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4 years ago

In which of the following ways are linear systems similar to quadratic systems? Select all that apply.

Mathematics

2 answers:

alex41 [277]4 years ago

8 0

The correct answers are:

A) Both can be solved by graphing;

C) Both can be solved by substitution; and

D) Both have solutions at the points of intersection.

Explanation:

Just as a system of linear equations can be solved by graphing, a system of quadratic equations can as well. We graph both equations. We then look for the intersection points of the graphs; these intersection points will be the solutions to the system.

We can also solve the system by substitution. If we can get one variable isolated, we can substitute this into the other equation to solve.

nikdorinn [45]4 years ago

7 0

Answer: A. Both can be solved by graphing.

C. Both can be solved by substitution.

D. Both have solutions at the points of intersection.

Step-by-step explanation:

Linear systems and quadratic systems both are solved by graphing and substitution.

We can graph both the systems. In linear systems , we get lines and in quadratic systems we get curves .

The points of intersection of two lines or curves are the solutions of the respectively system.

We can substitute the value of one variable in the equation find the solution .

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A. -22 B. 11 C. 13 D. 26

AlexFokin [52]

D. 26

$2n - 3 : 8$

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3 years ago

Suppose a bag of marbles has 4 green, 2 red, 5 yellow, 1 brown, and 7 blue marbles. What is the probability of picking a red mar

swat32

Assuming that each marble can be picked with equal probability, we notice that there is a total of

$4+2+5+1+7 = 19$

marbles, of which 2 are red.

So, the probability of picking a red marble is

$\dfrac{2}{19}$

In fact, as in any other case of (finite) equidistribution, we used the formula

$P(\text{event}) = \dfrac{\text{number of favourable cases}}{\text{number of all possible cases}}$

4 0

3 years ago

A lock has a two digit numeric code, with the numbers 1 through 6 available for each digit of the code. Note that numbers can be

Ede4ka [16]

There are 36 possible codes for this

8 0

2 years ago

If you have a rectangle with the width of 6 in and a length of 10 in. And you want to create a scaled copy using a scale factor

snow_tiger [21]

Answer:

The new side lengths will be 24 inches and 40 inches

Step-by-step explanation:

Here, given the width and length of a rectangle and we want to create a scaled copy using a scale factor of 4, we want to know the new side lengths

What we simply want to do here is to scale up or dilate the side lengths of the rectangle using a scale factor of 4

Hence, the new side lengths will be (4 * 6) inches and (10 * 4) inches

The new side lengths will be 24 inches and 40 inches

7 0

3 years ago

An auditorium with 52 rows is laid out where 6 seats comprise the first row, 9 seats comprise the second row, 12 seats comprise

ivanzaharov [21]

The number of seats per row generate an arithmetic sequence. Let $a_n$ denote the number of seats in the $n$ -th row. We're told that the number of seats increases by 3 per row, so we can describe the number of seats in a given row recursively by

$\begin{cases}a_1=6\\a_n=a_{n-1}+3&\text{for }2\le n\le52\end{cases}$

The total number of seats is given by the summation

$\displaystyle\sum_{n=1}^{52}a_n$

Because $a_n$ is arithmetic, we can easily find an explicit rule for the sequence.

$a_n=a_{n-1}+3$
$a_n=(a_{n-2}+3)+3=a_{n-2}+2\cdot3$
$a_n=(a_{n-3}+3)+2\cdot3=a_{n-3}+3\cdot3$
$\cdots$
$a_n=a_1+(n-1)\cdot3$

So the number of seats in the $n$ -th row is exactly

$a_n=6+(n-1)\cdot3=3+3n$

This means the total number of seats is

$\displaystyle\sum_{n=1}^{52}a_n=\sum_{n=1}^{52}(3+3n)=3\left(\sum_{n=1}^{52}1+\sum_{n=1}^{52}n\right)$

You should be familiar with the remaining sums. We end up with

$\displaystyle\sum_{n=1}^{52}a_n=3\left(52+\dfrac{52\cdot53}2\right)=4290$

3 0

3 years ago