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MissTica
2 years ago
15

9 cm

Mathematics
1 answer:
lions [1.4K]2 years ago
7 0

Answer:

I think it's 149cm

Step-by-step explanation:

I just doubled the numbers and multiplied them

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Find the area please​
Scilla [17]
144 I believe. You multiply length times width
5 0
3 years ago
Oomygawd plz help me with this math hw
aev [14]

Hi There!

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6 0
3 years ago
Find the number of four-digit numbers which are not divisible by 4?
SOVA2 [1]

Answer:

6750

Step-by-step explanation:

4 digit numbers are 1000,1001,1002,...,9999

let numbers=n

d=1001-1000=1

9999=1000+(n-1)1

9999-1000=n-1

8999+1=n

n=9000

now let us find the 4 digit numbers divisible by 4

4| 1000

______

| 250

4 |9999

  _____

  | 2499-3

9999-3=9996

so numbers are 1000,1004,1008,...,9996

a=1000

d=1004-1000=4

let N be number of terms

9996=1000+(N-1)4

9996-1000=(N-1)4

8996=(N-1)4

N-1=8996/4=2249

N=2249+1=2250

so number of 4 digit numbers not divisible by 4=9000-2250=6750

3 0
3 years ago
What is the base if the height is 6ft and the area is 14 ft on a triangle ​
insens350 [35]

Answer:

b = 4.66666667

Step-by-step explanation:

The equation for finding the area of a triangle is

0.5b x h = a

b = base

h = height

a = area

So I calculated 14 divided by 6

And got 2.3333333

And since that number is half of the base, I multiplied it by 2

2.333 x 2 ≈ 4.66666667

4 0
3 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
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