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Ivanshal [37]
3 years ago
13

Which of the following equations represent an exponential decay? Select all that apply

Mathematics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

yes

Step-by-step explanation:

the 3rd one

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it took Peter 3 hours to drive to his friend in Connecticut when he was driving with the average speed 54 mph.Find the constant
Vsevolod [243]

Answer:


Since k is constant (the same for every point), we can find k when given any point by dividing the y-coordinate by the x-coordinate.

so i assume it could be 162 sorry if i am wrong let me know if it is right or not

8 0
3 years ago
Factor the expression below
Mazyrski [523]

Answer:

(2x² + 5x + 2) = (2x + 1)(x +2)

Step-by-step explanation:

Given expression is (2x² + 5x + 2).

We have to factorize the given expression.

(2x²+ 5x + 2) = 2x² + 4x + x + 2

                     = 2x(x + 2) + 1(x + 2)

                     = (2x + 1)(x + 2)

So the factored form of the given expression will be (2x + 1)(x + 2).

6 0
3 years ago
Please helpppppppppppppp
Helen [10]

Answer:

90/23

Step-by-step explanation:

First, you want to start in the parentheses. 5-6=-1, then multiply it by -2. Since a negative multiplied by a negative is a positive, -2*-1=2. -3^2=-9. 2+8=10. Then -9 times 10= -90. Then, we move on to the bottom. 5--2=7. 4^2=16, and -2(2)=-4. So -5(7) + 16-4=-23. Now, the equation should look like this, -90/-23. Put the negatives together and you get 90/23.

6 0
3 years ago
What is the sum of i° + k°?
Brut [27]

Answer

1 + 32 + 243 + 1024 + .. + n5

Step-by-step explanation:

8 0
3 years ago
If <br>4x+(y+x)=180<br>what is x? what is y?​
uysha [10]

Step-by-step explanation:

4x+(y+x)=180\\\\\text{Solve for}\ x:\\\\4x+y+x=180\qquad\text{combine like terms}\\\\(4x+x)+y=180\qquad\text{subtract}\ y\ \text{from both sides}\\\\5x=180-y\qquad\text{divide both sides by 5}\\\\\dfrac{5x}{5}=\dfrac{180-y}{5}\\\\\boxed{x=\dfrac{180-y}{5}}\\\\\text{Solve for}\ y:\\\\5x+y=180\qquad\text{subtract}\ 5x\ \text{from both sides}\\\\\boxed{y=180-5x}

7 0
3 years ago
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