All categories

3 years ago

A stamp has the Length of 1/2 cm and a width of 1 1/2 is the area of the stamps

Mathematics

1 answer:

avanturin [10]3 years ago

3 0

Answer:

$Area = \frac{3}{4} cm^2$

Step-by-step explanation:

Given

$Length =\frac{1}{2}\ cm$

$Width = 1\frac{1}{2}\ cm$

Required

Determine the Area

Area is calculated as follows

$Area = Length * Width$

Substitute values for Length and Width

$Area = \frac{1}{2} * 1\frac{1}{2}$

Convert mixed fraction

$Area = \frac{1}{2} * \frac{3}{2}$

$Area = \frac{3}{4}$

Hence, the area is

$Area = \frac{3}{4} cm^2$

You might be interested in

Would you rather pay for 3 gifts and get 1 free, or Use a coupon for 20% off? Explain you answer.

N76 [4]

Answer:

I would just use the coupon

Step-by-step explanation:

Because why pay the whole price for three gifts just to get one for free when i could just use a coupon and get that one gifts 20% off the regular price

6 0

3 years ago

A) In a group of 60 students, 15 liked maths only, 20 liked science only and 5 did not like

Paladinen [302]

Part (i)

We have 60 students total, and 5 didn't like any of the two subjects, so that must mean 60-5 = 55 students liked at least one subject.

<h3>Answer: 55</h3>

=========================================================

Part (ii)

We have 15 who like math only, 20 who like science only, and 55 who like either (or both). Let x be the number of people who like both classes.

We can then say

15+20+x = 55

x+35 = 55

x = 55-35

x = 20

This means 20 people liked both subjects

<h3>Answer: 20</h3>

=========================================================

Part (iii)

There are 15 people who like math only, and 20 who like both. Therefore, there are 15+20 = 35 people who like math (and some of these people also like science)

<h3>Answer: 35</h3>

=========================================================

Part (iv)

We'll follow the same idea as the previous part. There are 20 people who like science only and 20 who like both subjects. That yields 40 people total who like science (and some of these people also like math).

<h3>Answer: 40</h3>

=========================================================

Part (v)

We'll draw a rectangle to represent the entire group of 60 students. This is considered the universal set. Inside the rectangle will be two overlapping circles to represent math (M) and science (S).

We'll have 15 go in circle M, but outside circle S to represent the 15 people who like math only. Then we have 20 go in circle S but outside circle M to show the 20 people who like science only. We have another copy of 20 go in the overlapped region between the circles. This is the 20 people who like both classes. And finally, we have 5 go outside both circles, but inside the rectangle. These are the 5 people who don't like either subject.

Note how all of the values in the diagram add up to 60

15+20+20+5 = 60

This helps confirm we have the correct values.

<h3>Answer: See the venn diagram below</h3>

3 0

3 years ago

Evaluate 9/g+2h+5 when g=3 and h=6

vladimir1956 [14]

To answer this question, simply input the values for g and h:

9/ (3) + 2 (6) + 5 = 9/ 3 + 12 + 5
= 9/ 20

Hope this helps!

6 0

3 years ago

Read 2 more answers

Clare is paid $90 for 5 hours of work. At thos rate how many seconds does i take for her to earn 25 cents?

natka813 [3]

We are given

Clare is paid $90 for 5 hours of work

Total money paid =$90

total hours of work =5 hours

so, firstly we will find rate of work

rate of work = ( total money paid)/( total hours of work)

now, we can plug values

and we get

The rate of work is

$=\frac{90}{5}$

$=18$ $/ hour

now, we have to find time for amount paid is 25 cents

so, amount paid =25 cents

100 cents =1$

so, amount paid =$ 0.25

Since, the rate is constant

so, we can take same rate here

now, we can use formula

rate of work = ( total money paid)/( total hours of work)

now, we can plug values

$18=\frac{0.25}{T}$

now, we can solve for T

$T=\frac{0.25}{18}$ hour

since, T is in hours

so, we can change into seconds

1 hour =3600seconds

so, we can plug

$T=\frac{0.25}{18}\times 3600$ seconds

$T=50$ seconds......................Answer

6 0

3 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

4 years ago