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3 years ago

Blitzkrieg is the term used to describe Germany's?

Mathematics

2 answers:

Andreyy893 years ago

5 0

“lightning” warfare with tanks used to end a fight swiftly.

Temka [501]3 years ago

3 0

Answer:

use of fast and strong armed forces (APEX)

Step-by-step explanation:

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Cual es la utilidad de la representacion a escala de diferentes distancias reales

svp [43]

Answer:

Scales are used to represent real distances and proportions into maps and planes, that way we can have a bigger perspective about something just having in paper.

One direct use of scales are seen in architecture plans, there, the architect uses scales to represent the construction he is designing. Also, the architect can have a more complete point of view about the principial elements on his work.

6 0

2 years ago

What is the simplified form of √144x^36

Nadusha1986 [10]

Answer:

The simplified form of $\sqrt{144x^{36}}$ is $12x^{18}$

Step-by-step explanation:

Given : $\sqrt{144x^{36}}$

We have to write the simplified form of $\sqrt{144x^{36}}$

Consider the given expression $\sqrt{144x^{36}}$

We know $\sqrt{144}=12$

and $\sqrt{x^{36}}=\sqrt{x^{18}\cdot x^{18}}$

Thus,

$\sqrt{144x^{36}}=\sqrt{12^2\cdot (x^{18})^2}$

Simplify, we have,

$=\sqrt{12^2\cdot (x^{18})^2}=12x^{18}$

Thus, The simplified form of $\sqrt{144x^{36}}$ is $12x^{18}$

6 0

2 years ago

Read 2 more answers

What is 7 1/5 written as a percent?

anastassius [24]

7 1/5 is 72 percent.

6 0

3 years ago

Read 2 more answers

What would x be in 5x=55

BabaBlast [244]

Answer:

5x=55

Divide both sides by 5 to solve for x

5x/5=55/5

x=11

8 0

3 years ago

Read 2 more answers

Without actual division prove that x4 +2x3 -2x2 +2x -3 is exactly divisible by<br> x2 +2x-3

Brut [27]

Answer:

Step-by-step explanation:

$\displaystyle\Large\boldsymbol{} x^4+2x^3-2x^2+2x-3= \\\\\\x^4+2x^2(x-1)+2x-\underbrace{2-1}_{-3}= \\\\\\x^4-1+2x^2(x-1)+2x-2 = \\\\\\(x^2-1)(x^2+1)+2x^2(x-1)+2(x-1) = \\\\\\\underline{(x-1)}(x+1)(x^2+1)+2x^2\underline{(x-1)} +2\underline{(x-1)} =\\\\\\(x-1)((x+1)(x^2+1)+2x^2+2)=\\\\\\(x-1)(x^3+x^2+2x^2+x+2+1)=\\\\\\(x-1)(x^3+3x^2+x+3) =\\\\\\(x-1)( \underline{(x+3)}x^2+\underline{(x+3)})=(x^2+1)\underbrace{(x-1)(x+3)}_{x^2+2x-3}= \\\\\\(x^2+1)(x^2+2x-3) : (x^2+2x-3)=x^2+1$

7 0

2 years ago