Please help!! Needed ASAP!!

Mathematics

1 answer:

yarga [219]2 years ago

3 0

Answer and Explanation:

If, on the other hand, the graph "flexes" or "flattens out" to some degree when it goes to cross the axis, then the zero is of a higher multiplicity; that is, it'll be of multiplicity three, five, or higher.

When a graph doesn't start at 0, it means that the graphs before it is not shown.

For number 4, I'd suggest Graph 3.

For number 3, I'd suggest graph 1.

For number 5, Write something along the lines of "Graph 2 does not show the car stopping slowly, it stops suddenly. Then it starts going really fast and stops suddenly again."

I hoped this helped.

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WILL MARK U BRAINLIEST

Lilit [14]

Answer:

Step-by-step explanation:

First, we should find the area of the trapezoid, and then subtract the area of the removed triangle in order to find the shaded area.

Area of the trapezoid

1) Area of the rectangle in the middle.

Base Length: 10

Height Length: 10

Area: 10 x 10 = 100

2. Area of the triangles on the side

Base Length: (14 - 10)/2 = 2

Height Length: 10

Area: 2 x 10 x 1/2 = 10

There are two triangles: 10 x 2 = 20

Area of the trapazoid: 100 + 20 = 120

Area of the triangle that's been removed

Base Length: 10

Height Length: 10 - 4 = 6

Area: 10 x 6 x 1/2 = 30

Shaded area

Area of the trapezoid - Area of the triangle

120 - 30 = 90

Area of the shaded region is 90.

4 0

1 year ago

On a coordinate plane, a shape is plotted with vertices of (3, 1), (0, 4), (3, 7), and (6, 4). what is the area of the shape if

QveST [7]

Let
A (3, 1)
B (0, 4)
C(3, 7)
D (6, 4)

step 1
find the distance AB
d=√[(y2-y1)²+(x2-x1)²]------> dAB=√[(4-1)²+(0-3)²]-----> dAB=√18 cm

step 2
find the distance CD
d=√[(y2-y1)²+(x2-x1)²]------> dCD=√[(4-7)²+(6-3)²]-----> dCD=√18 cm

step 3
find the distance AD
d=√[(y2-y1)²+(x2-x1)²]------> dAD=√[(4-1)²+(6-3)²]-----> dAD=√18 cm

step 4
find the distance BC
d=√[(y2-y1)²+(x2-x1)²]------> dBC=√[(7-4)²+(3-0)²]-----> dBC=√18 cm

step 5
find slope AB and CD
m=(y2-y1)/(x2-x1)
mAB=-1
mCD=-1
AB and CD are parallel and AB=CD

step 6
find slope AD and BC
m=(y2-y1)/(x2-x1)
mAD=1
mBC=1
AD and BC are parallel and AD=BC
and
AB and AD are perpendicular
BC and CD are perpendicular

therefore
the shape is a square wit length side √18 cm

area of a square=b²
b is the length side of a square
area of a square=(√18)²------> 18 cm²

the answer is
18 cm²

see the attached figure