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melamori03 [73]
3 years ago
10

Given an isosceles triangle with vertex angle 90°. You drop a line segment from the vertex to the opposite side, to intersect at

the midpoint. The segment you drew has a length of 3cm. Why & how is the length of one of the isosceles sides 3 square root of 2 and not 3?
Mathematics
1 answer:
qwelly [4]3 years ago
7 0
Basically, you are bisecting the vertex angle AND the base of a 90 45 45 triangle, which creates two side by side 90 45 45 triangles.
The legs of the original triangle = 3^2 + 3^2.
leg^2 = 18
leg = 3 * sqroot (2)


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find three consecutive odd integers such that the sum of the last two is 15 less than 5 times the first​
Viefleur [7K]

2n + 1 , 2n + 3 , 2n + 5

____________________________

( 2n + 5 ) + ( 2n + 3 ) = 5 × ( 2n + 1 ) - 15

4n + 8 = 10n + 5 - 15

4n + 8 = 10n - 10

Add both sides 10

4n + 8 + 10 = 10n - 10 + 10

4n + 18 = 10n

Subtract both sides 4n

4n - 4n + 18 = 10n - 4n

18 = 6n

Divide both sides by 6

18 ÷ 6 = 6n ÷ 6

n = 3

_________________________________

Thus that three numbers are :

2(3) + 1 , 2(3) + 3 , 2(3) + 5

6 + 1 , 6 + 3 , 6 + 5

7 , 9 , 11

And we're done ....

3 0
3 years ago
PLZZZZZZ NEED HELP!!!!!!!!
elena-14-01-66 [18.8K]
First you need to know what the annual or monthly income is.
4 0
3 years ago
A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes th
77julia77 [94]

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

A=A_{o} e^{-kt}  (1)

Where:

A=7 g is the quantity left after time t

A_{o}=12 g is the initial quantity

t=70 min is the time elapsed

k is the constant of decay for the material

So, firstly we need to find the value of k from (1) in order to move to the next part of the problem:

\frac{A}{A_{o}}=e^{-kt}  (2)

Applying natural logarithm on both sides of the equation:

ln(\frac{A}{A_{o}})=ln(e^{-kt})  (3)

ln(\frac{A}{A_{o}})=-kt  (4)

k=-\frac{ln(\frac{A}{A_{o}})}{t}  (5)

k=-\frac{ln(\frac{7 g}{12 g})}{70 min}  (6)

k=0.00769995 min^{-1}  (7)  Now that we have the value of k we can solve the other part of this problem: Find the time t for A=2 g.

In this case we need to isolate t from (1):

t=-\frac{ln(\frac{A}{A_{o}})}{k}  (8)

t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}  (9)

Finally:

t=232.697 min \approx 233 min

5 0
3 years ago
-2x+11+6x how to solve
max2010maxim [7]
Well to simplify it we combine the x's: 4x+11, then to solve for x: 

4x= -11 
x= -11/4 

:)
6 0
4 years ago
Find the amount in the account for the given principal, interest rate, time, and compounding period.
bagirrra123 [75]

Answer:

make a list of any five duties you generally follow

Step-by-step explanation:

think and write

5 0
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