How many 3/4 are in 1

Mathematics

2 answers:

AnnyKZ [126]3 years ago

8 0

Answer: There are $\dfrac{4}{3}$ numbers of $\dfrac{3}{4}$ in 1.

Step-by-step explanation: We are given to find the number of $\drac{3}{4}$ in 1.

Let n be the number of $\drac{3}{4}$ that are present in 1.

So, we must have

$n\times\dfrac{3}{4}=1\\\\\\\Rightarrow n=\dfrac{1}{\frac{3}{4}}\\\\\\\Rightarrow n=1\times\dfrac{4}{3}\\\\\\\Rightarrow n=\dfrac{4}{3}.$

Thus, there are $\dfrac{4}{3}$ numbers of $\dfrac{3}{4}$ in 1.

bazaltina [42]3 years ago

6 0

1 because the cant be anymore than 1 3/4 in the answer because 3/4 + 3/4 = 6/4 or 1 2/4. so i would take 1 and subtract 3/4 to 1/4 and 1/4 is less than 3/4

You might be interested in

What is the following simplified product? Assume x greater-than-or-equal-to 0

nlexa [21]

Answer:

$10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}$

Step-by-step explanation:

Remove perfect squares from under the radicals.

$(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})\\\\=(\sqrt{10x^4})(2\sqrt{15x^4}) +(\sqrt{10x^4})(\sqrt{3x^3}) -(x\sqrt{5x^2})(2\sqrt{15x^4}) -(x\sqrt{5x^2})(\sqrt{3x^3})\\\\=2\sqrt{150x^8}+\sqrt{30x^7}-2x\sqrt{75x^6}-x\sqrt{15x^5}\\\\=\boxed{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}$