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1 year ago

10

Which algae has larger mass at 3 hours , in kilograms

1 answer:

pychu [463]1 year ago

3 0

Answer:

i need more information on this

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With 1 ≤ |a| < 10 and N is an integer.

Elena-2011 [213]

So a times 10^n where 1≤|a|<10 and n is an integer is what i learned

so

a.
234=
2.34 times 100=
2.34 times 10²

b.
0.000052=
5.2 times 1/100000=
5.2 times 10⁻⁵

c.
-220000=
-2.2 times 100000=
-2.2 times 10⁵

d. ok, so just count the spaces moved to the left
-18.2 times 10⁻¹⁵

6 0

2 years ago

A box contains 75 red and 25 green light bulbs. it is found that 20 bulbs in the box are defective, of which 2 are red. what is

Afina-wow [57]

Answer:

15 bulbs are red.

Step-by-step explanation: Because 75 is 3 times as big as 25 so every 1 green light bulb 3 red light bulbs will be deflective.

8 0

2 years ago

I’ve asked 3 times and no one knows pls helpahh

BigorU [14]

<h3>Answer: x = 21</h3>

Explanation:

The two angles shown combine to 180 degrees because the angles form a straight line.

(3x+18)+(4x+15) = 180

7x+33 = 180

7x = 180-33

7x = 147

x = 147/7

x = 21

7 0

2 years ago

Draw as many different triangles as possible that have two sides of length 4 cm and a 45° angle. Clearly mark the side lengths a

Basile [38]

Student Verified! ✅

Answer:

The answer is 3!

5 0

1 year ago

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

Viktor [21]

Answer:

Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

$T(t)=230 -e^{-kt}$

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

$138=230 -Ae^{-k\times 18}$

$\Rightarrow -Ae^{-18k}=138-230$

$\Rightarrow Ae^{-18k}=92$ .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

$184=230-Ae^{-k\times 36}$

$\Rightarrow Ae^{-36k}=230-184$

$\Rightarrow Ae^{-36k}=46$ .......(2)

Dividing (2) by (1)

$\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}$

$\Rightarrow e^{-18k}=\frac{46}{92}$

Taking ln both sides

$ln e^{-18k}=ln\frac{46}{92}$

$\Rightarrow -18k =ln (\frac12)$

$\Rightarrow -18k= ln1-ln2$

$\Rightarrow k= \frac{ln2 }{18}$

Putting the value k in equation (1)

$Ae^{-18\frac{ln2}{18}}=92$

$\Rightarrow A e^{ln2^{-1}}=92$

$\Rightarrow A.2^{-1}=92$

$\Rightarrow \frac{A}{2}=92$

$\Rightarrow A= 92 \times 2$

⇒A= 184.

Therefore $k= \frac{ln2 }{18}$ , A=184

7 0

2 years ago