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nikdorinn [45]
2 years ago
6

Can someone please help me???????????????????

Mathematics
2 answers:
Vinil7 [7]2 years ago
8 0

Answer:

C

Step-by-step explanation:

jasenka [17]2 years ago
7 0

Answer:

The answer is A

Step-by-step explanation:

The equation is d=4t

The slope of this equation would be 4

Every new point on the graph is up 4, right 1

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Simplify completely quantity of x squared plus 7 x minus 30 all over quantity of x plus 10. x2 + 4 x2 ? 4 x ? 3 x + 3
GalinKa [24]

This looks like: \frac{x^2+7x-30}{x+10}

To solve this problem, you'll want to use the AC method to factor x^2+7x-30.

After factoring your expression will look like this:

\frac{(x-3)(x+10)}{x+10}

Cancel out the x+10 because they are common factors, so you are just left with the simple x-3.

Answer is:

x - 3

4 0
3 years ago
Read 2 more answers
What is the answer help
Alex Ar [27]

Answer:

900 batteries

Step-by-step explanation:

First we must find how many batteries does 1 calculator need

we can do this by dividing 160 by 40

which will give us 4 batteries per calculator

then we multiply the amount of calculators we have with 4

225×4 = 900 so we need 900 batteries

4 0
2 years ago
Please help for brainliest answer
erma4kov [3.2K]

Answer:

874.6

Step-by-step explanation

Profit = Sales - Costs

Profits by day:

Mon:

225 - (225*0.12) -9 = 189.0

Tue

260 - (260*0.12) -9 = 219.9

Wed

230 - (230*0.12)-9 = 193.4

Thu

180 - (180*0.12) - 9 = 149.4

Fri

150 - (150*0.12) -9 = 123.0

Total:

189+219.9+193.4+149.4+123 = 874.6

6 0
3 years ago
Help please I need my extra credit ​
masha68 [24]

Answer: For a test? Nah, thats cheating

Step-by-step explanation: Just kidding, the answer is B...

8 0
2 years ago
A population of bacteria is introduced into a culture. the number of bacteria P can be modeled by P=500(1+4t/(50+t^2 )) where t
Dennis_Churaev [7]
P(t)=500(1+4t/(50+t^2 ))

P'(t) = 500 [(50+t^2).4 - 4t.2t]/(50+t^2)^2

  by the quotient rule

 500 (-4t^2 + 200)/(t^2 + 50)^2

Hence

 P'(2) = 500 . (-16 + 200)/54^2 ~= 31.6
6 0
3 years ago
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