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snow_lady [41]
3 years ago
10

Please help! Will give Brainliest!

Mathematics
2 answers:
nataly862011 [7]3 years ago
4 0

I think it's the 3rd one going down. The A(n) = 1.5 • (0.74)n-1 ; 0.33 cm

KIM [24]3 years ago
3 0

Answer: Second option is correct.

Step-by-step explanation:

Since we have given that

Height of ball = 1.5 meters = 150 cm

Each curved path has 74% of the height of the previous path.

so, r = 74% = 0.74

Since initial height is n = 1.

As it form Geometric series.

So, the general equation would be

A(n)=ar^{n-1}\\\\A(n)=150(0.74)^{n-1}

We need to find the height of the ball at the top of the sixth path .

So, n= 6,

So, it becomes,

A(6)=150(0.74)^{6-1}=150(0.74)^5=33.285\approx 33.29\ cm

Hence, Second option is correct.

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the sides of a triangle are in the ratio 3:4:5 what is the length of each side if the perimeter of the triangle is 90cm
avanturin [10]
P of triangle = a + b = c = 90

<span>ratio 3:4:5
so a = 3x, b = 4x , c = 5x

3x + 4x + 5x = 90
12x = 90
x = 90/12 = 7.5

a = 3x = 3 * 7.5 = 22.5
b = 4x = 4 * 7.5 = 30
c = 5x = 5 * 7.5 = 37.5

Answer:
1st side = 22.5
2nd side = 30
3rd side = 37.5

Double check: 22.5 + 30 + 37.5 = 90 (perimeter)</span>
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The number N of bacteria present in a culture at time t (in hours) obeys the equation N = 1000e^0.01t After how many hours will
jasenka [17]

Answer: N(t) = (2^t)*1500

Step-by-step explanation:

Let's define the hour "zero" as the initial population.

So if N(t)  is the number of bacteria after t hours, then:

N(0) = 1500.

Now, we know that the population doubles every hour, so we will have that after one hour, at t = 1

N(1) = 2*1500 = 3000

after two hours, at t = 2.

N(2) = 2*(2*1500) = (2^2)*1500

After three hours, at t = 3

N(3) = 2*(2^2)*1500 = (2^3)*1500

So we already can see the pattern, the number of bacteria after t hours will be:

N(t) = (2^t)*1500

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Solve the equation 3/4 X+-2X=-1/4+1/2X+5​
Vaselesa [24]

Answer:

x = -19/7 = -2.714

Step-by-step explanation:

Step  1  :

           1

Simplify   —

           2

Equation at the end of step  1  :

   3             1   1

 ((—•x)-2x)-(((0-—)+(—•x))+5)  = 0

   4             4   2

Step  2  :

           1

Simplify   —

           4

Equation at the end of step  2  :

   3             1  x

 ((—•x)-2x)-(((0-—)+—)+5)  = 0

   4             4  2

Step  3  :

Calculating the Least Common Multiple :

3.1    Find the Least Common Multiple

     The left denominator is :       4

     The right denominator is :       2

       Number of times each prime factor

       appears in the factorization of:

Prime

Factor   Left

Denominator   Right

Denominator   L.C.M = Max

{Left,Right}

2 2 1 2

Product of all

Prime Factors  4 2 4

     Least Common Multiple:

     4

Calculating Multipliers :

3.2    Calculate multipliers for the two fractions

   Denote the Least Common Multiple by  L.C.M

   Denote the Left Multiplier by  Left_M

   Denote the Right Multiplier by  Right_M

   Denote the Left Deniminator by  L_Deno

   Denote the Right Multiplier by  R_Deno

  Left_M = L.C.M / L_Deno = 1

  Right_M = L.C.M / R_Deno = 2

Making Equivalent Fractions :

3.3      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)2   and  (y2+y)/(y+1)3  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

  L. Mult. • L. Num.      -1

  ——————————————————  =   ——

        L.C.M             4

  R. Mult. • R. Num.      x • 2

  ——————————————————  =   —————

        L.C.M               4  

Adding fractions that have a common denominator :

3.4       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

-1 + x • 2     2x - 1

——————————  =  ——————

    4            4  

Equation at the end of step  3  :

   3                 (2x - 1)    

 ((— • x) -  2x) -  (———————— +  5)  = 0

   4                    4        

Step  4  :

Rewriting the whole as an Equivalent Fraction :

4.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  4  as the denominator :

        5     5 • 4

   5 =  —  =  —————

        1       4  

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

4.2       Adding up the two equivalent fractions

(2x-1) + 5 • 4     2x + 19

——————————————  =  ———————

      4               4  

Equation at the end of step  4  :

   3                (2x + 19)

 ((— • x) -  2x) -  —————————  = 0

   4                    4    

Step  5  :

           3

Simplify   —

           4

Equation at the end of step  5  :

   3                (2x + 19)

 ((— • x) -  2x) -  —————————  = 0

   4                    4    

Step  6  :

Rewriting the whole as an Equivalent Fraction :

6.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  4  as the denominator :

         2x     2x • 4

   2x =  ——  =  ——————

         1        4  

Adding fractions that have a common denominator :

6.2       Adding up the two equivalent fractions

3x - (2x • 4)     -5x

—————————————  =  ———

      4            4

Equation at the end of step  6  :

 -5x    (2x + 19)

 ——— -  —————————  = 0

  4         4    

Step  7  :

Adding fractions which have a common denominator :

7.1       Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

-5x - ((2x+19))     -7x - 19

———————————————  =  ————————

       4               4    

Step  8  :

Pulling out like terms :

8.1     Pull out like factors :

  -7x - 19  =   -1 • (7x + 19)

Equation at the end of step  8  :

 -7x - 19

 ————————  = 0

    4    

Step  9  :

When a fraction equals zero :

9.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

 -7x-19

 —————— • 4 = 0 • 4

   4  

Now, on the left hand side, the  4  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

  -7x-19  = 0

Solving a Single Variable Equation :

9.2      Solve  :    -7x-19 = 0

Add  19  to both sides of the equation :

                     -7x = 19

Multiply both sides of the equation by (-1) :  7x = -19

Divide both sides of the equation by 7:

                    x = -19/7 = -2.714

One solution was found :

                  x = -19/7 = -2.714

Processing ends successfully

plz mark me as brainliest :)

8 0
3 years ago
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