What is the distance between (5, -6) and (2, -6) on a coordinate plane?

Mathematics

1 answer:

Ann [662]3 years ago

7 0

Answer:

3 units is the answer..

hope it helps

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You give 1/2 up a $46,000 salary to go back to school for 5 years. The total cost for going to school is $22,000. How much money

Ket [755]

The individual needs to earn $74,750 salary per year in order to recover the investment in 4 years.

What salary amount makes the individual as they were without a salary cut?

By the time the individual resumes back for work after 5 years, they need to earn their previous salary of $46,000 and recover the lost salaries, which is half of $46,000 lost in the last 5 years.

Total salaries lost=1/2*$46,000*5years

Total salaries lost=$115,000

Because the salaries lost needs to be recovered over 4 years, the extra earnings over and above the $46,000 in 4 years is the total salaries lost divided by 4 years

extra salary per year=$115,000/4

extra salary per year=$28,750

Total salary now=prior salary+ extra salary per year

Total salary now=$46,000+$28,750

Total salary now=$74,750

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2 years ago

A soda company had sales of $35,107 million in 2010 and $45,992 million in 2014. Use the Midpoint Formula to estimate the sales

Dima020 [189]

Answer:

40,550

Step-by-step explanation:

40, 550 is the mod point

8 0

3 years ago

If T: (x, y) → (x + 6, y + 4), then T-1: (x,y) → _____.

alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

$\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]$

The inverse of the matrix is

$\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]$

So, the Inverse Transformation is given by

$T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$