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3 years ago

What must be a factor of the polynomial function f(x) graphed on the coordinate plane below?

Mathematics

1 answer:

Phoenix [80]3 years ago

4 0

Answer:

Option B

A factor is x-1

Step-by-step explanation:

To find the factors of the polynomial, we will have to look for the roots of the polynomial.

Whenever a polynomial is plotted on a graph, it is important to note that the root of the polynomial refers to the points where the line cuts the x-axis.

From our curve given in the question, the roots of the polynomial are:

x = 1 and x = 6

Let us consider the first root

x = 1

from this, we can get x-1 =0

We will have to take only this as our answer because we are looking for only one of the factors.

Hence, a factor is x-1 and also our correct answer.

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What is 9 - g ≥ 3; g = 5

Yanka [14]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

The following table shows the times of trains between Juno and Odiham.

Luba_88 [7]

Answer:

(a) The earliest time that Ferdo can be back at Juno station is 12:15

Step-by-step explanation:

From the given table, we have;

The starting time for train departure from Juno = 07:15

The frequency of train departures from Juno = Every 20 minutes

The time of departure, 't₁' of first train that departs from Juno after 08:30 is given as follows;

The train departure times from Juno = 07:15, 07:35, 07:55, 08:15, 08:35

Therefore;

The time at which the first train departs from Juno after 08:30 = 08:35

The time it takes a train to travel from Juno to Odiham = 0:805 - 07:15 = 50 minutes

The time of arrival of the train Ferdo takes from Juno at Odiham = 08:35 + 50 minutes = 09:25

The number of hours Ferdo spends at Odiham = 2 hours

The time after which Ferdo departs from Odiham = 09:25 + 02:00 = 11:25

The time of departure of the train from Odiham after 11:25 = 08:05 + 20 minutes × 10

Therefore, the time of departure of the train from Odiham after = 11:25

The time it takes on the trip back to Juno from Odiham = 50 minutes

The earliest time that Ferdo can be back at Juno station = 11:25 + 50 minutes = 12:15

8 0

3 years ago

Select the correct answer. Which number has a repeating decimal form? A. B. C. D.

lianna [129]

Answer:

0.1727

Step-by-step explanation:

this is the correct answer because g00gle is a thing

4 0

3 years ago

Ways the answer to 12x-3y=3 and y=4x-1

Ede4ka [16]

Well first you want to be trying to solve for the same variable in both problems. Lets solve for Y.
This means we must make both equations say " Y="

first we subtract 12x from both sides of "12x-3y=3" which leaves us with
-3y=3-12x and y=4x-1.

you are basically going to merge the 2 problems by adding "1" to both problems and then subtracting "4x" from both as well as subtracting "y" from both. This gets rid of a variable in one problem and essentially puts in to the other in the process.

This leaves you with -4y=4-16x
Can you solve from there?

4 0

4 years ago

Math geometry help please show work thanks

Licemer1 [7]

<h3>Answer: Tony is correct</h3>

========================================================

Explanation:

There are number of ways to go about this, but let's pick on the sides marked in the diagram below (circled in red and blue). I picked on those sides specifically because they are on either side of the marked angles (A and Q).

They help form the proportion below, in which we'll show later on its a false statement.

AE/PQ = AB/QR

24/(14.5) = 18/(11.5)

24*11.5 = 14.5*18

270 = 261

We arrive at a false statement at the end, so the original proportion is false. Furthermore, it means the figures are <u>not</u> similar.

--------------

We could try the other way around and try to see if

AE/QR = AB/PQ

is true or not

Let's find out

24/(11.5) = 18/(14.5)

24*14.5 = 11.5*18

348 = 207

We see this is false as well. The gap is even bigger than before, so the first scenario was slightly close to having similar polygons. But neither case works perfectly. So that's why we don't have similar polygons and why Tony is correct. It's not enough to look at the angles only when dealing with polygons that have more than 3 sides. If we were just focused on triangles, then Jacenta would be correct.

4 0

3 years ago