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Katena32 [7]
2 years ago
11

Please help I need to pass this test

Mathematics
1 answer:
Gemiola [76]2 years ago
3 0
Ok it’s circular prisms of the purple blue
You might be interested in
A survey reveals that the sales of smartphones is increasing considerably. The average annual sales of smartphones, in million u
EleoNora [17]

Answer:

when the survey was initially conducted the average sales of smartphones was 1.7 million units

Step-by-step explanation:

The average annual sales of smartphones is given by function

P(n)= 1.7 (b)^n

Exponential growth function is given by

f(x)= ab^x

Where a= initial value

b = growth factor , b= 1+r where 'r' is the growth rate

x is the time period

Now we compare f(x) with P(n)

the value of a= 1.7 that is the initial value.

So we can say, when the survey was initially conducted the average sales of smartphones was 1.7 million units

8 0
3 years ago
3 On Monday, three hundred ninety-three students went on a trip to the zoo. All eight buses were
Kazeer [188]

Answer:

48

Step-by-step explanation:

First you would subtract 9 from 393 which is 384 then you would divide the 384 remaining student among the 8 busses which is 48.

Hope this helps.

8 0
3 years ago
Who won the race between the two balls of string
motikmotik
The higher one I think
7 0
3 years ago
There are 150 adults and 225 children at a zoo. If the zoo makes a total of $5100 from the entrance fees, and of an adult and a
Sonja [21]
Y+x=31, 150x+225y=5100. 
y=-x+31
150x+225(-x+31)=5100
-75x=-1875
x=25 (adult tickets are $25)
31-25= 6 (child tickets are $6)
3 0
3 years ago
Read 2 more answers
Which solution to the equation 1/x-1=x-2/2x^2-2 is extraneous?
olga2289 [7]
<span>An extraneous solution is a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the original problem.

Given the equation:
\frac{1}{x-1} = \frac{x-2}{2x^2-2}  \\  \\ 2x^2-2=(x-1)(x-2)=x^2-3x+2 \\  \\ x^2+3x-4=0 \\  \\ (x-1)(x+4)=0 \\  \\ x-1=0 \ or \ x+4=0 \\  \\ x=1 \ or \ x=-4

From the equation, it is obtained that x = 1 and x = -4 are the solutions of the equation.

Now, we substitute x = 1 into the equation as follows:
</span><span>\frac{1}{1-1} = \frac{1-2}{2(1)^2-2}  \\  \\  \frac{1}{0} = \frac{-1}{2-2}  \\  \\  \frac{1}{0} = \frac{-1}{0}

As can be see, for x = 1, the equation is undefined.

Now, we substitute x = -4 into the equation as follows:
\frac{1}{-4-1} = \frac{-4-2}{2(-4)^2-2} \\  \\  \frac{1}{-5} = \frac{-6}{2(16)-2} = \frac{-6}{32-2} = \frac{-6}{30} \\  \\ - \frac{1}{5} =- \frac{1}{5}

It can be seen that x = -4 is a valid solution of the orignal equation.

Therefore, x = 1 is an extraneous solution to the equation
\frac{1}{x-1} = \frac{x-2}{2x^2-2}</span>
8 0
3 years ago
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