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2 years ago

8, 12, 16,... Find the 31st term.

Mathematics

2 answers:

Ainat [17]2 years ago

7 0

<em>Hope I helped. :)</em>

<em />

<em>- Valenteer</em>

hoa [83]2 years ago

6 0

The answer to this question is 124
Hope this helped:)

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Pls help my test is timed!! Which of the following scatterplots would have a trend line with a negative slope?

swat32

the first/upper one

the other point cloud is too compact to leave a clear clue about any slope

8 0

2 years ago

If Smith is guilty, then Jones is innocent. If Jones is innocent, then Smith is guilty. Write each biconditional statement in ex

Volgvan

Answer:

The required biconditional statement is: If Smith is guilty, if and only if Jones is innocent.

Step-by-step explanation:

From the provided statement.

The statement is: If Smith is guilty, then Jones is innocent.

The converse is: If Jones is innocent, then Smith is guilty.

The combination of a conditional statement and its converse is called biconditional statement.

The biconditional statement contains if and only if phrase between two part of the statement.

Which means the statement and converse both are true.

Therefore, the required biconditional statement is: If Smith is guilty, if and only if Jones is innocent.

6 0

3 years ago

If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?

Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

$(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2$

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

5 0

3 years ago

Help meeee with this questionnnnnnnnnn!!!!!!!!!!!!

Dafna1 [17]

Given:

The scale factor between two circles is $\dfrac{2x}{5y}$ .

To find:

The ratio of their areas.

Solution:

We know that, all circles are similar.

The ratio of the areas of similar figures is equal to the ratio of squares of their corresponding sides or equal to the square of ratio of their corresponding sides.

The scale factor is the ratio of the corresponding sides.

Ratio of areas of circles = Square of scale factor between two circles

$\text{Ratio of areas of circles}=\left(\dfrac{2x}{5y}\right)^2$

$\text{Ratio of areas of circles}=\dfrac{(2x)^2}{(5y)^2}$

$\text{Ratio of areas of circles}=\dfrac{4x^2}{25y^2}$

Therefore, the correct option is D.

3 0

3 years ago

Cindy claims that 4^x/12^x=1/3

pav-90 [236]

A whole number that would support Cindys claim would be 2 because if u do the ,math it would be 8/24 which would be .333 repeating which is simplify to 1/3 and i do not know a number that would not work

4 0

2 years ago