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Crank
3 years ago
10

What is the common difference, d, of the sequence?

Mathematics
2 answers:
NeTakaya3 years ago
7 0

Answer:

d=11

Step-by-step explanation:


mrs_skeptik [129]3 years ago
4 0
D is equal to 11. From -15 to -4, the common difference is 11. From -4 to 7, the common difference is 11, and so on.
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Help ;-; me please;-;
Diano4ka-milaya [45]

Answer: 13 is c im pretty sure and 14 is a im think hope fully it helps

Step-by-step explanation:

8 0
2 years ago
Barbara, Mark, and Carlos participated in the third heat of “The Big Race”. Barbara thought she could win with a 3 meter head st
steposvetlana [31]
Distance to travel = 20 m.
Let us determine the time, t, that each participant took to complete 20 m.

Barbara:
Average speed = (3 m)/(2 s) = 1.5 m/s
t = 20/1.5 = 13.3 s

Mark:
t =  5 s

Carlos:
y = x + 1
where
y = 20 m
x = time, t
Therefore
t =x = y - 1 = 20 - 1 = 19 s

Summary:
Barbara:  13.3 s
Mark:       5 s
Carlos:     19 s

Answer:
The winner is Mark
7 0
3 years ago
How to solve 2x2+50=100
Ugo [173]

Answer:

It would be 54.

Step-by-step explanation:

It would be 2 times 2, which is four, and then you would add by 50.

2 x 2 = 4

4 + 50 = 54

So, it will not be 100, it will be 54.

4 0
3 years ago
Read 2 more answers
Show that the equation x^3+6x-5=0 has a solution between x=0 and x=1
Mnenie [13.5K]

Answer:

Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.

Step-by-step explanation:

The equation f(x) given is:

f(x) = x^3+6x-5

For x = 0. the value of the expression is:

f(0) = 0^3+0-5\\f(0) = -5

For x = 1, the value of the expression is:

f(1) = 1^3+6-5\\f(1)=2

Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.

In other words, there is at least one solution for the equation between x=0 and x=1.

6 0
2 years ago
What does n^9/n^5 equal
vovangra [49]

Answer:

n^4

Step-by-step explanation:

Rule:

\dfrac{a^m}{a^n} = a^{m - n}

Your problem:

\dfrac{n^9}{n^5} = n^{9 - 5} = n^4

4 0
3 years ago
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