All categories

4 years ago

Find the reciprocal of 3/5

2 answers:

8 0

The reciprocal of 3/5 is 5/3.

To get the reciprocal of a fraction, just turn it upside down.

A fraction has a numerator, top number, and a denominator, bottom number.
To get the reciprocal of a fraction, just swap over the numerator and denominator.

If whole numbers, to get the reciprocal of whole number, just divide the whole number by 1 to make it into a fraction, then get the reciprocal of that fraction.

xz_007 [3.2K]4 years ago

6 0

The reciprocal is flipping it over and making the denominator which is 5 the numerator. So just switch the top and bottom numbers to get 5/3.

You might be interested in

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

makkiz [27]

Answer:

9a + 6b and 5c - 3d

Step-by-step explanation:

(a)

4a + 7b + 5a - b ← collect like terms

= (4a + 5a) + (7b - b)

= 9a + 6b

(b)

6c + 4d - c - 7d ← collect like terms

= (6c - c) + (4d - 7d)

= 5c - 3d

3 0

3 years ago

Read 2 more answers

What's 1/8+1/4.Make sure your answer fully reduced

ArbitrLikvidat [17]

Answer:

3/8

Step-by-step explanation:

1/8+1/4=1/8+2/8=3/8

6 0

3 years ago

Read 2 more answers

An international company has 26,900 employees in one country. If this represents 26.2% of the company's employees, how many empl

9966 [12]

There are 7048 employees in total

8 0

3 years ago

Translate the sentence into an equation<br>two more than the product of a number and 8 is 3.

Bingel [31]

Answer: 8x + 2 = 3

Step-by-step explanation: "two more than" means "add 2" which is +2

The product of 8 and a (unknown) number is 8x

Use the = sign in place of "is"

3 0

3 years ago

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

Lilit [14]

Check the picture below.

$\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}$

$\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1$

7 0

3 years ago

Read 2 more answers