1). 140 tomatoes
7/5= 1.4
1.4(100)=140
2). 41 people
58/1.4= 41.4
Answer:
m = -2 and b = -1/3
m = -2 and b = -2/3.
Step-by-step explanation:
y = -2x - 1/3
y = -2x - 2/3
This system has no solutions because y cannot be 2 different values at once.
Hello!
You can use the Pythagorean Theorem to solve this
c is the hypotenuse which is the side across the right angle
a and b are the other sides
Put in the values you know
Square the numbers
Add
Take the square root of both sides
15 = c
The answer is 15
Hope this helps!
So you are seeing how much time it'll take so you solve for "t", the time.
<span>So you take the formula A=Pe^(rt) </span>
<span>A=2000 because it's the end value </span>
<span>P=20 because it's the starting value </span>
<span>r=.85 since 85%=.85 and .85 is the rate </span>
<span>Plug the values in and you get 2000=20e^(.85t) </span>
<span>What you do is you divide by 20 so you get 100=e^(.85t) </span>
<span>Take the natural logarithm of both sides 'cause of e and a natural log is written as ln so you get </span>
<span>ln 100=.85t ln e and because you can use the power rule you end up with .85t ln e and </span>
<span>ln e=1 so you have ln 100 = .85t so you divide by .85 so (ln 100)/.85=t and t=5.4178472776331 </span>
<span>hours </span>
<span>3. Exponential decay: </span>
<span>A= Pe^(rt) </span>
<span>where </span>
<span>A is the final amount </span>
<span>P is the initial value </span>
<span>r is rate of decay </span>
<span>t is time (years) </span>
<span>Let's say x is the initial amount then (1/2)x=xe^(32r) </span>
<span>I used x because the value isn't given but anyway division by x would give you 1/2=e^(32r) </span>
<span>Take the ln of both sides so ln 1/2=32r ln e and then ln e=1 so ln 1/2=32r. </span>
<span>Divide both sides by 32 and you'd get (ln 1/2)/32=r and r= -0.021660849392498 </span>
<span>4. Another depreciation question. </span>
<span>Each year the item retains 88% of its last-year value. </span>
<span>Solve: 250,000(0.88)^x = 100,000 </span>
<span>0.88^x = 0.4 </span>
<span>x = [log0.4]/[log0.88] </span>
<span>x = 7.168 years </span>
No, Crammer’s Rule isn’t always
applicable when trying to solve a system of linear equations because let’s say
for example, i<span>f the determinant of the coefficient matrix is 0,
then Cramer's rule cannot be applied. This
usually happens there’s no solution or an infinite number of solutions. </span><span>In
linear algebra, </span>Cramer's rule<span> is
an explicit formula for the solution of a system of linear equations with as
many equations as unknowns, valid whenever the system has a unique solution.</span>
I am hoping that this answer has satisfied your query and it will be
able to help you in your endeavor, and if you would like, feel free to ask
another question.