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3 years ago

Choose the proper tools with the correct endpoints then click and drag the endpoints to the correct location.

Mathematics

1 answer:

qwelly [4]3 years ago

6 0

X < = 1 or x > 8

closed circle on 1, shaded to the left
open circle on 8, shaded to the right

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Please simplify y(-16+2x)=

Liula [17]

The answer is -16y + 2yx

4 0

2 years ago

HELP PLEASE!

wel

Answers:

$\text{Perimeter} = 14\sqrt{5}\\\\\text{Area} = 50\\\\$

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

Work Shown:

$L = 2\sqrt{5} = \text{length}$

$W = 5\sqrt{5} = \text{width}$

P = perimeter

$P = 2*(L+W)\\\\P = 2*(2\sqrt{5}+5\sqrt{5})\\\\P = 2*(7\sqrt{5})\\\\P = 14\sqrt{5}\\\\$

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

A = area

$A = L*W\\\\A = (2\sqrt{5})*(5\sqrt{5})\\\\A = (2*5)(\sqrt{5}*\sqrt{5})\\\\A = 10\sqrt{5*5}\\\\A = 10\sqrt{25}\\\\A = 10*5\\\\A = 50\\\\$

3 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

2 years ago

Escribevel numero decimal que corresponde alos puntos intermedios que indican las flechas

NeTakaya

Answer:

cualeeeeeeeeeeeeeeeeeeeeees fffffleeeeeccccchhhhaaaasssss

8 0

2 years ago

Autumn cut out a triangular piece of fabric to sew on her comforter. The area of the triangle is 48 square inches, and the heigh

Step2247 [10]

$A=\dfrac{1}{2}ah\\
2A=ah\\
a=\dfrac{2A}{h}\\\\
a=\dfrac{2\cdot48}{12}\\
a=4 \Rightarrow D$

8 0

2 years ago