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liq [111]
2 years ago
14

A runner moves 8 miles every 1 minute. Write and graph an equation in two variables that shows the relationship between time and

distance
Mathematics
1 answer:
Mila [183]2 years ago
5 0

Answer:

what grade is this?

Step-by-step explanation:

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use distributive property to simplify the expression 3 (8x + 2y) I need help with my algebra homework someone please help.
inysia [295]

6 \times (4 \times x + y)
4 0
3 years ago
2x2 − 4x = 0<br> A. <br> 0, -4<br> B. <br> 0, -2<br> C. <br> 0, 2<br> D. <br> 2, 4
leonid [27]

Answer:

C

Step-by-step explanation:

We see that we can factor out 2x from the equation! So our equation becomes 2x(x-2) = 0.

Since the only way for things to multiply to equal to zero is if at least one of the numbers is 0.

So if x-2 is the 0, then x = 2.

If 2x is the 0, then x = 0.

So our answer is C. 0,2

5 0
3 years ago
Please tell me the answer
kirill115 [55]

Answer:

<h2>(-∞,∞) Bottom answer</h2>

I hope this helps you with your algebra

5 0
3 years ago
Read 2 more answers
Find the 15th term in the following<br> arithmetic sequence :<br> 1, 6, 11, 16,
Sunny_sXe [5.5K]

Answer:

71

Step-by-step explanation:

This arithmetic sequence starts at 1 and increases by 5 every time. The formula for this sequence could be expressed as A(n) = 1 + 5(n-1).

Since you're finding the 15th term, substitute n = 15 into the formula.

A(15) = 1 + 5((15)-1) = 1 + 5(14) = 1 + 70 = 71

6 0
3 years ago
Read 2 more answers
The half-life of carbon 14 is years. How much would be left of an original -gram sample after 2,292 years? (To the nearest whole
lozanna [386]

Answer:

A = 0.75 gram or 1 gram

Step-by-step explanation:

The half-life of carbon 14 is years. How much would be left of an original -gram sample after 2,292 years? (To the nearest whole number).

We can use the following formula for half-life of ^{14}C to find out how much is left from the original sample after 2,292 years:  

A = A_{0}e^{-0.000124t}

where:

<em>A</em> is the amount left of an original gram sample after <em>t</em> years, and  

A_{0} is the amount present at time <em>t</em> = 0.

The half-life of  ^{14}C  is the time <em>t</em> at which the amount present is one-half the amount at time <em>t </em>= 0.

If 1 gram of  ^{14}C is present in a sample,  

 Solve for A when t = 2,292:  

Substituting A_{0}  = 1 gram into the decay equation, and we have:  

A = A_{0}e^{-0.000124t}

A = A_{0}e^{-0.000124(2,292)}

A = 0.75 g or 1 g  

6 0
2 years ago
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