Which set or sub set does \large \frac{24}{8} belong?

Find the Surface Area of the right triangular prism.

Elis [28]

It’s always gonna be 60 Units this is just an known fact

8 0

2 years ago

A line passes through the point (-6,7) and has a slope of -2 write an equation in point slope form for this line

Anuta_ua [19.1K]

Answer:

Step-by-step explanation:

(-6,7) & m = -2

point slope form of line : y - y₁ = m (x-x₁)

y - 7 = (-2) (x - [-6] )

y - 7 = (-2) (x + 6)

y - 7 = -2x - 2 * 6

y - 7 = -2x - 12

2x + y = -12 +7

2x + y = -5 0r y = -2x - 5

3 0

3 years ago

Round to the nearest whole number:<br> a) 19.3 b) 127.8

enot [183]

Answer:

19 and 128

Step-by-step explanation:

19.3 is closer to 19 than 20 and 127.8 is closer to 128 than 127

7 0

2 years ago

Read 2 more answers

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

Josie spent 4/7 of ten money in her purse on some books, and the rest of the money on 18 markers. Each marker cost $1.20. How mu

NeTakaya

Let's assume money in Josie's purse to begin with as 'x'

There are 18 markers

Each marker cost $1.20

so, we can find total cost of markers

Total cost of markers = ( total number of markers)*(price of each marker)

so, we can plug value

Total cost of markers is

$=18\times 1.20$

$=21.6$ $

now, we are given

Josie spent 4/7 of the money in her purse on some books

the rest of the money on 18 markers

so, rest money is

$=x-\frac{4}{7} x$

$=\frac{3}{7} x$

and this rest money is total cost of markers

so, we can set them equal and then we can solve for x

$\frac{3}{7} x=21.6$

$7\cdot \frac{3}{7}x=21.6\cdot \:7$

$3x=151.2$

$x=50.4$

So, total money in Josie's purse in beginning is $50.4..........Answer

8 0

3 years ago