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disa [49]
3 years ago
8

I need the answer fast plzzz

Mathematics
2 answers:
aleksley [76]3 years ago
7 0
The answer is a because it matches the histogram
Alexandra [31]3 years ago
5 0

Answer:

A

Step-by-step explanation:

compare the numbers

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Jillian walked 0.5 miles before she started jogging at an average pace of 5 miles per hour. The equation d = 0.5 + 5t can be use
Mekhanik [1.2K]

Answer: independent variable: time (t). Dependent variable: distance (d).

Step-by-step explanation: an independent variable is a variable whose variation does not depend on that of another. In the given problem, the independent variable is time, because time will pass by no matter what (without depending in any other variable). The dependent variable in this case is the distance, because how far Jillian goes, depends on how much time she expends walking and jogging.

8 0
3 years ago
Read 2 more answers
PLZ HELP I AM FAILING MATH!!
Yuliya22 [10]

Answer:

4

Step-by-step explanation:

Given algebraic expression: 6x^3y+7x^2+5x+46x3y+7x2+5x+4

We know that the constants are the terms in the algebraic expression that contain only numbers.  

In the given expression only last term has only numerical value and no variable, rest of them have variable x.

Therefore, the constant term in the given algebraic expression  6x^3y+7x^2+5x+46x3y+7x2+5x+4 is 

6 0
3 years ago
Can someone please help if you can you are amazing and awesome
Ainat [17]

Answer:

domain(-4,6) and range(-1, 8)

Step-by-step explanation:

domain is an imput on x axis, and range on y

4 0
3 years ago
PLS HELP
Travka [436]

Answer:

A

Step-by-step explanation:

I did it on a test too

6 0
2 years ago
The half-life of cesium-137 is 30 years. Suppose we have a
VMariaS [17]
(a) If y(t) is the mass (in mg) remaining after t years, then y(t) = y(0) e^{kt}=100e^{kt}.

y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}

(b) y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}

(c)
100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}
 
3 0
3 years ago
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