/3

Step-by-step explanation:

3/5 in one means 1 ÷ 3/5

To divide the fraction you have to use multiplication and flip the numerator and denominator

1/1 × 5/3 = 5/3

exis [7]3 years ago

7 0

Answer:

What is 3/5 of a whole?

3 / 5 = 0.6

Step-by-step explanation:

If you insist on converting 3/5 to a whole number, the best we can do is round the decimal number up or down to the nearest whole number. 3/5 rounded up to the nearest whole number is 1 and 3/5 rounded down to the nearest whole number is 0.

You might be interested in

Find four consecutive even integers such that if the sum of the first and third is multiplied

Alenkasestr [34]

Answer:

Step-by-step explanation:

-8-10,-12-14

-8 plus -12 is -20

-20 times 5 is -100

-14 times 8 is -112

-100 is 12 more than -112

6 0

2 years ago

Are ADEF and ARPQ congruent?

kari74 [83]

Answer:

B. Yes. ΔDEF can be mapped to ΔRPQ by a 180° rotation about the origin followed by a translation 2 units down.

Step-by-step explanation:

^^^just did it on edg and got it right.

8 0

3 years ago

If Samantha can pay off her loan in 36 months at a 10% interest rate rather than in 48 months at a 12% interest rate, how much m

Ymorist [56]

Hi there
The simple interest formula is
I=prt
I interest changes
P amount of the loan 6000
R interest rate
T time( number of months/12 months)

The interest in 36 months at a 10%
I=6,000×0.1×(36÷12)=1,800
The interest in 48 months at a 12%
6,000×0.12×(48÷12)=2,880
she will save
2,880−1,800=1,080

Good luck!

7 0

3 years ago

Find the sum of 0.333......and 0.444....

SashulF [63]

Answer:

0.777......

Step-by-step explanation:

0.333...... + 0.444......

= 0.777......

Hope it helps :)

Please mark my answer as the brainliest

7 0

2 years ago

Read 2 more answers

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

2 years ago