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2 years ago

Which of the following graphs show a proportional relationship?choose all that apply

Mathematics

2 answers:

ziro4ka [17]2 years ago

5 0

Answer:

Step-by-step explanation:

Mrac [35]2 years ago

4 0

Answer:

the first one coz the all the cordinates of the graph have the same number on both axis' we can take example of 2,2 3,3 5.5,5.5 etc.

whereas the other graph starts itself on the why axis instead at the origin.

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Which terrorist group has been a major force in the conflicts in Iraq, Syria, and Yemen?

gizmo_the_mogwai [7]

D. Al- Qaeda
Hope this helped ;)

5 0

2 years ago

Help me on number 12 13 14 and 15

aleksandr82 [10.1K]

12. 1.625 [terminating]; 13. 0.83 [bar notation over 3 (repeating)]; 14. 900 cm = 9 m; 15. 0.23 cm = 2.3 mm

Repeating decimals are parts of decimals that have repetitive digits; terminating decimals are decimals whose digits end.

Whether you are using Metric or Imperial, you have to determine whether you are going from a small unit to a big unit or vice versa. Then perform your operation. So, in exercise 14, the smaller unit is centimeters, so you would be going from big to small. Exercise 15 has you going from small to big.

There are centimeters in one meter, so multiply 9 by to get 900 centimeters.

There are 10 millimeters in one centimeter, so divide 2.3 by 10 simply by moving the decimal point ONCE to the left [Power of 10].

small to BIG → Division

BIG to small → Multiplication

I am joyous to assist you anytime.

5 0

2 years ago

Here are some values of sequence Q. Write a recursive definition for the sequence.

Rashid [163]

Answer: Q(n) = Q(n - 1) + 2.5

Step-by-step explanation:

We have 3 values of the sequence Q(n)

These values are:

Q(1) = 3

Q(3) = 8

Q(7) = 18

I would think that this is a geometric sequence.

Remember that the equation for the n-th term of a geometric sequence is:

A(n) = A(1)*r^(n-1)

where r is a constant, and A(1) is the first term of the sequence.

If we rewrite the terms that we know of Q(n) in this way we get:

Q(3) = Q(1)*r^(3 - 1) = 3*r^2 = 8

Q(7) = Q(1)*r^(7 - 1) = 3*r^6 = 18

Then we have two equations:

3*r^2 = 8

3*r^6 = 18

We should see if r is the same for both equations:

in the first one we get:

r^2 = 8/3

r = (8/3)^(1/2) = 1.63

and in the other equation we get:

r^6 = 18/3

r = (18/3)^(1/6) = 1.34

Then this is not a geometric sequence.

Now let's see if this is an arithmetic sequence.

The n-th term of an arithmetic sequence is written as:

A(n) = A(1) + (n - 1)*d

where d is a constant.

If we write the terms of Q(n) that we know in this way we get:

Q(3) = Q(1) + (3 - 1)*d = 3 + 2*d = 8

Q(7) = Q(1) + (7 - 1)*d = 3 + 6*d = 18

We need to see if d is the same value for both equations.

in the first one we get:

3 + 2*d = 8

2*d = 8 - 3 = 5

d = 5/2 = 2.5

In the second equation we get:

3 + 6*d = 18

6*d = 18 - 3 = 15

d = 15/6 = 2.5

d is the same for both terms, then this is an arithmetic sequence.

An arithmetic sequence is a sequence where the difference between any two consecutive terms is always the same value (d)

Then the recursive relation is written as:

A(n) = A(n - 1) + d

Then the recursive relation for Q is:

Q(n) = Q(n - 1) + 2.5

4 0

2 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$