Can someone help me with this? Its fairly easy but i genuinely dont remember

HELP ASAP!! Lesson 4: Perimeters and Areas of Similar Figures

ziro4ka [17]

Part 1)
area of triangle=b*h/2
b=2.9 cm
h=9 cm
so
area=2.9*9/2-----> 13.05 cm²

the answer part 1) is 13.05 cm²

Part 2)
area of the figure=area of rectangle+area of triangle

find the area of rectangle
area rectangle=b*h
b=12 ft
h=?
tan 45=1
tan 45=h/(22-10)----------> h/10=1-------> h=10 ft
area of rectangle=12*10-----> 120 ft²

area of triangle=b*h/2
b=10 ft
h=10 ft
area of triangle=10*10/2----> 50 ft²

area of the figure=120+50----> 170 ft²

the answer Part 2) is 170 ft²

Part 3)
<span>the area of kite is half the product of the diagonals.
</span>Area=(d1*d2)/2
d1=5+5---> 10 ft
d2=16+8----> 24 ft
area=(10*24)/2----> 120 ft²

the answer Part 3) is 120 ft²

Part 4)
the area of rhombus is half the product of the diagonals.
Area=(d1*d2)/2
d1=6+6---> 12 m
d2=6+6----> 12 m
area=(12*12)/2----> 72 m²

the answer part 4) is 72 m²

Part 5)

area of the figure=6*area of one triangle

area of triangle=b*h/2
b=4 cm
is an equilateral triangle
applying the Pythagoras theorem
h²=4²-2²-----> h²=12-----> h=2√3 cm

area of triangle=4*2√3/2----> 4√3 cm²

area of hexagon=6*(4√3)----> 24√3 cm²

the answer Part 5) is 24√3 cm²

3 0

3 years ago

Identify all the sets to which the number belongs. Choose from Rational Number, Irrational number, whole number, and or a intege

vaieri [72.5K]

From that list of choices, 0.1233 is only a rational number.

3 0

4 years ago

Read 2 more answers

a student achieved the following grades on his tests :87,95 ,76 and 88 percent . he wants 85 percent or better overall . what gr

m_a_m_a [10]

Answer:

Step-by-step explanation:

87+95+76+88+x/5=85

346+x=85x5

x=425-346

x=79

4 0

3 years ago

Read 2 more answers

A pillow that usually costs $20 is marked down 45%. Table Use the table to find the new price of the pillow. $9 $11 $29 $33

sattari [20]

It would now cost $11. 45% of 20 is 9, therefore, you would subtract 9 from 20 (since it is getting marked down).

6 0

3 years ago

Read 2 more answers

A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of

Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

the angle of elevation is Ф = 45° and
the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

the angle of elevation is Ф = 45° and
the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So, y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0

2 years ago