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3 years ago

50 points!!

Mathematics

2 answers:

V125BC [204]3 years ago

6 0

Answer:

the answer is 2

Step-by-step explanation:

I completed the test

Daniel [21]3 years ago

4 0

Answer:

iykvhchjgmh.Algal blooms have dramatic effects on water chemistry, most notably pH and dissolved oxygen (DO). When algae remove carbon dioxide during photosynthesis they raise the pH by increasing the level of hydroxide. The opposite reaction occurs during respiration when carbon dioxide is produced lowering hydroxide and lowering the pH.

Step-by-step explanation:

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A student has $78 in his checking account. If the student spends $12 on lunch, which number could be combined with the amount le

Alborosie

Answer:

-66

Step-by-step explanation:

There will be $66 left after the $12 spent on lunch. 66 + -66 = 0

7 0

4 years ago

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Help please asap today is the deadline

Korvikt [17]

The last one

1. 3-p
2. 8 (divid sign) d +6
3. 11+9*4
4. 4 (divide sign) 2
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7 0

3 years ago

Which of the following is equal to √^3√5 5^1/3 5^1/6 5^2/3 5^3/2

svlad2 [7]

I think it’s number 3

5 0

3 years ago

NEED ANSWER NOW PLEASE!

Len [333]

For this case, the first thing you should do is the following sum:
income = Annual Salary + benefits
Substituting values we have:
income = (48500) + (2500 + 5000)
income = 56000
Answer:
She can expect a total amount of income with benefits from:income = 56000
(option 4)

7 0

3 years ago

Situation:

Gala2k [10]

Answer:

5.5 days (nearest tenth)

Step-by-step explanation:

Given formula:

$\sf N=N_0e^{-kt}$

$\sf N_0$ = initial mass (at time t=0)
N = mass (at time t)
k = a positive constant
t = time (in days)

Given values:

$\sf N_0$ = 11 g
k = 0.125

Half-life: The time required for a quantity to reduce to half of its initial value.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

$\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}$

Therefore, the substance's half-life is 5.5 days (nearest tenth).

Learn more about solving exponential equations here:

brainly.com/question/28016999

5 0

2 years ago

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