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2 years ago

Z/m − z/n =1, if m≠n

Mathematics

1 answer:

kondaur [170]2 years ago

4 0

Answer:

m= zn/n+z or n= - zm/m-z

Step-by-step explanation:

hope this helps

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A ballroom in a hotel has the shape shown. The manager needs to determine how much carpet is needed to recarpet this area. All a

Mars2501 [29]

Area of a rectangle=length x width

You have two rectangles:

First rectangle (the biggest):
length=9 yd
width=12 yd-6 yd=6 yd

Area of the biggest rectangle=(9 yd)(6 yd)=54 yd²

Second rectangle (the smallest)
length=6 yd
width=3 yd

Area of the smallest rectangle=(6 yd)(3 yd)=18 yd²

Area of the ballroom=area of the biggest rectangle + area of the smallest rectangle.

area of the ballroom=54 yd² +18 yd²=72 yd²

7 0

2 years ago

Each rectangle below has an area of 4 cm? Each square has an area of 2 cm?

Andre45 [30]

Answer:

Literally any even number that’s not two or four

Step-by-step explanation:

I need a layout

4 0

2 years ago

Find the measure of the missing angle using the triangle angle sum theorem

Greeley [361]

Answer:

m∠60

Step-by-step explanation:

30 + 90 = 120

180-120=60

7 0

2 years ago

Read 2 more answers

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

2 years ago

Read 2 more answers

Plss solve 29 for me.with the steps<br> plsssßs I am begging u

lakkis [162]

That means that the first rectangle and the first triangle ave athe same perimiter

so therefor
perimiter=sides added up
rectangle has 4 sides bu 2 is given so double it since the other side is same legnth
rec=x+x+4
multiply 2
2x+2x+8=4x+8=perimiter
this is equal to perimiter of triangle which is
x+9+x+5+x=3x+14
therefor
4x+8=3x+14
subtract 3x from both sides
x+8=14
subtract 8 from both sides
x=6
first one x=6

second rectangle and second triangle
s+7 and 3s
2 sides
2s+14+6s=p
8s+14=p

traignel=2s+12+2s+12+2s+12=6s+36=p
they are equal so
8s+14=6s+36
subtract 6s from both sides
2s+14=36
subtract 14 from both sides
2s=22
divide both sideds by 2
s=11

the values are
pair one:
x=6

pair 2:
s=11

5 0

2 years ago