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3 years ago

If michelle walked 5 meters, how many inches did she walk

Mathematics

2 answers:

DerKrebs [107]3 years ago

4 0

1 meter = 39.370 inches.....so 5 meters = (5)(39.370) = 196.85 inches

nikitadnepr [17]3 years ago

4 0

196.85 inches Michelle walked

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Brainliest, 35 points

Hitman42 [59]

Answer:

13 1/2

Step-by-step explanation:

9 ÷ 2/3

Copy dot flip

9 * 3/2

27/2

Change from an improper fraction to a mixed number

2 goes into 27 13 times with 1 left over

13 1/2

5 0

3 years ago

The redwood national park is home to some of the largest trees in the world. Hyperion is the tallest tree in the park, with a he

vladimir1956 [14]

Answer:

$h= 126.33\ yd$

Step-by-step explanation:

Let h be the height of the tree.

Given:

Height of the tree = 379 feet

We need to find the height of the tree in yards.

Solution:

From the given statement, Hyperion is the tallest tree in the park, with a height of approximately 379 feet,

We need to convert the height of the tree from feet to yard. So, We divide the height of the tree by three for yard.

For one feet

$h = \frac{1}{3}\ yd$

For 379 feet

$h = \frac{379}{3}\ yd$

$h= 126.33\ yd$

Therefore , height of the tree $h= 126.33\ yd$

3 0

3 years ago

Solve for x. a = bx A) (-b)\a B) b - a C) (-a) + b D) (-a)/b

Andreyy89

I doubt the options you've mentioned here.
The correct answer should be x=a/b

3 0

3 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

What is the absolute value of -8 and 8

chubhunter [2.5K]

The absolute value is 8 because the absolute value is opposite of -8.

5 0

3 years ago

Read 2 more answers