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DIA [1.3K]
3 years ago
11

Given two similar triangles, how does one use proportions to find the lengths of one of the sides of one of the triangles? Answe

r in complete sentences, and give an example of how proportions can work.
Mathematics
1 answer:
Murrr4er [49]3 years ago
8 0
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ As they are similar triangles, there will be a scale factor between them. You may be given one pair of values on the triangle. Divide the larger value by the smaller value in order to find the scale factor between the triangles. For example, if the larger triangle has a length of 6cm and the smaller triangle has a length of 3cm, the scale factor would be 2. You would then multiply the given side by the scale factor to get the length.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

<u></u>

↬ ʜᴀɴɴᴀʜ ♡

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Between 2005 and 2008, the average rate of inflation was about3.4 per year. If a cart of groceries cost $140 in 2005, what did i
Elan Coil [88]
The formula is
A=p (1+r)^t
A future value?
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T time 2008-2005=3 years
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A=154.77 round your answer to get
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3 years ago
What is the sum of -5 and -5?<br> -10<br> 10<br> 0
Kipish [7]

Answer:

-10

Step-by-step explanation:

5 0
3 years ago
What is the mean of this set 18,21,20,14,17
konstantin123 [22]

Answer: 18

Step-by-step explanation:

Add them all up

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90/5=18

6 0
3 years ago
Read 2 more answers
there were 50 cars parked at the airport car rental lot. Of those 50 cars, 10% were red. How many cars were red?
solmaris [256]
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8 0
3 years ago
Read 2 more answers
Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

5 0
3 years ago
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