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2 years ago

Find the are of the trapezoid

Mathematics

2 answers:

snow_lady [41]2 years ago

6 0

Answer:

Step-by-step explanation:

52√3 ft²

butalik [34]2 years ago

6 0

Answer:

120 ft.

Step-by-step explanation:

I did it by adding 15 and 15, then dividing that by 2, and finally multiplying that by the height, which is 8, to get 120.

Hope this helps :)

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A pie can be cut into more than seven pieces by making four diameter cuts.

n200080 [17]

I'm pretty sure it's true

7 0

3 years ago

Read 2 more answers

If AB=BC,AB=4=4x-2,andBC=3x+3,fine the length of AB

laila [671]

Answer:

Step-by-step explanation:

First, I'm assuming AB=4=4x-2 was a typo and it's supposed to be AB = 4x - 2

AB=BC

AB = 4x - 2 BC = 3x + 3

4x - 2 = 3x + 3

Solve for x Add 2 to each side

4x - 2 = 3x + 3

4x - 2 + 2 = 3x + 3 + 2

4x = 3x + 5 Subtract 3x from each side.

4x - 3x = 3x- 3x + 5

4x - 3x = 5

x = 5

Now plug back in to the original equations

AB = 4x - 2 BC = 3x + 3

AB = 4 (5) - 2 BC = 3(5) + 3

AB = 20 - 2 BC = 15 + 3

AB = 18 BC = 18

So AB is 18

3 0

3 years ago

The perimeter of an equilateral triangle is 13 inches more than the perimeter of a square, and the side of the triangle is 6 in

ss7ja [257]

Answer:

The side of the triangle is 11 inches

Step-by-step explanation:

please kindly see the attached files for explanation

8 0

3 years ago

Pls pls pls help

Natasha_Volkova [10]

Answer:

Parallelograms I, II, and IV

Step-by-step:

Area of parallelograms:

I. A=3*5=15 units squared

II. A=5*3=15 units squared

III. A=4*4=16 units squared

IV. A=5*3=15 units squared

So, parallelograms I, II, and IV have the same area of 15 units squared.

5 0

3 years ago

Graph y = x2 + 2. Identify the vertex of the graph. Tell whether it is a minimum or maximum. (0, 2); maximum (0, 2); minimum (2,

miskamm [114]

Answer:

$(0,2)$ ; minimum

Step-by-step explanation:

Given:

The function is, $y=x^{2}+2$

The given function represent a parabola and can be expressed in vertex form as:

$y=(x-0)^{2}+2$

The vertex form of a parabola is $y=(x-h)^{2}+k$ , where, $(h,k)$ is the vertex.

So, the vertex is $(0,2)$ .

In order to graph the given parabola, we find some points on it.

Let $x=-2,y=(-2)^{2}+2=4+2=6$

$x=-1,y=(-1)^{2}+2=1+2=3$

$x=0,y=(0)^{2}+2=0+2=2$

$x=2,y=(2)^{2}+2=4+2=6$

$x=1,y=(1)^{2}+2=1+2=3$

So, the points are $(-2,6),(-1,3),(0,2),(1,3),(2,6)$ .

Mark these points on the graph and join them using a smooth curve.

The graph is shown below.

From the graph, we conclude that at the vertex $(0,2)$ , it is minimum.

8 0

3 years ago