All categories

3 years ago

Can someone please help me with this question

Mathematics

1 answer:

aniked [119]3 years ago

6 0

2/9 divided by 1/2. invert the second fraction and multiply

2/9 * 2/1 = 4/9

You might be interested in

The value of $\sqrt{73}$ is between two positive, consecutive integers. What is the product of these two integers?

AnnyKZ [126]

Answer:

Step-by-step explanation:

What are the two closest square numbers to 73?

64 and 81

$\sqrt{64} \leq \sqrt{73} \leq \sqrt{81}\\8\leq \sqrt{73} \leq 9\\8*9 = 72$

3 0

3 years ago

The shapes below are similar. What does x have to be?

laila [671]

I think it is 3.5
if 8 is divided by 4, you get two since they are the same shape, 14 should be divided by 4 as well which is 3.5

7 0

3 years ago

Read 2 more answers

4. 700 J of work is used to push an object for 12 seconds. What is the power of this machine?

s2008m [1.1K]

Answer:

58.3333333333

Step-by-step explanation:

3 0

3 years ago

Plsss Help!!! Will mark brainiest!! Due soon!!

Vladimir [108]

Answer:

√41

Step-by-step explanation:

Top is the formula, and the bottom would be how to plug in the points.

4 0

3 years ago

Read 2 more answers

Sarah is a computer engineer and manager and works for a software company. She receives a

daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = $a$

Given that:

$a_4=129\\a_{10}=207$

Formula for $n^{th}$ term of an Arithmetic Progression is given as:

$a_n=a+(n-1)d$

Where $d$ will represent the number of projects increased every year.

and $n$ is the year number.

$a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)$

Subtracting (2) from (1):

$78 = 6d\\\Rightarrow d =13$

By equation (1):

$129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90$

Number of projects in the first year = 90

b)

Number of projects in the twelfth year =

$a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233$

Each project pays $500

Earnings in the twelfth year = 233 $\times$ 500 = $116500

Sum of an AP is given as:

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938$

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 $\times$ 1938 = $969000

8 0

3 years ago