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3 years ago

Solve the equation. a² = 100 Enter the correct answer in the boxes. a = 10 or a = ?

1 answer:

7 0

A = 10 or a = -10

When you square a number, you're essentially multiplying it by itself.

Recall that multiplying two negative numbers will result in a positive numbers. Knowing this, we know that -10 times -10 equals 100.

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Represent each of these relations on {1, 2, 3, 4} with a matrix (with the elements of this set listed in increasing order). a) {

Irina-Kira [14]

Answer:

for a a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

$\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]$

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

$\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]$

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

$\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]$

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

$\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]$

Step-by-step explanation:

in matrix, arrays are placed in rows , which represents the horizontal sides from left to right, while arrays in the column are placed vertically from top to bottom. Here, we placed the arrays in a 4x4 matrix

for a a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

$\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]$

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

$\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]$

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

$\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]$

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

$\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]$

7 0

3 years ago

A rectangular park of length 60 m breadth 50 m encloses w volleyball court of length 18 m and breadth 10 m. Find the area of the

Zanzabum

Given

A rectangular park of length 60 m breadth 50 m encloses with volleyball court of length 18 m and breadth 10 m.

To find:

The area of the park excluding the court at the rate of Rs 110 per square meter.

Solution:

Area of a rectangle is:

$Area=length \times breadth$

Area of whole park is:

$A_1=60 \times 50$

$A_1=3000$

Area of volleyball court is:

$A_2=18 \times 10$

$A_2=180$

Now, the area of the park excluding the court is:

$A=A_1-A_2$

$A=3000-180$

$A=2820$

Therefore, the area of the park excluding the court is 2820 square meter.

4 0

3 years ago

At high school basketball game, the athletic department sold 250 tickets and brought in 837.50. Adult tickets cost $5 and studen

Tpy6a [65]

Answer:

The system of equations is:

$a+s = 250\\5a+2.50s = 837.50$

Here a is the number of adult tickets and s is the number of student tickets

Step-by-step explanation:

We will use the given statements to make a system of equations for the given scenario.

First of all we have to decide which variables to be used.

Let a be the number of adult tickets and

Let s be the number of student tickets

So according to the statement that the department sold 250 tickets in total, the equation will be:

$a+s = 250$

This is equation one of the system of equations

According to the statement that cost of adult ticket is $5, total cost of adult tickets will be: $5a$

According to the statement that cost of student ticket is $2.50, total cost of student tickets will be: $2.50s$

The total earning was $837.50 so the second equation will be:

$5a+2.50s = 837.50$

Hence,

The system of equations is:

$a+s = 250\\5a+2.50s = 837.50$

Here a is the number of adult tickets and s is the number of student tickets

7 0

3 years ago

Here is your questions

viva [34]

Answer:

Step-by-step explanation:

strategy: find the distance of the two points, take half of that, b/c that will be the radius, Then use the area of a circle formula to find the answer

P1 = (1,-1) in the form (x1,y1)

P2= ((3,3) in the form (x2,y2)

dist = sqrt [ (x2-x1)^2 + (y2-y1)^2 ]

dist = sqrt [ 3 -1 )^2 + 3-(-1)^2 ]

dist = sqrt [ 2^2 + 4^2 ]

dist = sqrt [ 4 + 16 ]

dist = sqrt [20]

r = $\sqrt{20}$ /2

area of circle = $\pi$ $r^{2}$

area of circle = $\pi$ ( $\sqrt{20}$ /2)^2

area of circle = $\pi$ ( $\sqrt{20}$ /2)( $\sqrt{20}$ /2)

area of circle = $\pi$ (20 / 4)

area of circle = $\pi$ 5

area of circle = 5 $\pi$

3 0

3 years ago

Can anyone please help me with the last question I’m struggling!

iren2701 [21]

The equation is x=-6

Step-by-step explanation:

In standard form the equation is y=4

And this is a horizontal line so the perpendicular line will be a vertical line of the form x=a

so that is why the equation is x=-6

5 0

3 years ago

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