8-7x- x2 x+5/х+8 9x -9<br><br><br> Every step on how you got your answer.

Diff. / Original amount= % of increase or decrease

V125BC [204]

Answer:

% increase = $\frac{\text{ Increased number -Original number }}{\text{Original number}} \times 100\%$

% decrease = $\frac{\text{ Original number - Decreased number }}{\text{Original number}} \times 100\%$

Step-by-step explanation:

Percentage Increase:

Let there be an original number. If it increased to a certain value we can calculate the percentage increase value with the help of following formula:

% increase = $\frac{\text{ Increased number -Original number }}{\text{Original number}} \times 100\%$

Percentage Decrease:

Let there be an original number. If it decreases to a certain value we can calculate the percentage decrease with the help of following formula:

% decrease = $\frac{\text{ Original number - Decreased number }}{\text{Original number}} \times 100\%$

6 0

3 years ago

Complete the proof. PROVE:△VWY≅△UXY

svetoff [14.1K]

Answer:

<h3>D.) SAS or Side angle Side</h3>

osjdjaidbsojaanod

8 0

2 years ago

Read 2 more answers

How to find degrees in triangle using trig?

nordsb [41]

Remember SOHCAHTOA

Sine- opposite/hypotenuse
Cosine- adjacent/hypotenuse
Tangent- opposite/adjacent

Look at the angle (x) you are trying to figure out and pick which one of these it is.

In your calculator, do the equation first to get a decimal

Then press whichever one you chose on the calculator, so if you got like 0.2 by doing opposite/adjacent for tangent, you would type in 0.2 and then the TAN button

hope this makes sense!!

8 0

2 years ago

PLEASE HELP 10 POINTS

abruzzese [7]

A) 70 pools in total
B) 5:2

5 0

2 years ago

Situation:A researcher in North America discoversa fossile that contains 65% of its originalamount of C-14..-ktN=NoeNo inital am

zheka24 [161]

SOLUTION

We have been given the equation of the decay as

$\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}$

So we are looking for the time

Plugging the values into the equation, we have

$\begin{gathered} N=N_0e^{-kt} \\ 0.65N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.65N_0}{N_0} \\ e^{-0.0001t}=0.65 \end{gathered}$

Taking Ln of both sides, we have

$\begin{gathered} ln(e^{-0.000t})=ln(0.65) \\ -0.0001t=ln(0.65) \\ t=\frac{ln(0.65)}{-0.0001} \\ t=4307.82916 \end{gathered}$

Hence the answer is 4308 to the nearest year

8 0

5 months ago

8-7x- x2 x+5/х+8 9x -9 Every step on how you got your answer.