Problem is in the attached image. Thanks in advance!

Mathematics

1 answer:

lutik1710 [3]3 years ago

3 0

Answer: it will take 2.5 hours for both people to meet after person B leaves.

Step-by-step explanation:

The distance between the house of person A and his cousin's house is 234 miles.

At the point where they meet, they would have travelled 234 miles.

Person A travelled at an average speed of 44 miles per hour.

Let t represent the time spent by Person A in travelling from his house to the point where they meet.

Distance = speed × time

Distance covered by person A would be

44 × t = 44t

One and half hours later, person B leaves his house to visit person A, travelling at an average speed of 62 miles per hour. This means that total time spent by person B in travelling to the point where they meet is (t - 0.5) hours. Distance covered would be

62(t - 0.5) = 62t - 31

Since the total distance covered is 234 miles, then

44t + 62t - 31 = 234

106t = 234 + 31 = 265

t = 265/106 = 2.5 hours

You might be interested in

In the figure below, lines 1 and 2 are parallel. Lines 3 and 4 are also parallel.

attashe74 [19]

I don’t see a diagram, but I assume that you must use these things.

Opposite interior angles. (Which are equal)
Same side interior angles (which are supplementary or they add up to 180)
Corresponding angles (which are equal)
Vertical angles (which are equal)

5 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

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