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spin [16.1K]
3 years ago
10

In a laboratory The Count of bacteria in a certain experiment was increasing at the rate of 2.5 % per hour find the bacteria at

the end of 2 hours if the count wasn't really 5,06,000.​
Mathematics
1 answer:
lesantik [10]3 years ago
3 0

Answer:

Прости я не понямайу по англискем

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7 over 8 plus 4 over nine
Natasha2012 [34]
(7/8) + (4/9) = 1.3194
6 0
3 years ago
The graph shows the best-fit regression model for a set of data comparing the number of hours spent hiking and the number of mil
Naddik [55]

Answer:

C

Step-by-step explanation:

Edgen

3 0
2 years ago
Read 2 more answers
Find the HCF and LCM of 4x^2 - 36 ,2x^2 - 12x + 18 and 2x^2 + x - 21​
olga_2 [115]

Answer:

(x-3), 4 (x - 3)^2 (x + 3) (2 x + 7)

Step-by-step explanation:

Factor all the expressions,

1st expression= 4x^2 - 36=4(x^2-9)=4(x+3)(x-3)

2nd expression=2x^2 - 12x + 18 =2(x^2-6x+9)=2 (x - 3)^2=2(x-3)(x-3)

3rd expression=2x^2 + x - 21​=(x - 3) (2 x + 7)

HCF=Commo factor=(x-3)

LCF=Common factor*Remaining factor=4(x+3)(x-3)(x-3) (2 x + 7)=4 (x - 3)^2 (x + 3) (2 x + 7)

4 0
2 years ago
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
Find the measures of the angles of a triangle if the measure of one angle is twice the measure of a second angle and the third a
guapka [62]

Answer:

<h2><em> 38°, 66° and 76°</em></h2>

Step-by-step explanation:

A triangle consists of 3 angles and sides. The sum of the angles in a triangle is 180°. Let the angle be <A, <B and <C.

<A + <B + <C = 180° ...... 1

If the measure of one angle is twice the measure of a second angle then

<A = 2<B ...... 2

Also if the third angle measures 3 times the second angle decreased by 48, this is expressed as <C = 3<B-48............ 3

Substituting equations 2 and 3 into 1 will give;

(2<B) + <B + (3<B-48) = 180°

6<B- 48 = 180°

add 48 to both sides

6<B-48+48 = 180+48

6<B = 228

<B = 228/6

<B =38°

To get the other angles of the triangle;

Since <A = 2<B  from equation 2;

<A = 2(38)

<A = 76°

Also <C = 3<B-48 from equation 3;

<C = 3(38)-48

<C = 114-48

<C = 66°

<em>Hence the measures of the angles of the triangle are 38°, 66° and 76°</em>

4 0
3 years ago
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