If m ABC = 125°, and mZABD = 949, then m2DBC = [? ]°

I don’t understand the problem or question

skad [1K]

Answer:

Part A

a. Could be a parallelogram but cannot be a rectangle

Part B

c. 100°

Step-by-step explanation:

Part A

The polygon constructed by Andrew = Four sided polygon having no right angles

From the given four sided polygon (quadrilateral) options, the options which can have no right angles are the parallelogram, and rhombus

The rectangle, and square both have 4 right angles

Therefore, the polygon can be either a parallelogram or a rhombus, but cannot be either a rectangle or square

The correct option is therefore, 'His polygon could be a parallelogram but cannot be a rectangle'

Part B

The given parameters of the rhombus EFGH are;

Angle E = 3·x + 5, angle H = 4·x

The characteristics of a rhombus are;

The sum of the interior angles of a rhombus = 360°

The opposite angles of a rhombus are equal

The sum of the adjacent angles = 180° (The adjacent angles are supplementary)

Therefore, in the rhombus EFGH, we have;

Angle E, and angle H are supplementary angles

∴ Angle E + angle H = 180°

Substituting the values of angle E and angle H gives;

3·x + 5 + 4·x = 180°

7·x = 180° - 5 = 175°

x = 175°/7 = 25°

Angle H = 4·x

∴ Angle H = 4 × 25° = 100°

Angle H = 100°

Angle F = angle H = 100°

Therefore, angle F = 100°

4 0

3 years ago

A school wishes to enclose its rectangular playground using 480 meters of fencing.

Harlamova29_29 [7]

Answer:

Part a) $A(x)=(-x^2+240x)\ m^2$

Part b) The side length x that give the maximum area is 120 meters

Part c) The maximum area is 14,400 square meters

Step-by-step explanation:

The picture of the question in the attached figure

Part a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x

we know that

The perimeter of the rectangular playground is given by

$P=2(L+W)$

we have

$P=480\ m\\L=x\ m$

substitute

$480=2(x+W)$

solve for W

$240=x+W\\W=(240-x)\ m$

Find the area of the rectangular playground

The area is given by

$A=LW$

we have

$L=x\ m\\W=(240-x)\ m$

substitute

$A=x(240-x)\\A=-x^2+240x$

Convert to function notation

$A(x)=(-x^2+240x)\ m^2$

Part b) What side length x gives the maximum area that the playground can have?

we have

$A(x)=-x^2+240x$

This function represent a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the length that give the maximum area that the playground can have

Convert the quadratic equation into vertex form

$A(x)=-x^2+240x$

Factor -1

$A(x)=-(x^2-240x)$

Complete the square

$A(x)=-(x^2-240x+120^2)+120^2$

$A(x)=-(x^2-240x+14,400)+14,400$

$A(x)=-(x-120)^2+14,400$

The vertex is the point (120,14,400)

therefore

The side length x that give the maximum area is 120 meters

Part c) What is the maximum area that the playground can have?

we know that

The y-coordinate of the vertex represent the maximum area

The vertex is the point (120,14,400) -----> see part b)

therefore

The maximum area is 14,400 square meters

Verify

$x=120\ m$

$W=(240-120)=120\ m$

The playground is a square

$A=120^2=14,400\ m^2$

8 0

3 years ago

Is this right??? if not helpp

Nonamiya [84]

Yupz u b right cuz 8+9 is ur answer Don’t be so hard on yourself take it easy

7 0

2 years ago

What are the zeros of (x-2)(x^-9)

Nostrana [21]

I suspect you may have slipped up as you copied the question. Exactly as you wrote it, it has a single zero, at x= 2 .

8 0

3 years ago

Question 5 of 10

Alona [7]

I don’t understand?????

5 0

1 year ago