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2 years ago

7

Manuel is also in Mr. Whittaker’s science class. He observed that the water level rose 1.8 millimeters per year, which matches t

he average annual trend. This occurred for 2.2 years. Explain how to determine the total variation in water level.

2 answers:

Savatey [412]2 years ago

8 0

Answer:

To determinethe total variation in water level you would multiply 1.8 by 2.2 (3.96) and I can check my answer by rounding 2.2 to 2 and 1.8 to 2 and multiply that which equals 4 and 4 is close to 3.96

Step-by-step explanation:

Degger [83]2 years ago

3 0

Answer:

yoi could determine by multiyplying

Step-by-step explanation:

because if you do you'll find the easy way out on dividing then multiplying

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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

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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 $\left[\begin{array}{c}24&-8\end{array}\right]$

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1 year ago

What is the perimeter of the trapezoid with vertices Q(8, 8), R(14, 16), S(20, 16), and T(22, 8)? Round to the nearest hundredth

uysha [10]

Answer:

The perimeter of the trapezoid is $38.25\ units$

Step-by-step explanation:

we know that

The perimeter of the trapezoid is the sum of its four side lengths

so

In this problem

$P=QR+RS+ST+QT$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$Q(8, 8), R(14, 16), S(20, 16),T(22, 8)$

step 1

Find the distance QR

$Q(8, 8), R(14, 16)$

substitute the values in the formula

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$d=\sqrt{(8)^{2}+(6)^{2}}$

$d=\sqrt{100}$

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step 2

Find the distance RS

$R(14, 16), S(20, 16)$

substitute the values in the formula

$d=\sqrt{(16-16)^{2}+(20-14)^{2}}$

$d=\sqrt{(0)^{2}+(6)^{2}}$

$d=\sqrt{36}$

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step 3

Find the distance ST

$S(20, 16),T(22, 8)$

substitute the values in the formula

$d=\sqrt{(8-16)^{2}+(22-20)^{2}}$

$d=\sqrt{(-8)^{2}+(2)^{2}}$

$d=\sqrt{68}$

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step 4

Find the distance QT

$Q(8, 8),T(22, 8)$

substitute the values in the formula

$d=\sqrt{(8-8)^{2}+(22-8)^{2}}$

$d=\sqrt{(0)^{2}+(14)^{2}}$

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step 5

Find the perimeter

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GaryK [48]

Answer:

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(1p, 20p), (50p, 2p), (50p, 20p), (2p, 20p)

Explanation:

The possible combinations are:

1. Assuming the first coin is 10p:

(10p, 1p)
(10p, 50p)
(10p, 2p)
(10p, 20p)

2. Asuming the first coin is 1p

Do not count (1p, 10p) as it is the same combination as (10p, 1p)

(1p, 50p)
(1p, 2p)
(1p, 20p)

3. Assuming the first coin is 50p:

Do not count (50p, 10p) nor (50p, 1p) as they are the same combinations (10p, 50p) and (1p, 50p) counted earlier:

(50p, 2p)
(50p, 20p)

4. Assuming the first coin is 2p:

The only new combination is:

(2p, 20p)

5. All the combinations with 20p have already been listed.

Therefore:

There are 4 + 3 + 2 + 1 = 10 different combinations

The list of different combinations is:

(10p, 1p), (10p, 50p), (10p, 2p), (10p, 20p), (1p, 50p), (1p, 2p),

(1p, 20p), (50p, 2p), (50p, 20p), (2p, 20p)

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