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3 years ago

X = y - 3

1 answer:

7 0

If you would like to solve the system of equations, you can do this using the following steps:

x = y - 3
x + 3y = 13 ... y - 3 + 3y = 13
___________
<span>y - 3 + 3y = 13
</span>4y = 13 + 3
4y = 16
y = 4

<span>x = y - 3 = 4 - 3 = 1

(x, y) = (1, 4)
</span>
The correct result would be A) (1, 4).

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How to solve 7*8=_*_+_*_

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3 years ago

Mr Smith's art class took a bus trip to an art museum. The bus averaged 65 miles per hour on the highway and 25 miles per hour i

Leya [2.2K]

Let x be the distance traveled on the highway and y the distance traveled in the city, so:
$\left \{ {{x+y=375} \atop { \frac{1}{65}x+ \frac{1}{25}y =7}} \right.$

Now, the system of equations in matrix form will be:
$\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right] \left[\begin{array}{ccc}x&\\y&\end{array}\right] = \left[\begin{array}{ccc}375&\\7&\end{array}\right]$

Next, we are going to find the determinant:
$D= \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}$
Next, we are going to find the determinant of x:
$D_{x} = \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8$

Now, we can find x:
$x= \frac{ D_{x} }{D} = \frac{8}{ \frac{8}{325} } =325mi$

Now that we know the value of x, we can find y:
$y=375-325=50mi$

Remember that time equals distance over velocity; therefore, the time on the highway will be:
$t_{h} = \frac{325}{65} =5hours$
An the time on the city will be:
$t_{c} = \frac{50}{25} =2hours$

We can conclude that the bus was five hours on the highway and two hours in the city.

8 0

3 years ago

What is the answer to 2-2×3+3

andreyandreev [35.5K]

2 - 2 x 3 + 3
= 2 - 6 + 3
= -4 + 3
= -1

7 0

3 years ago

Explain why the function is discontinuous at the given number

zaharov [31]

Calculator can help you

6 0

4 years ago

Quadrilateral PEST has vertices (-1, -5), (8, 2), (11, 13), and (2, 6), respectively. Classify the quadrilateral as a square, rh

Alchen [17]

Answer:

The figure PEST is a rhombus

Step-by-step explanation:

* Lets talk about the difference between all these shapes

- At first to prove the shape is a parallelogram you must have one

of these conditions

# Each two opposite sides are parallel OR

# Each two opposite sides are equal in length OR

# Its two diagonals bisect each other

- After that to prove the parallelogram is:

* A rectangle you must have one of these conditions

# Two adjacent sides are perpendicular to each other OR

# Its two diagonals are equal in length

* A rhombus you must have one of these conditions

# Two adjacent sides are equal in length OR

# Its two diagonals perpendicular to each other OR

# Its diagonals bisect its vertices angles

* A square you must have two of these conditions

# Its diagonals are equal and perpendicular OR

# Two adjacent sides are equal and perpendicular

* Now lets solve the problem

∵ The vertices of the quadrilateral PEST are

P (-1 , -5) , E (8 , 2) , S (11 , 13) , T (2 , 6)

- Lets find the slope from each two points using this rule :

m = (y2 - y1)/(x2 - x1), where m is the slope and (x1 , y1) , (x2 , y2)

are two points on the line

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ m of PE = (2 - -5)/(8 - -1) = 7/9

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ m of ES = (13 - 2)/(11 - 8) = 11/3

- Let (x1 , y1) is (11 , 13) and (x2 , y2) is (2 , 6)

∴ m of ST = (6 - 13)/(2 - 11) = -7/-9 = 7/9

- Let (x1 , y1) is (2 , 6) and (x2 , y2) is (-1 , -5)

∴ m of TP = (-5 - 6)/(-1 - 2) = -11/-3 = 11/3

∵ m PE = m ST = 7/9

∴ PE // ST ⇒ opposite sides

∵ m ES = m TP = 11/3

∴ ES // TP ⇒ opposite sides

- Each two opposite sides are parallel

∴ PEST is a parallelogram

- Lets check if the parallelogram can be rectangle or rhombus or

square by one of the condition above

∵ If two line perpendicular , then the product of their slops = -1

- Lets check the slopes of two adjacent sides (PE an ES)

∵ m PE = 7/9

∵ m ES = 11/3

∵ m PE × m ES = 7/9 × 11/3 = 77/27 ≠ -1

∴ PE and ES are not perpendicular

∴ PEST not a rectangle or a square (the sides of the rectangle and

the square are perpendicular to each other)

- Now lets check the length of two adjacent side by using the rule

of distance between two points (x1 , y1) and (x2 , y2)

d = √[(x2 - x1)² + (y2 - y1)²]

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ PE = √[(8 - -1)² + (2 - -5)²] = √[9² + 7²] = √[81 + 49] = √130 units

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ ES = √[(11 - 8)² + (13 - 2)²] = √[3² + 11²] = √[9 + 121] = √130 units

∴ PE = ES ⇒ two adjacent sides in parallelogram

∴ The four sides are equal

* The figure PEST is a rhombus

4 0

3 years ago