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

Mathematics

1 answer:

ziro4ka [17]3 years ago

3 0

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

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Please explain with working!!! Find the set of values of x that satisfy the inequality 9x^2-15x<6

Oksi-84 [34.3K]

Answer:

$-1/3$

Step-by-step explanation:

When solving a quadratic inequality, first solve it normally like you would for a normal quadratic equation. We have:

$9x^2-15x$

Ignore the less than sign and replace it with an equal sign and solve the quadratic for its zeros:

$9x^2-15x=6$

Subtract 6 from both sides:

$9x^2-15x-6=0$

Divide everything by 3:

$3x^2-5x-2=0$

Factor. Find two numbers that equal (3)(-2)=-6 that add up to -5.

-6 and 1 works. Thus:

$3x^2-6x+x-2=0\\3x(x-2)+1(x-2)=0\\(3x+1)(x-2)=0$

Find the x using the Zero Product Property:

$3x+1=0 \text{ or }x-2=0\\x=-1/3\text{ or }x=2$

Now, we need to replace the equal signs with symbols again. To do so, we need to test which symbol to place. Let's do the first zero first.

So, the first zero is:

$x=-1/3$

Assume that the correct symbol is >. Thus,

$x>-1/3$

Now, pick any number that is greater than -1/3. I'll pick 0 since it's the easiest. Now, plug 0 back into the very original inequality. If it works, then the sign is correct, if it doesn't, then simply use the opposite one. Therefore:

$9x^2-15x$