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Kitty [74]
3 years ago
15

There are 144 leaves on a tree outside of school. Each week the tree loses 8 leaves. How many leaves will be remaining on the tr

ee after 6 weeks
Mathematics
1 answer:
olga_2 [115]3 years ago
4 0

Answer:

96

Step-by-step explanation:

Day 1: 144-8= 136

Day 2: 136-8= 128

Day 3: 128-8= 120

Day 4: 120-8= 112

Day 5: 112-8= 104

Day 6: 104-8=96

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Wilma works at a bird sanctuary and stores birdseed in plastic containers. She has 2 small containers that hold 8 pounds of bird
Amanda [17]

Answer:

 60 lbs left over

Step-by-step explanation:  Since the small containers hold 8 lbs each you multiply 8 by 2 since there are 2 containers. Same for the large, multiply 9 by 12 because each container holds 12 lbs and there is 9 containers. Then, you subtract 4 from that total because she already used 4 lbs.

x=(2*8)+(9*12)-4 Multiply

x=16+108-4 Simplify

x=120

So, before she fills the bird feeders she has 120 lbs of bird seed. There are 30 bird feeders and since they hold 2 lbs each you multiply 30 and 2. Then you subtract that from the 120 lbs because she is filling the bird feeders with that seed.

120- (30*2)

120-60= 60 lbs left over

3 0
3 years ago
Simplify the expression.<br><br> Thanks yall, I'll give brainliest to the right answer
Alik [6]

Answer:

ok.

Step-by-step explanation:

3 0
2 years ago
Please help quick please
Vedmedyk [2.9K]

Answer:

the second one brotha

Step-by-step explanation:

7 0
3 years ago
PLZ HELP WILL GIVE BRAINS :)
kipiarov [429]
It would be 75 cents per song, I hope you get a good grade!
5 0
2 years ago
WILL GIVE BRAINLIEST: What is the local minimum value of the function g(x)=x^4-5x^2+4? (Round answer to the nearest hundredth)
IrinaK [193]

Answer:

Step-by-step explanation:

you can find where the first derivative is 0 to find the critical points

g'(x) = ( x^4 -5x^2 +4)' = 4x³-10x

g'(x) =0, make y =0 to find find where g'(x) is 0

4x³-10x =0 , factor 2x

2x(2x²-5)= 0 , each factor must be 0

2x= 0, so x= 0

2x²-5 =0, so x = ±√5/2

we now have 3 critical points -√5/2, 0, and √5/2

make intervals (-∞, -√5/2), (-√5/2, 0) , (0, √5/2) and (√5/2, +∞)

pick a point to test on each interval: -2, -1, 1 and 2 for example, and

calculate g'(x) = 4x³-10x at those points

for x= -2 we have 4(-2)³-10(-2) = -12 , negative number, decrease

for x= -1 we have 4(-1)³-10(-1) =6, positive number, increase

for x= 1 we have 4(1)³-10(1) = -6, negative number, decrease

for x= 2 we have 4(2)³-10(2) = 12, positive number, increase

we went from a decrease to an increase on intervals (-∞, -√5/2), (-√5/2, 0) so x= - √5/2 ≈ -1.58 is a minimum

we went from a decrease to an increase on intervals (0, √5/2), (√5/2, +∞) so

x= √5/2 ≈ 1.58 is a minimum as well

4 0
2 years ago
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