All categories

3 years ago

What is 3/4 of 4/5. Plz help

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

6 0

Answer:

3/5

Step-by-step explanation:

We want to find 3/4 of 4/5.

The keyword of tells us that we need to multiply. Therefore:

$\frac{3}{4}\times\frac{4}{5}$

Multiply:

$=12/20=3/5$

So, 3/4 of 4/5 is 3/5.

We can confirm this by converting them to decimals.

4/5 is 0.8.

3/4 is 0.75.

0.75 or 75% of 0.8 is indeed 3/5 or 0.6.

You might be interested in

5/6 - 1/2 = ? A. 1/4 B. 1/3 C. 3/4 D. 2/3

zheka24 [161]

5. . 1
---- - -----
6. . 2

5. . 3
----- - -----
6. . 6

5-3
-------
6

2
-----
6

= 1/3

B

3 0

3 years ago

Solve x<3 and x≥−1 and write the solution in interval notation. If there is no solution, type ∅ .

bearhunter [10]

Shown in the Figure below

Step-by-step explanation:

In this exercise, we have the following system of inequalities:

$x$

The word and means intersection, so both inequalities must be true at the same time. We can graph this system as indicated in the figure below. Indeed, there is intersection between -1 and 1. Also, the number -1 is included in the solution because $x\geq -1$ and this is represented by the closed circle. On the other hand, the number 3 is not included in the solution because $x$ and this is represented by the open circle.

<h2>Learn more:</h2>

Inequalities: brainly.com/question/12261425

#LearnWithBrainly

7 0

3 years ago

The arithmetic sequence from the previous problem is 8. 7, 7. 3, 5. 9, 4. 5, 3. 1,. . . . What is the common difference, d, of t

o-na [289]

I think 8 is again maybe its true

3 0

3 years ago

Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x

svetlana [45]

Hello,

$sec(x)= \dfrac{1}{cos(x)} \\

cosec(x)= \dfrac{1}{sin(x)} \\

sec(x)*cosec(x)*(tg(x)+cotg(x))=\dfrac{1}{cos(x)}* \dfrac{1}{sin(x)}*( \frac{sin(x)}{cos(x)} +\frac{cos(x)}{sin(x)})\\

= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\

= \dfrac{1}{sin^2x*cos^2x} \\

$
==============================================================
$2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\

=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\

=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\

= \dfrac{(sin^2x+cos^2x)^2}{sin^2x*cos^2x}} \\

= \dfrac{1}{sin^2x*cos^2x}} 
$

8 0

3 years ago

Read 2 more answers

Please help and show work on paper please

jasenka [17]

You learn more from failure than from success. Don’t let it stop you. Failure builds character.

8 0

3 years ago