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REY [17]
3 years ago
9

Please help me I will give brainliest

Mathematics
2 answers:
Talja [164]3 years ago
8 0

Answer:

Answer in short, 1.32 x 10^5 is your answer.

Step-by-step explanation:

Take the first number, or 136000 and subtract it by 4000 (the second number.

solmaris [256]3 years ago
4 0

Answer:

1.32*10^5

Step-by-step explanation:

1.36*10^5 = 136,000

4*10^3= 4,000

136,000-4,000= 132,000

1.32*10^5=132,000

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A middle school took all of its sixth grade students on a field trip to see a play at a theater that has 2000 seats. The student
almond37 [142]

Answer:

1300 middle school students.

Step-by-step explanation:

We can turn the words into this expression:

2000 x 65/100

And that expression simplifies to 1300, so 1300 students.

3 0
3 years ago
If the area of a circle is 64 square meters then what is the radius
s2008m [1.1K]

Answer:

Radius = 4.51

Step-by-step explanation:

7 0
3 years ago
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Please help me let me know how you got the answer
inessss [21]

Area of building is 3,84,000m^3 .

<u>Step-by-step explanation:</u>

Here we have , a figure in given below picture where we have to find volume of this office building , Let's find out:

If we look closely we see that it's basically cuboid with dimensions 50 m by 110 m by 70 m . But there's one small cuboid sliced off from this cuboid with dimensions 50 m by 10 m by (110-90) m i.e.  50 m by 10 m by 20 m .

Area of building = Area of larger cuboid - Area of smaller cuboid

Area of cuboid = length(width)(breadth)

⇒ Area of building = Area of larger cuboid - Area of smaller cuboid

⇒ Area = (50(110)(70))m^3 - (50(10)(20))m^3

⇒ Area = 3,85,000m^3 - 10,000m^3

⇒ Area = 3,84,000m^3

Therefore, Area of building is 3,84,000m^3 .

3 0
3 years ago
2a^2 + -1(2-a)+3b ; a = 5 , b = 0
Andreas93 [3]
The answer is 53
——————————-
8 0
3 years ago
Read 2 more answers
Prove that sin^2A/cos^2A + cos^2A/sin^2A = 1/cos^2A*sin^2A - 2
jenyasd209 [6]

Answer:

prove that:

Sin²A/Cos²A + Cos²A/Sin²A = 1/Cos²A Sin²A - 2

LHS = \frac{Sin^2A}{Cos^2A} + \frac{Cos^2A}{Sin^2A}

Cos

2

A

Sin

2

A

+

Sin

2

A

Cos

2

A

= \begin{lgathered}= \frac{Sin^4A + Cos^4A}{Cos^2A . Sin^2A}\\\\Using\: a^2 + b^2 = (a+b)^2 - 2ab\\\\a = Cos^2A \: \& \:b = Sin^2A\\\\= \frac{(Sin^2A + Cos^2A)^2 - 2Sin^2A Cos^2A}{Cos^2A Sin^2A} \\\\Sin^2A + Cos^2A = 1\\\\= \frac{1 -2Sin^2A Cos^2A}{Cos^2A Sin^2A}\end{lgathered}

=

Cos

2

A.Sin

2

A

Sin

4

A+Cos

4

A

Usinga

2

+b

2

=(a+b)

2

−2ab

a=Cos

2

A&b=Sin

2

A

=

Cos

2

ASin

2

A

(Sin

2

A+Cos

2

A)

2

−2Sin

2

ACos

2

A

Sin

2

A+Cos

2

A=1

=

Cos

2

ASin

2

A

1−2Sin

2

ACos

2

A

\begin{lgathered}= \frac{1}{Cos^2A Sin^2A} - 2\\\\= RHS\end{lgathered}

=

Cos

2

ASin

2

A

1

−2

=RHS

LHS=RHS

4 0
3 years ago
Read 2 more answers
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