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2 years ago

The measure of ∠1 = 55 62.5 110

Mathematics

2 answers:

elena-s [515]2 years ago

8 0

360 - 250 = 110
m<1 = 110/2 = 55

answer
55

sergiy2304 [10]2 years ago

7 0

both lines are 5 so both outsides are 125 degrees each = 250 degrees

a full circle = 360

so 360-250 = 110 degrees between the 2 lines

angle 1 would be half that so 110/2 = 55

angle 1 = 55 degrees

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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

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

1 year ago

What is the distance between (3, 5.25) and (3, –8.75)? 6 units 8.25 units 11.75 units 14 units

evablogger [386]

By using distance formula :

$\text{Distance formula,} \bold{ \boxed{ Distance = \sqrt{( x_{2} -x_{1})^{2}+(y_{2} - y_{1})^{2}) }}}$

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$\bold{Taking \: \: \: x_{1}=3 \: \: , \: \: x_{2}= 3 \: \: , \: \: y_{1}= 5.25 \: \: , \: \: y_{2}= -8.75}$

On applying formula, we get

$Distance = \sqrt{ ( x_{2}-x_{1})^{2}+(y_{2}-y_{1})^2} \\ \\ \\ Distance = \sqrt{ ( 3 - 3 )^{2} + ( - 8.75 - 5.25 )^{2}} \\ \\ \\ Distance = \sqrt{ ( 0 )^{2} + ( - 14)^{2}} \\ \\ \\ Distance = \sqrt{ ( - 14 )^{2}} \\ \\ \\ Distance = \sqrt{ 14^{2}} \:\:\:\:\:\:\:\:\:\:\: \: \: \: \: \: \: \: \: | \bold{ ( - 14 )^{2} = 14^{2}} \\ \\ \\ Distance = {14}^{2 \times \frac{1}{2} } \\ \\ \\ Distance = {14}^{1} \\ \\ \\ Distance = 14 \: units$

Hence, Option D is correct.

4 0

2 years ago

Read 2 more answers

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

1 year ago

Jean lives about 10 miles from the college where she plans to attend a 10-week summer class. There are two main routes she can t

jeka57 [31]

Answer:

We would advise her to choose Country Route because in this route time consistent between 15 and 18 minutes.

Step-by-step explanation:

We are given that Jean lives about 10 miles from the college where she plans to attend a 10-week summer class. There are two main routes she can take to the school, one through the city and one through the countryside.

Jean sets up a randomized experiment where each day she tosses a coin to decide which route to take that day. She records the following data for 5 days of travel on each route.

Country Route - 17, 15, 17, 16, 18 (in minutes)

City Route - 18, 13, 20, 10, 16 (in minutes)

Now, we have to decide which route is better for Jean to go to college.

As we can see from the data that the Country route timings is consistent between 15 to 18 minutes which means most of the times she will reach college between these minutes only.

While on the other hand, we can observe that City route timings are very much consistent as it has a low value of 10 minutes and high value of 20 minutes which means Jean can't be sure that at which time she will reach college.

Hence, we would advise her to choose Country route.

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2 years ago

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melomori [17]

Answer:

d.There is insufficient evidence to conclude that the quality and price of a car are associated. There were ten cars used in the sample.

Step-by-step explanation:

Hello!

You have two variables X₁: quality score of a car and X₂: the price of a car.

It was analyzed id there is an association between the quality and the price.

The null hypothesis of a Spearman's rank correlation test is:

H₀: There is no association between the quality and the price of cars.

The researcher failed to reject the null hypothesis which means that there is no association between the variables of interest.

The sample size is listed in the output n= 10 consumer reports.

I hope you have a SUPER day!

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