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Rufina [12.5K]
3 years ago
7

HELPpppp MeeeeE PLZ!!!!!!!!

Mathematics
1 answer:
Art [367]3 years ago
7 0

Answer:

answer is a

Step-by-step explanation:

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Factor this expression
Natasha_Volkova [10]

Answer:

6(3x−4y)

Step-by-step explanation:

6 0
1 year ago
Each week, a cook purchases 12 pounds of butter. During the last year, the cook has paid as little as $23.04 and at much as $29.
Tatiana [17]

Each week, a cook purchased 12 LBS. of Butter:

During the Last year:  (12 Months):

Cook Paid:

Little:   $23.04

Much:  $29.40, For Butter he or she purchased in a week:

Question: is: what is the Difference between, the Greatest price per pound, and the least price per pound of butter the cook paid within the last year?


EQUATION:

Least Paid / 12  =====>  23.04 /12

Most Paid / 12  ======>    29.40 / 12



Divide:

23.04 / 12    =     1.92 / LB

29.40 / 12     =     2.45 / LB

Now Subtract:

2.45   -     1.92

Answer   ======>    0.53 is the difference, between the greatest price per round, and least price per round of butter the cook would have paid within the last year.






Hope that helps!!!                                : )

4 0
3 years ago
A system consisting of one original unit plus a spare can function for a random amount of time X. If the density of Xis given (i
eimsori [14]

Answer:

0.2397

Step-by-step explanation:

Given data:

fX(x) = Cxe^(−x/2) if x >0 0 otherwise, where C > 0 is some constant.

The probability that the system functions for at least 5 months

= 0.2397

attached below is a detailed solution

3 0
2 years ago
A rectangular garden must have a perimeter of 150 feet and an area of at least 1200 square feet. Describe the possible lengths o
kipiarov [429]

Answer: 51.86 and 23.14

Step-by-step explanation:

The rectangular garden must have a perimeter of 150 feet

Perimeter of rectangular garden =

2l + 2w=150 -----------1

The rectangular garden must have a perimeter of 150 feet and an area of at least 1200 square feet.

Area of rectangular garden =

l×w= 1200 feet^2 -----------2

From equation 2, l=1200/w

Put l=1200/w in equation 1

2× 1200/w + 2w = 150

(2400/w) +2w = 150

(2400+2w^2)/w =150

2400+2w^2= 150w

2w^2- 150w+2400=0

Using the general formula

w = [-b+-√(b^2-4ac)]/2a

a = 2, b =-150, c=2400

w =[--150+/-√(-150^2-4×2×2400)]/2×2

=[150+/-√(22500-19200)]/4

=[150+/-√3300)]/4

=(150+57.45)/4 or (150-57.45)/4

w= 207.45/4 or 92.55/4

w= 51.86 or w= 23.14

l = 1200/51.86 or l= 1200/23.14

l = 23.14. or l= 51.86

For an area of at least 1200ft^2

The dimensions are 51.86 and 23.14

4 0
3 years ago
Find the length of arc ac in terms of pi
aniked [119]
=theta/360°×2pi ×r
theta=180-60=120°
r=24÷2=12

120/360 × 2 × 22/7 × 12=
1/3 × 2 × 22/7 × 12=
176/7=
25.14


3 0
3 years ago
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