What is 14% as a fraction in simplest form

The number of bags of mulch you need to cover a planting area varies jointly with the area to be mulched (a) in sq ft and the de

Lena [83]

Answer:

22 2/9

Step-by-step explanation:

When z "varies jointly" with x and y, it can be described by the formula

z = kxy

Here, we have bags of mulch (n) varying jointly with area (a) and depth (d), both in feet. The given information can let us find the value of k.

n = kad

10 = k·(120)(1/4)

10/30 = k = 1/3 . . . . . divide by the coefficient of k

Now, we can fill in the other values of interest.

n = (1/3)·(200)·(1/3) = 200/9

n = 22 2/9

You need 22 2/9 bags of mulch to cover 200 ft² to a depth of 4 inches.

_____

Comment on the problem

This problem requires the formula be written with both area and depth expressed in feet, yet it gives depth in inches. The formula can also be written using depth in inches. In that case, k = 1/36.

8 0

3 years ago

The jars of peanut butter that I buy cost $7.20 each and normally contain 500 grams. At present, however, each jar contains an e

Alex_Xolod [135]

Answer:

$0.54

Step-by-step explanation:

Given:

The jars of peanut butter that I buy cost $7.20 each and normally contain 500 grams.

At present, however, each jar contains an extra 20% free.

And additionally, the supermarket has a "buy 3 jars get another one free''.

Question asked:

By how much is the cost per 100 grams of peanut butter less than it normally is if I take full advantage of the supermarket's offer today?

Solution:

Cost of a jars of peanut = $7.20

Normally, weight of peanut butter in one packet = 500 gram

At supermarket offer, each jar contains an extra 20% free, means a packet of peanut butter contains = 500 + 20% of 500 $=500+500\times \frac{20}{100} =500+100=600\ gram$

And extra benefit = buy 3 jars get another one free, means we will get 4 jars by just giving price of only 3 jars.

That means, total weight of peanut butter you are getting by giving 3 packet's price = Each jar weigh $\times$ number of jars

= 600 $\times$ 4 = 2400 gram

Cost of each packet will be same as previously as during the offer; $7.20

Cost of 3 packet's peanut (2400 gram) = $7.20\times3=\$21.6$

Now, we will calculate cost per 100 grams of peanut butter :

During offer:

Cost of 2400 grams of peanut butter = $21.6

Cost of 1 gram of peanut butter = $\frac{21.6}{2400}$

Cost of 100 gram of peanut butter = $\frac{21.6}{2400}\times100=\frac{2160}{2400} =\$0.9$

Normal day:

Cost of 500 grams of peanut butter = $7.20

Cost of 1 gram of peanut butter = $\frac{7.20}{500}$

Cost of 100 gram of peanut butter = $\frac{7.20}{500}\times100=\frac{720}{500} =\$1.44$

We found that cost of 100 gram of peanut butter in normal day is $1.44 while cost of 100 gram of peanut butter during offer is $0.9 means cost per 100 gram during offer is ($1.44 - $0.9 = )$0.54 less than cost during normal day.

Conclusion:

Therefore, cost per 100 grams of peanut butter is $0.54 less than it is normally if you take full advantage of the supermarket's offer today.

7 0

3 years ago

Can someone please help me!

marishachu [46]

Answer:

for the first one it would be y=3x-4 the second one is y=x-3

Step-by-step explanation:

graph the one with the x behind it one the x axis and the one without a letter on the y axis

3 0

3 years ago

Ebi, Jose, Derell, and Asami measured their heights. Ebi's height was 2.5 cm greater than Jose's height. Jose's height was 3.1 c

irga5000 [103]

Start with assigning each person with a variable to represent their height

Ebi: e
Jose: j
Derell: d
Asami: a

Ebi'd height was 2.5 cm greater than Jose's height

j + 2.5 = e

Jose's height was 3.1 cm greater than Derell's

d + 3.1 = j

Derell's height is 0.4 cm less than Asami's height

a - 0.4 = d

Ebi is 162.5 cm tall

e = 162.5

So, plug in 162.5 into any of the above equations were there is a variable of e

j + 2.5 = e

j + 2.5 = 162.5

Subtract 2.5 from both sides of the equation

j = 160 cm

Jose's height is 160 cm

Now, plug in 160 into any of the above equations where there is a j

d + 3.1 = j

d + 3.1 = 160

Subtract 3.1 from both sides of the equation

d = 156.9 cm

Derell's height 156.9 cm

so, plug in 156.9 into any of the above equations where there is a d

a - 0.4 = d

a - 0.4 = 156.9

Add 0.4 on both sides of the equation

a = 157.3 cm

Asami's height is 157.3 cm

7 0

3 years ago

Read 2 more answers

Hey! Can someone check my answers?

Andrew [12]

The greatest common factor is the biggest number taken from the values.

Q1. The answer is A. 5y^6

$20 y^{9} +5 y^{6}= 4*5 y^{9}+5 y^{6}$
Since $x^{a}* x^{b} =x^{a+b}$ , then $x^{9}= x^{3}* x^{6}$

Back to our expression:
$4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+ 5y ^{6}*1=5 y^{6} (4y ^{3} +1)$
The greatest common factor is thus $5 y^{6}$

Q2. The answer is D. 12xy^2

$12x y^{5}+60 x^{4} y^{2} -24 x^{3} y^{3}=12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}$
We will use the rule $x^{a}* x^{b} =x^{a+b}$ to factorise the exponents:
$12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)$
The greatest common factor is thus $12xy^{2}$

3 0

4 years ago