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3 years ago

Please help quickly. Will mark brainlist!

Mathematics

1 answer:

Ronch [10]3 years ago

8 0

Answer:

i'm not sure but i think its 9

i'm soooooo sorry if its wrong

Step-by-step explanation:

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Thank you in advance for the help !!

babymother [125]

The Answer is C, my good friend.

4 0

2 years ago

Arrange the fractions from least to greatest. 3/5 7/3. -3/4. 7/5. -1/4

zhannawk [14.2K]

1/4 3/5 3/4 7/5 7/3 is the correct order well at least i think it is but i hope it helps

5 0

3 years ago

Which statement about the angle measures is true?

sattari [20]

Answer:

Option (4)

Step-by-step explanation:

By the property of exterior angle of a triangle,

"Exterior angle of a triangle is equal to the sum of two opposite interior angles."

In the triangle ABC,

∠ACD is an exterior angle and ∠BAC and ∠ABC are the opposite interior angles.

m∠ACD = m∠BAC + m∠ABC

95° = m∠BAC + m∠ABC

Therefore, Option (4) will be the correct option.

5 0

3 years ago

Haylie and Justin went camping with their families for 7 days. They each took their own bottles of water. Haylie drank 6 bottles

trasher [3.6K]

Haylie took 48 bottles of water on the trip.

4 0

2 years ago

When a bactericide is added to a nutrient broth in which bacteria are growing, the bacteria population continues to grow for a

baherus [9]

Answer:

a) 1296 bacteria per hour

b) 0 bacteria per hour

c) -1296 bacteria per hour

Step-by-step explanation:

We are given the following information in the question:

The size of the population at time t is given by:

$b(t) = 9^6 + 6^4t-6^3t^2$

We differentiate the given function.

Thus, the growth rate is given by:

$\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t$

a) Growth rates at t = 0 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}$

b) Growth rates at t = 3 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}$

c) Growth rates at t = 6 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}$

3 0

3 years ago