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Ulleksa [173]
3 years ago
11

. A colony of bacteria triples every hour. If

Mathematics
1 answer:
ikadub [295]3 years ago
3 0

Answer:

a)60, b)14580, c)20*3^n

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2. Carissa also has a sink that is shaped like a half-sphere. The sink has a volume of 3617 in? One day, her
Juliette [100K]

Answer:

863 cups

Step-by-step explanation:

step 1

Find the volume of the conical cup

The volume of the cone (cup) is equal to

V=\frac{1}{3}\pi r^{2}h

we have

r=2/2=1\ in ----> the radius is half the diameter

h=4\ in

assume

\pi =3.14

substitute

V=\frac{1}{3}(3.14)(1^{2})4=4.19\ in^3  

step 2

Find out how many cups of water must Carissa scoop out  of the sink

Divide the volume of the sink by the volume of the cup

so

\frac{3,617}{4.19}= 863\ cups

8 0
3 years ago
Number of
jekas [21]

Answer:

The constant rate of change is 3

Step-by-step explanation:

Given

x   y

2   6

4   12

6   18

8   24

Required

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Represent the constant rate of change with k.

k is calculated using:

k = \frac{y}{x}

When y = 24; x = 8

So, we have:

k = \frac{24}{8}

k = 3

7 0
2 years ago
Solve each quadratic equation by factoring <br> X+3=24x
melomori [17]
X+3=24x :)))))))))))))))))))))
4 0
3 years ago
Read 2 more answers
7(x-2) - 3x + 5 = -2 (5x-4) + 4<br> Please help asap! - xxx
Oxana [17]

Answer:

x=

2

3

​

=1.500

Step-by-step explanation:

(((7•(x-2))-3x)+5)-((0-2•(5x-4))+4)  = 0

STEP

2

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Equation at the end of step 2

 ((7 • (x - 2) -  3x) +  5) -  (12 - 10x)  = 0

STEP

3

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STEP

4

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Pulling out like terms

4.1     Pull out like factors :

  14x - 21  =   7 • (2x - 3)

Equation at the end of step

4

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 7 • (2x - 3)  = 0

STEP

5

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Equations which are never true

5.1      Solve :    7   =  0

This equation has no solution.

A  non-zero constant never equals zero.

Solving a Single Variable Equation:

5.2      Solve  :    2x-3 = 0

Add  3  to both sides of the equation :

                     2x = 3

Divide both sides of the equation by 2:

                    x = 3/2 = 1.500

One solution was found :

x = 3/2 = 1.500

5 0
3 years ago
Read 2 more answers
If f(x) is differentiable for the closed interval [−4, 0] such that f(−4) = 5 and f(0) = 9, then there exists a value c, −4 &lt;
Pavel [41]
\bf \textit{mean value theorem}\\\\&#10;f'(c)=\cfrac{f(b)-f(a)}{b-a}\qquad &#10;\begin{cases}&#10;a=-4\\&#10;b=0&#10;\end{cases}\implies f'(c)=\cfrac{f(0)-f(-4)}{0-(-4)}&#10;\\\\\\&#10;f'(c)=\cfrac{9-5}{0+4}\implies f'(c)=\cfrac{4}{4}\implies f'(c)=1
4 0
3 years ago
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