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3 years ago

What is the 70th term of the sequence that begins -3,1,5,9?

Mathematics

1 answer:

VashaNatasha [74]3 years ago

3 0

The 70th term is 277
Hope this helps

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Ricardo was mixing blue paint and yellow paint in a ratio of 3:5 to make green paint. He wants to make 40 liters of green paint.

Ivanshal [37]

Answer:

15L:40L

Step-by-step explanation:

6 0

3 years ago

Find the perimeter of the figure to the nearest hundredth. Sorry to ask ANOTHER question on this stuff but I'm doing an assignme

WARRIOR [948]

Answer:

I did my best, but it has been a while. I really hope that this helps

Step-by-step explanation:

so you can find the perimeter of a circle

-----The perimeter of a circle is π × d where d is the diameter so half of that is 1/2 π × d. But If you are going to walk around a semicircle, you need to go halfway around the circle and then across a diagonal to get back to where you started, so the perimeter is 1/2 π × d + d.

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2 years ago

Translate: "The product of 8 and a number, b, is taken away from 24"

goblinko [34]

24-8b
24-(8*b)
i don’t know the exact way it is written on your test but it is something along those lines

7 0

2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

3 years ago

Solve the equasion<br><br>2y-10y

Tamiku [17]

It’s -8y hope this help

3 0

3 years ago