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3 years ago

Can someone help me What one

Mathematics

1 answer:

Sav [38]3 years ago

5 0

Answer:

try the second one sorry if its wrong report me if it is

Step-by-step explanation:

hope this help

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The midpoint of segment AB is (4, 2). The coordinates of point A are (2, 7). Find the coordinates of point B.

erastovalidia [21]

Answer:

<h3>The option B) is correct</h3><h3>Therefore the coordinate of B $(x_2,y_2)$ is (6,-3)</h3>

Step-by-step explanation:

Given that the midpoint of segment AB is (4, 2). The coordinates of point A is (2, 7).

<h3>To Find the coordinates of point B:</h3>

Let the coordinate of A be $(x_1,y_1)$ is (2,7) respectively
Let the coordinate of B be $(x_2,y_2)$
And Let M(x,y) be the mid point of line segment AB is (4,2) respectively
The mid-point formula is

Now substitute the coordinates int he above formula we get
$(4,2)=(\frac{2+x_2}{2},\frac{7+y_2}{2})$
Now equating we get

$4=\frac{2+x_2}{2}$ $2=\frac{7+y_2}{2}$

Multiply by 2 we get Multiply by 2 we get

$4(2)=2+x_2$ $2(2)=7+y_2$

$8=2+x_2$ $4=7+y_2$

Subtracting 2 on both

the sides Subtracting 7 on both the sides

$8-2=2+x_2-2$ $4-7=7+y_2-7$

$6=x_2$ $-3=y_2$

Rewritting the above equation Rewritting the equation

$x_2=6$ $y_2=-3$

<h3>Therefore the coordinate of B $(x_2,y_2)$ is (6,-3)</h3><h3>Therefore the option B) is correct.</h3>

7 0

3 years ago

ihelp me unerstand > i convert 8/25 into decimal >32/100 - and because 100 had two zeros. I place the decimal over 2 place

agasfer [191]

8/25....." / " simply means division

so 8 divided by 25 = 0.32

7 0

3 years ago

Explain or show how you know that 600:450, 60:45, and 4:3 are all equivalent.

saw5 [17]

600:450 divided by 10 = 60:45
60:45 divided by 15 = 4:3
so 4:3 x 15 = 60:45 x 10 = 600:450

5 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

How do you rewrite 1 2/3 using twelfths

Setler79 [48]

The answer is 1 8/12

6 0

3 years ago