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3 years ago

Where is the head-quarter of Microsoft Corporation?

Mathematics

2 answers:

natka813 [3]3 years ago

8 0

Answer:

A straight angle.

Step-by-step explanation:

MArishka [77]3 years ago

6 0

Answer:

The Headquater of Microsoft Corporation is somewhere in the Redmond Area in Washington, USA $.$

Explore more:

Do you know something more about Microsofts?

The founder of Microsoft is Bill Gates and Paul Allen. With this, Bill Gates became the youngest billionaire of all time.

Hope you have seen the logo. It was designed by Paul Allen.

The 'O' is known as Blibbet in the logo of Microsoft.

As the name says, Microsoft is a leading global company made for developing softwares and applications.

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Using the elimination method, what would be the first step? x-4y=6 5x+y=8

enyata [817]

X-4y=6. 5x+y=8

minus X from both sides

4y=6. 4x+y=8

minus y from both sides

3y=6. 4x=8

y=2

X=-2

7 0

3 years ago

You already owe $4.32 in overdue rental fees and are returning a movie that is 4 days late. Now you owe $6.48. How much is the d

mr_godi [17]

6.48-4.32=2.16
2.16/4=.54
answer 54 cents

4 0

3 years ago

In a triangle HIJ, i = 6.6 inches, j = 6.4 inches and Angle,H=62º. Find the length of h, to the

sweet [91]

Answer:

7.6 inches

Step-by-step explanation:

We solve the above question using the Law of Cosines.

The formula is given as:

h² = i² + j² - 2ij Cos H

h = √i² + j² - 2ij Cos H

From the above question:

i = 6.6 inches

j = 6.4 inches

Length h = ?

Angle H=62º.

Hence:

h= √6.6² + 6.4² - 2 × 6.6 × 64 × Cos 62

h = √43.56 + 40.96 - 84.48 × Cos 62

h = 7.64292 inches

Approximately to the nearest tenth, Length h = 7.6 inches

8 0

2 years ago

What is the solution to x2 = 225?

Blababa [14]

Answer:

x = ± 15

Step-by-step explanation:

Given

x² = 225 ( take the square root of both sides )

x = ± $\sqrt{225}$ = ± 15

That is x = - 15 0r x = 15

6 0

3 years ago

Read 2 more answers

If f(x)= 2x^5-3x^3+2x^2+1, then f(1)=? A. -1 B. 0 C. 1 D. 2 E. 4

Ivahew [28]

Answer:

Step-by-step explanation:

f(x)= 2x^5-3x^3+2x^2+1

Let x=1

f(1)= 2(1)^5-3(1)^3+2(1)^2+1

= 2*1 -3*1+2(1) +1

= 2-3+2+1

8 0

3 years ago

Read 2 more answers