Ask question

Ask question

All categories

3 years ago

9

In a School two flowerbeds as given in the figure are available for planting. If cost of planting flowers in smaller flowerbed i

s rs 5000 ,find the cost of planting The Other flowerbed at the same rate

1 answer:

Gala2k [10]3 years ago

8 0

? im sorry i cant help im still in middle school

You might be interested in

$7.80/hour = ____ cents/minute? a. 4.7 b. 13 c. 8.8 d. 780

g100num [7]

1$ = 100 cents, 1 hour = 60 minutes

$7.80/hour = (7.8*100) cents / (1*60)minutes

= 780 cents/60 minutes

= 13 cents/minute

<span>Option b.</span>

7 0

3 years ago

Read 2 more answers

The value from the set {10, 11, 12, 13} that makes the equation 2(x − 5) = 12 true is <br>.

Diano4ka-milaya [45]

Answer:

x = 11

Step-by-step explanation:

2(x - 5) = 12 ← divide both sides by 2

x - 5 = 6 ( add 5 to both sides )

x = 11

4 0

2 years ago

Read 2 more answers

Have a wonderful day lvu<3 free pts...... also if you go for biden plz dont come in saying trump sucks or sum like that ive b

Morgarella [4.7K]

✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽

➷ $455+446= 901$

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ May ♡

4 0

2 years ago

Use the quadratic formula to find the solutions to the equation.<br> 3x2 - 10x + 5 = 0

Nitella [24]

Answer: A

Step-by-step explanation:

3x² - 10x + 5 = 0

a = 3 b = -10 c = 5

Δ = b²- 4ac = 100 - 4.3.5 = 40 > 0 => have 2 solutions

=> x = $\frac{10+\sqrt{40} }{6}$ or x = $\frac{10-\sqrt{40} }{6}$

3 0

3 years ago

Read 2 more answers

John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m

Dmitry [639]

Answer:

The minimum possible initial amount of fish: $52$

Step-by-step explanation:

Let's start by saying that

$x$ = is the initial number of fishes

John:

When John arrives:

he throws away one fish from the bunch

$x-1$

divides the remaining fish into three.

$\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}$

takes a third for himself.

$\dfrac{x-1}{3} + \dfrac{x-1}{3}$

the remaining fish are expressed by the above expression. Let's call it John

$\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}$

and simplify it!

$\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}$

When Joe arrives:

he throws away one fish from the remaining bunch

$\text{John} -1$

divides the remaining fish into three

$\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}$

takes a third for himself.

$\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

the remaining fish are expressed by the above expression. Let's call it Joe

$\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

and simiplify it

$\text{Joe}=\dfrac{2}{3}(\text{John}-1)$

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

$\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)$

$\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}$

When James arrives:

We're gonna do this one quickly, since its the same process all over again

$\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}$

$\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)$

$\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}$

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say $n$ ), a third is taken away by James. the remaining bunch would be $\frac{n}{3}+\frac{n}{3}$

hence we've expressed the last pile in terms of n as well. Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

$\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

$\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

$x=52$

This is minimum possible amount of fish before John threw out the first fish

8 0

2 years ago