All categories

2 years ago

A sequence is defined by the following: an = 4n – 1 What is a1?

Mathematics

2 answers:

Alexandra [31]2 years ago

6 0

an = 4n – 1

a1 means n=1

a1 = 4(1) -1

a1 = 4-1

a1 =3

Answer: 3

grigory [225]2 years ago

4 0

Answer:

Explain Your Answer:

You might be interested in

Ethan lives 2 5/12 miles from school. Marquis lives 7/12 mile closer to school than ethan. How far does marquis live from school

goblinko [34]

Answer:

1 5/6 miles

Step-by-step explanation:

first, we take the amout ethan lives from school, and turn it into a mixed number

2 5/12 = 29/12

marquis lived 7/12 mile closer

so, we have to subtract that from how far away ethan lives

29/12 - 7/12 = 22/12

22/12 simplified is 1 10/12 or 1 5/6

7 0

2 years ago

You saved 300 to spend over summer. You decide to budget 50 to spend each week. Write and equation that represents the situation

Anestetic [448]

Y=300-50x
Y= total $
X= number of weeks

5 0

2 years ago

Read 2 more answers

Which vector matrix represents the reflection of the vector <-1,5> across the line x = y

Arturiano [62]

Answer:

$V = \left[\begin{array}{ccc}5&-1\end{array}\right]$

Step-by-step explanation:

We want to reflect this 2x1 vector on the line y = x.

To make this reflection we must use the following matrix:

$R=\left[\begin{array}{cc}0&1\\1&0\\\end{array}\right]$

Where R is known as the reflection matrix on the line x = y

Now perform the product of the vector <-1,5> x R.

$\left[\begin{array}{ccc}-1\\5\end{array}\right]x\left[\begin{array}{ccc}0&1\\1&0\end{array}\right]\\\\\\\left[\begin{array}{ccc}-1(0) +5(1)&-1(1)+5(0)\end{array}\right]\\\\\\\left[\begin{array}{ccc}5&-1\end{array}\right]$

The vector matrix that represents the reflection of the vector <-1,5> across the line x = y is:

$V = \left[\begin{array}{ccc}5&-1\end{array}\right]$

6 0

3 years ago

The options are (will/will not) and (above/below)

Svetradugi [14.3K]

Answer:

Step-by-step explanation:

The answer are will not and below

4 0

2 years ago

Read 2 more answers

4a/3-b/4=6<br> 5a/6+b=13<br> Solve linear system by elimination

kondor19780726 [428]

$\bf \cfrac{4a}{3}-\cfrac{b}{4}=6
\qquad \qquad 
\cfrac{5a}{6}+b=13
\\\\\\
\textit{let us remove the denominators off those folks}\\
\textit{by multiplying the first one by 12, both sides}\\
\textit{and the second one by 6, both sides, thus}
\\\\\\
12\left( \cfrac{4a}{3}-\cfrac{b}{4} \right)=12(6)\implies 16a-4b=72
\\\\\\
6\left( \cfrac{5a}{6}+b \right)=6(13)\implies 5a+6b=78$

$\bf \textit{now, let's do the elimination}
\\\\

\begin{array}{llll}
16a-4b=72&\boxed{\times 3}\implies &48a-\underline{12b}=216\\\\
5a+6b=78&\boxed{\times 2}\implies &10a+\underline{12b}=156\\
&&--------\\
&&58a+0\quad=372
\end{array}$

solve for "a", once you get "a", plug it back into either equation to get "b"

3 0

2 years ago