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4 years ago

In the triangles, and .

Mathematics

2 answers:

Ratling [72]4 years ago

8 0

The correct answers are:

#1) Angle G is smaller than angle P; #2) The three sides have the same length; #3) The path along XY is longer than the path along ZY, and the path along ZY is longer than the path along AB.

Explanation:

#1) Since the measure of HK is smaller than the measure of MN, this means the angle opposite HK will be smaller than the angle opposite MN. This means ∠G is smaller than ∠P.

#2) Since all of the angles are congruent, all of the sides will be congruent as well.

#3) The side opposite the largest angle will be longest; the side opposite the smallest angle will be shortest. This means the side opposite the 36° angle, AB, will be smallest, and the side opposite the 79° angle, XY, will be largest. ZY will have a length between these two.

lana66690 [7]4 years ago

4 0

Answer:

B. Angle G is smaller than angle P.

Step-by-step explanation:

on edgenuity 2020

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The semicircle shown at left has center X and diameter W Z. The radius XY of the semicircle has length 2. The chord Y Z has leng

gladu [14]

${\bold{\red{\huge{\mathbb{QUESTION}}}}}$

The semicircle shown at left has center X and diameter W Z. The radius XY of the semicircle has length 2. The chord Y Z has length 2. What is the area of the shaded sector formed by obtuse angle WXY?

$\bold{ \red{\star{\blue{GIVEN }}}}$

RADIUS = 2

CHORD = 2

RADIUS --> XY , XZ , WX

( BEZ THEY TOUCH CIRCUMFERENCE OF THE CIRCLES AFTER STARTING FROM CENTRE OF THE CIRCLE)

$\bold{\blue{\star{\red{TO \: \: FIND}}}}$

THE AREA OF THE SHADED SECTOR FORMED BY OBTUSE ANGLE WXY.

$\bold{ \green{ \star{ \orange{FORMULA \: USED}}}}$

AREA COVERED BY THE ANGLE IN A SEMI SPHERE

$AREA = ANGLE \: \: IN \: \: RADIAN \times RADIUS$

$\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}$

Total Area Of The Semi Sphere:-

$AREA = \pi \times radius \\ \\ AREA = \pi \times 2 = 2\pi$

Area Under Unshaded Part .

Given a triangle with each side 2 units.

This proves that it's is a equilateral triangle which means it's all angles r of 60° or π/3 Radian

So AREA :-

$AREA = \frac{\pi}{3} \times radius \\ \\ AREA = \frac{\pi}{3} \times 2 \\ \\ AREA = \frac{2\pi}{3}$

$\green{Now:- } \\ \green{ \: Area \: Under \: Unshaded \: Part }$

Total Area - Area Under Unshaded Part

$Area= 2\pi - \frac{2\pi}{3} \\ Area = \frac{6\pi - 2\pi}{3} \\ Area = \frac{4\pi}{3} \: \: ans$

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3 years ago

What is 0.675 in its simplest form?

Musya8 [376]

27 over 40 (as a fraction)

WORK:
.675 is 675 over 1000
5 goes into both of those so
675/5=135 1000/5=200
135/5=27
200/5=40

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3 years ago

NEED HELP ASAP!!! TEN POINTS!

Bogdan [553]

The answer it’s letter D

7 0

3 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: