Change the following mixed number to an improper fraction. Reduce if possible. 2 4/40

Kevin has mowed 40% of the lawn. If he has been mowing for 20 minutes, how long will it take him to mow the rest of the grass?

Katen [24]

It will take 30 minutes more for remaining grass. As he mowed 40% in 20 minutes, it shows that he mows twice percentage of grass in the time calculated in minutes. So it will take 30 minutes for remaining 60% grass.

3 0

3 years ago

2(a + 3) = −4 i need help i don't understand

kkurt [141]

Answer: a = -5

Step-by-step explanation:

2(a+3)=-4

Distribute

2a+6=-4

subtract the six with the four and you get -10

-6-4=-10

now divide 2a with -10

you get a= -5

you can plug the answer in

3 0

3 years ago

Read 2 more answers

What is x and y math help

Readme [11.4K]

<h3>Answer:</h3>

20 cans of cola
10 cans of root beer

<h3>Step-by-step explanation:</h3>

x and y are whatever you want them to be.

It can be convenient for solving a problem like this to use x and y to represent what the problem is asking for: the number of cans of cola and the number of cans of root beer. It is also convenient (less confusing) to use those variable names in the same order that the nouns of the problem are named:

... x = # of cans of cola

... y = # of cans of root beer

Then the problem statement tells you ...

... x + y = 30 . . . . . . . 30 cans total were bought

... x = 2y . . . . . . . . . . the number of cans of cola is twice the number of cans of root beer

_____

This set of equations is nicely solved by substitution: use the second equation to substitute for x in the first.

... (2y) +y = 30 . . . . . put 2y where x was

... 3y = 30 . . . . . . . . collect terms

... y = 10 . . . . . . . . . divide by 3

... 2y = x = 20

You're not done yet. You need to answer the question the problem asks.

Jared bought 20 cans of cola and 10 cans of root beer.

_____

Comment on x and y

You customarily see x and y as the variables of a problem. Personally, I like to use variables that remind me what they stand for. In this problem, I might use "c" for cans of cola and "r" for cans of root beer. Then when I've found the solution, I know exactly how it relates to what the question is asking.

Always start by writing down what the variables stand for (as we did here). Sometimes, this is called writing a Let statement: Let x = number of colas; let y = number of root beers.

Comment on problems of this type

When a proportional relationship is given between the items in a sum (2 cola cans for every root beer can), it is often convenient to work the problem in terms of groups of items. Here, a group of 3 items can consist of 2 cola cans and 1 root beer can. Then 30 items will be 10 groups, so 10 root beers and 20 colas. The problem is solved even before you can name the variables.

Even when the relationship isn't exactly proportional, you can add or subtract the extras and still work the problem this way. Had we said colas numbered 3 more than twice as many root beers, we could have our groups of 3 total 27 (30 less the 3 extra), giving 9 root beers and 21 colas (3 + 2·9).

8 0

3 years ago

Mrs. Rodrigues was seling two kinds of raffle tickets for $1.50 and $3.50. The number of 1.50 tickets sold was 5 more than three

disa [49]

Step-by-step explanation:

पति ऑफिस से घर आया... और खाना खाने बैठा

खाते-खाते उसने अपनी पत्नी से कहा कि

"खाना ठीक नहीं है, कोई टेस्ट नहीं आ रहा है।"

पत्नी चुपचाप उठी, और उसने कॉरपोरेशन मेंं कॉल किया और एम्बुलेंस को बुला ली

और कहा..

"इन्हे टेस्ट नहीं आ रहा है.."

एम्बुलेंस पति को ले गयी और उसे क्वॉरंटाइन कर दिया

अस्पताल में पत्नी ने पति को फोन किया और पूछा: अब आया स्वाद??

यहाँ तक की कहानी सबने सुनी होगी। अब देखिये कहानी में नया मोड़:

⬇️

.

उधर पति को पूछा गया कि

"आपके संपर्क में कौन-कौन आया था ?"

पति ने बड़ी ही शांति से कहा..

- मेरी पत्नी

- मेरा ससुर

- मेरी सास

- मेरा साला

- मेरी साली

बस अब ये सब भी अस्पताल के बिस्तर पर बैठे-बैठे पति को घूर रहे हैं!

Moral of the story is

बेजुबान को ज्यादा सताना भी ठीक

5 0

3 years ago

Read 2 more answers

What is the sum of the first 51 consecutive odd positive integers?

Angelina_Jolie [31]

We call:

$a_{n}$ as the set of the first 51 consecutive odd positive integers, so:

 $a_{n} = \{1, 3, 5, 7, 9...\}$

Where:
$a_{1} = 1$
$a_{2} = 3$
$a_{3} = 5$
$a_{4} = 7$
$a_{5} = 9$
and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

 $a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}$

Then:

 $a_{n} = a_{1} + (n-1)d$

So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

 $S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k$

Therefore, we need to find:
 $a_{k} = a_{51}$

Given that $a_{1} = 1$ , then:

$a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1$

Thus:
$a_{k} = a_{51} = 2(51)-1 = 101$

Lastly:

$S_{51} = ( \frac{1 + 101}{2} ).51 = 2601$

4 0

3 years ago