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3 years ago

Trench Warfare Define in own words

Mathematics

2 answers:

Andrei [34K]3 years ago

4 0

Answer:

Trench warfare was a technique used to fight in World War 1. As they could hide/defend there. Get supplie there. They could even have medical areas there, but the problem was a stalemate as that was such a thing that no one attacks and they stay without moving on in the battle.

Leto [7]3 years ago

3 0

Hey bestie here’s a definition: a form of land warfare in which military trenches are used as part of occupied battle lines.

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I want to know x=,y=,z=

My name is Ann [436]

Y=29

x+y+78=180
x+29+78=180
x=180-78-29
x=73

z=180-73
z=107

x=73 degrees
y=28 degrees
z=107 degrees

3 0

3 years ago

Find the quotient and renainder When X⁴ - 3X³ + 2x-5 as divided by x²-x+1.

OverLord2011 [107]

Answer:

Answer is in picture

Step-by-step explanation:

Hope it is helpful....

8 0

3 years ago

In a plane, points P and Q are 20 inches apart. If point R is randomly chosen from all the points in the plane that are 20 inche

Karolina [17]

Answer:

E. $\frac{2}{3}$

Step-by-step explanation:

Consider a circle having centre P and radius 20 inches,

So, the area of the circle,

$A=\pi (20)^2\text{ square inches}$

Also, suppose points Q and R are on the circumference,

i.e. PQ = PR = 20 inches,

If QR = 20 inches,

So, triangle PQR is an equilateral triangle,

⇒ m∠RPQ = 60°,

Now, suppose R' is another point on the circle,

Such that, ΔPQR ≅ Δ PQR',

⇒ m∠QPR' = 60°,

Thus, minor angle, m∠RPR' = m∠RPQ + m∠QPR' = 60° + 60° = 120°,

⇒ major angle, m∠RPR' = 360° - 120° = 240°,

So, the area of the circle where a point on the circumference is closer to P than it is to Q

$=\frac{240^{\circ}}{360^{\circ}}\pi (20)^2$

$=\frac{2}{3}\pi (20)^2$

Hence, the probability that a point is closer to P than it is to Q = $\frac{\frac{2}{3}\pi (20)^2}{\pi (20)^2}$

$=\frac{2}{3}$

i.e. OPTION E is correct.

7 0

3 years ago

12 yds

Anton [14]

Answer:

176 square yards

Step-by-step explanation:

The picture of the question in the attached figure N 1

we know that

The area of the walkway around the rectangular pool, is equal to the area of two trapezoids (#1 and #2), plus the area of two smaller rectangles (#3 and #4)

see the attached figure N 2 to better understand the problem

step 1

Find the area of the two trapezoids (#1 and #2)

$A=2[\frac{1}{2}(b_1+b_2)h]$

simplify

$A=(b_1+b_2)h$

we have

$b_1=12\ yd\\b_2=4+12+4=20\ yd\\h=4\ yd$

substitute

$A=(12+20)(4)=128\ yd^2$

step 2

Find the area of the two smaller rectangles (#3 and #4)

$A=2[LW]$

we have

$L=4\ yd\\W=6\ yd$

substitute

$A=2[(4)(6)]=48\ yd^2$

step 3

Find the area of the walkway around the rectangular pool

$128+48=176\ yd^2$

8 0

4 years ago

Read 2 more answers

Only answer if your 100% sure What equation is shown by the graph

devlian [24]

Hello!

The equation for a line is y = mx + b

m is the slope
b is the y-intercept

The y-intercept is -3

y = mx - 3

To find the slope you count how many times up and over it goes from point to point

It goes up one over one

$y = \frac{1}{1} x - 3$

Simplify

The answer is y = x - 3

Hope this helps!

5 0

3 years ago