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4 years ago

What is the perimeter of this rectangle?

Mathematics

1 answer:

BlackZzzverrR [31]4 years ago

7 0

The length of the rectangle would be X2 - X1
The height of the rectangle would be Y2-Y1

Perimeter = 2 X length + 2 times height

The perimeter would be 2(X2-X1) + 2(Y2-Y1)

The first choice would be the correct answer.

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How many solutions does the system of equations have?<br> y=-2x+3<br> y = x² - 6x +3

Mashutka [201]

Answer:

2 solutions

Step-by-step explanation:

y = - 2x + 3 → (1)

y = x² - 6x + 3 → (2)

substitute y = x² - 6x + 3 into (1)

x² - 6x + 3 = - 2x + 3 ( subtract - 2x + 3 from both sides )

x² - 4x = 0 ← factor out x from each term on the left side

x(x - 4) = 0

equate each factor to zero and solve for x

x = 0

x - 4 = 0 ⇒ x = 4

substitute these values into (1)

x = 0 : y = - 2(0) + 3 = 0 + 3 = 3 ⇒ (0, 3 )

x = 4 : y = - 2(4) + 3 = - 8 + 3 = - 5 ⇒ (4, - 5 )

the 2 solutions are (0, 3 ) and (4, - 5 )

6 0

2 years ago

Given that tan theta = square root of 3 over 4, and 0 < theta < pi over 2, what is the exact value of cot theta?

Natali5045456 [20]

Answer:

$\frac{4\sqrt{3} }{3}$

Step-by-step explanation:

4sqrt3/3

5 0

3 years ago

The stopping distance of a car is modeled by the function d = 0.05r(r + 2) where d is the stopping distance of the car measured

adell [148]

Answer:

The car was moving approximately at a speed of 37.74 miles per hour.

Step-by-step explanation:

We are given the following in the question:

The stopping distance of a car is given by

$d = 0.05r(r + 2)$

where d is the stopping distance in feet and r is speed of the car in miles per hour.

The stopping distance is 75 feet, we have to find the speed of the car,

Putting d = 75 in the equation, we get,

$75 = 0.05r(r + 2)\\r^2 + 2r = 1500\\r^2 + 2r-1500 = 0\\\\r =\dfrac{-2\pm \sqrt{4-(4)(1)(-1500)}}{2}\\\\r\approx -39.74,37.74$

Thus, the car was moving approximately at a speed of 37.74 miles per hour.

3 0

3 years ago

3) Solve each equation using the quadratic formula. Show<br> a. x2 – 3x – 10 = 0

Naya [18.7K]

Answer:

Step-by-step explanation:

The quadratic formula of a quadratic equation

$ax^2+bx+c=0\\\\\text{If}\ b^2-4ac>0\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\text{If}\ b^2-4ac=0\ \text{then the equation has one solution}\ x=\dfrac{-b}{2a}$

$\text{If}\ b^2-4ac$

We have:

$x^2-3x-10=0\to a=1,\ b=-3,\ c=-10\\\\b^2-4ac=(-3)^2-4(1)(-10)=9+40=49>0\\\\x=\dfrac{-(-3)\pm\sqrt{49}}{2(1)}=\dfrac{3\pm7}{2}\\\\x=\dfrac{3-7}{2}=\dfrac{-4}{2}=-2\\or\\x=\dfrac{3+7}{2}=\dfrac{10}{2}=5$

5 0

3 years ago

How many time does 8 go into 27

iogann1982 [59]

It is: 27/8 = 3 with a remainder of 3

4 0

4 years ago

Read 2 more answers