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3 years ago

9

What is the area of a triangle.

2 answers:

Firdavs [7]3 years ago

8 0

Answer:

The area of a triangle is the base times height times 1/2. Please mark brainliest!!!!

ololo11 [35]3 years ago

6 0

Answer:

1/2 x b x h

Step-by-step explanation:

The area of a triangle is

A half x base x height

= 1/2 x b x h

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Find the minimum value of the function f(x) = x2 + 5x – 6

Klio2033 [76]

So the easiest method to find the vertex (the minimum in this case) to do this is to find the axis of symmetry, then plug it into the function.

Firstly, the equation to find the axis of symmetry is $x=\frac{-b}{2a}$ , with b = x coefficient and a = x^2 coefficient. The equation equation can be solved as such:

$x=\frac{-5}{2*1}\\\\x=\frac{-5}{2}\\\\x=-2.5$

Since the vertex falls on the axis of symmetry, we know that the x-coordinate of the vertex is -2.5. Now to solve for the y-coordinate, plug in x with -2.5 and solve as such:

$f(-2.5)=(-2.5)^2+5*(-2.5)-6\\ f(-2.5)=6.25+(-12.5)-6\\ f(-2.5)=-12.25$

Now putting it all together, our minimum value (vertex) is (-2.5,-12.25).

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3 years ago

The coordinates of the verticals of a rectangle are (-3,4), (7,2), (6,-3), & (-4,-1). What is the perimeter of the rectangle

natali 33 [55]

Answer:

30.594

Step-by-step explanation:

To find perimeter we need to find the length of all the sides

The coordinates are A(-3,4), B(7,2), C(6,-3), D (-4,-1)

To find the length of side , we use distance formula

$Distance = \sqrt{(x2-x1)^2+(y2-y1)^2}$

Side $AB =\sqrt{(7--3)^2+(2-4)^2} = \sqrt{100+4}= \sqrt{104}=10.198$

Side $BC =\sqrt{(6-7)^2+(-3-2)^2} = \sqrt{1+25}= \sqrt{26}=5.099$

Side $CD =\sqrt{(-4-6)^2+(-1--3)^2} = \sqrt{100+4}= \sqrt{104}=10.198$

Side $DA =\sqrt{(-3--4)^2+(4--1)^2} = \sqrt{1+25}= \sqrt{26}=5.099$

To find perimeter we add all the sides

10.198 + 5.099 + 10.198 + 5.099= 30.594

the perimeter of the rectangle= 30.594

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