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2 years ago

Olivia counted the number of ladybugs on each plant in her garden, then made the graph below.

Mathematics

1 answer:

Leni [432]2 years ago

5 0

Answer:

A) Roses , C) Alfalfa

Step-by-step explanation:

Each ladybug symbol = 5 ladybugs

Roses: 35 ladybugs

Lettuce: 15 ladybugs

Alfalfa: 25 ladybugs

Grape vines: 10 ladybugs

10 ladybugs go from lettuce to alfalfa. You end up with:

Roses: 35 ladybugs

Lettuce: 5 ladybugs

Alfalfa: 35 ladybugs

Grape vines: 10 ladybugs

Roses and alfalfa end up with 35 ladybugs each.

Answer: A) Roses , C) Alfalfa

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A contaminant is leaking into a lake at a rate of R(t) = 1400e0.06t gallons/h. Enzymes have been added to the lake that neutrali

ruslelena [56]

Answer:

Step-by-step explanation:

Rate of leakage, R(t) = 1400 e^0.06t gallons/h

fraction remains , S(t) = e^(-0.32t)

initial contaminant = 1000 gallon

gallons contaminant present after t hour is S(t) R(t)

G(t) = S(t) R(t)

$G(t) = 1400 e^{0.06t}\times e^{-0.32t}$

$G(t) = 1400 e^{- 0.26t}$

Put t = 18 hours

$G(t) = 1400 e^{- 0.26\times 18}$

Taking log on both the sides

ln G = ln 1400 - 0.26 x 18

ln G = 7.244 - 4.68

ln G = 2.564

G = 13 gallons

5 0

2 years ago

A recipe for guacamole uses 12% lime juice, if a batch contains 0.75 cup of lime juice, how large is the batch of guacamole?

FinnZ [79.3K]

0.75/0.12=6.25

the batch is 6.25 cups

8 0

2 years ago

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What is 1.532 rounded to the nearest hundredth

Anni [7]

Answer:

1.53 would be the answer

Step-by-step explanation:

If you are rounding to the nearest hundredth you are rounding to the 2nd decimal place. Since the next number is under 5, we would just drop it.

8 0

2 years ago

54.6g of sugar is needed to make 7 cakes. How much sugar is needed for 12 cakes?<br> |

Elena L [17]

Answer:

153.8 grams of sugar is needed

Step-by-step explanation:

5 0

2 years ago

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Which of the following is the expansion of (3c + d2)6?

vovangra [49]

Answer: Option A) $729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$ is the correct expansion.

Explanation:

on applying binomial theorem, $(a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r$

Here a=3c, $b=d^2$ and n=6,

Thus, $(3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r$

⇒ $(3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6$

⇒ $(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}$

⇒ $(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}$

⇒ $(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$

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2 years ago

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