What is the solution to 34x + 95 = 3 (14X +9)
2 answers:
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Answer:
x = 14
y = 4
Explanation:
Ok so, just from looking at the two triangles i can tell they're congruent right triangles. I used different colors to show which sides of the triangle correspond and are equal to each other in my attatched photo.
So the side thats equal to x is the same length as the side that's equal to y+10 on the other triangle.
So we can write the equation x = y + 10.
Using this same method, the side that's equal to x + 2 is the same length as the side that's equal to 4y on the other triangle.
So, we can write the equation 4y = x + 2.
Now we have the equations you could rewrite to be in slope- intercept form so they're easier to graph. But a graphing calculator online would plot it just fine.
If you graph these two equations they'll intersect at the solution ( 14, 4 ). I'll include the graph in my images as well.
To check your answer, you can plug in x and y and see if the triangle sides end up being the same length. I did and it was correct.
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:
1) Equation given:
← 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:
3) Solving for α (remember α is λ)
αt ≈ 0.693
⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for 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:
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:
3) Solving for α (remember α is λ)
αt ≈ 0.693
⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ