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brilliants [131]
2 years ago
7

If you know the answers please let me know

Mathematics
1 answer:
lisov135 [29]2 years ago
4 0

Answer:

I believe its A C D

Step-by-step explanation:

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If f(x) = 3x to the 2nd+ 1 and g(x) = 1-x, what is the value of (f-g)(2)?
melisa1 [442]

 

\displaystyle\\f(x)=3x^2+1\\g(x)=1-x\\\\(f-g)(x)=f(x)-g(x)=3x^2+1-(1-x)=3x^2+1-1+x=\boxed{3x^2+x}\\\\(f-g)(2)=f(2)-g(2)=3\cdot2^2+2=3\cdot4+2=12+2=\boxed{14}




3 0
3 years ago
Help me, please! For 20 pts
emmasim [6.3K]

Answer:

First box ⇒ 3.6 x 10^-1

Second box ⇒ 3.1 x 10^6

Third box ⇒ 5.3 x 10^-6

Fourth box ⇒ 4.2 x 10^-1

Step-by-step explanation:

I used a calculator and solved it :)

6 0
2 years ago
Read 2 more answers
Find two positive numbers whose product is 36 and whose sum is a minimum.
alukav5142 [94]
Let one of the numbers be x. The other number cab then be represented as 36-x (x+36-x = 36).
The product can then be represented as y = x(36-x) or y=36x-x2

The maximum or minimum is always on the axis of symmetry which has the formula x=-b/2a.
In our case, the axis of symmetry is -36/-2, so x=18.

If one number is 18 and the 2 numbers add to 36, the other number is 18 as well.
So the 2 numbers are 18 and 18 and the maximum product is 324,
8 0
2 years ago
One pipe can empty a tank 5 times faster than another pipe can fill the tank. Starting with a full tank, if both pipes are turne
xxTIMURxx [149]

Answer:

40 hours

Step-by-step explanation:

Let's say that the slower pipe can fill x amount of the tank in one hour. It adds x amount to the tank every hour. Therefore, we can say, for the slower pipe,

1 hour = x amount

Then, the faster pipe empties the tank 5 times faster than the slower pipe fills it, so it removes 5 times the amount that the smaller pipe puts in, so for the faster pipe,

1 hour = -5x amount (negative to symbolize removing).

For the problem at hand, the tank starts at full, or 100%=1. It ends empty, or at 0% = 0. After 10 hours, if we only account for the slower pipe, we add x amount to the tank every hour, so we add 10 times that total, resulting in

1 + 10 * x as the ending result if we don't include the faster pipe. Then, the faster pipe removes 5x every hour, so in 10 hours, it removes 50x, so we have

1+10 * x - 50 * x as the final amount of stuff in the tank, which is equal to 0. Therefore, we have

1 + 10 * x - 50 * x = 0

1 - 40 * x = 0

add 40*x to both sides to isolate the x and its coefficient

1 = 40 * x

divide both sides by 40 to isolate x

x= 0.025

Therefore, the slower pipe adds 0.025, or 1/40 = 2.5% to the tank every hour. We want to figure out how long it would take for the slower pipe to fill up an empty tank, or turn it from 0% to 100% full.

Because the slower pipe adds 2.5% of the tank every hour, we can say that over y hours, it fills up

2.5% * y amount of the tank. We want to figure out how many hours it would take to make it 100% (we need to add 100% of the tank in the problem), so we can say

2.5% * y = 100%

divide both sides by 2.5% to isolate y

100%/2.5% = y = 40

5 0
3 years ago
What expression can be used for estimating 868/28
ivann1987 [24]
<span>31 would be the quotient.
The most simplified fraction would be 217/7.
Hope that's the answer you're looking for...

</span>
3 0
3 years ago
Read 2 more answers
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