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kherson [118]
4 years ago
12

PLEASE HELP ME LOL

Mathematics
1 answer:
FrozenT [24]4 years ago
6 0
If the gardener want the length to be 14 feet longer than it is wide, then L=w+14.  We are given an area of 112.  The area formula for a rectangle is A=L*W.  Filling in accordingly, 112 = (w+14)(w).  Distributing we have 112=w^2+14w.  We are looking for the dimensions of the garden, so we have to solve for w.  To do that we have to factor. Set it equal to 0 to get the quadratic w^2+14w-112=0.  Plug that into the quadratic formula to get w values of 5.69 and -19.69.  The 2 things in math that will never EVER be negative are time and distance/length so we will not use the negative value.  Therefore, our width is 5.69.  Our length of w+14 is then 5.69+14 is 19.69.  There's our dimensions:  width is 5.69 and length is 19.69
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Reduced form of 8/10.
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Answer:

4/5

Step-by-step explanation:

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Remember, SF= new/old. On shape 2, the side with the length 3 corresponds to the length of 9 on shape 1. (SF=3/9)=0.3333333....
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Assume that T is a linear transformation. Find the standard matrix of T. T: R^3 right arrow R^2 , T(e 1) =(1,2), and T(e2 ) =( -
irina [24]

Answer:

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

Step-by-step explanation:

Given

T:R^3->R^2

T(e_1) = (1,2)

T(e_2) = (-4,6)

T(e_3) = (2,-6)

Required

Find the standard matrix

The standard matrix (A) is given by

Ax = T(x)

Where

T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]

Ax = T(x) becomes

Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]

The x on both sides cancel out; and, we're left with:

A = [T(e_1)\ T(e_2)\ T(e_3)]

Recall that:

T(e_1) = (1,2)

T(e_2) = (-4,6)

T(e_3) = (2,-6)

In matrix:

(a,b) is represented as: \left[\begin{array}{c}a\\b\end{array}\right]

So:

T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]

T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]

T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]

Substitute the above expressions in A = [T(e_1)\ T(e_2)\ T(e_3)]

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

Hence, the standard of the matrix A is:

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

6 0
3 years ago
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Answer:

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Lorico [155]

Answer:

9.38cm

Step-by-step explanation:

you have a right triangle. use pathagorean

x^2 + 9^2 = 13^2

x^2 + 81 = 169

x = √88

x = 9.38

4 0
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