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2 years ago

Drag the numbers to label the points plotted on the number line.

Mathematics

2 answers:

ivolga24 [154]2 years ago

8 0

Answer:

The first box should be - $\frac{10}{5}$ ;

The 2nd box should be -1.25 ;

The 3d box should be -0.5 ;

The 4th box should be 0.75 ;

The 5th box should be 3/2 which is the last box left.

leva [86]2 years ago

5 0

Answer:

First Box: -10/5

Second Box = -1.25

Third Box = -0.5

Fourth Box = 0.75

Fifth Box = 3/2

Step-by-step explanation:

NOTE: I AM SAYING THIS FROM LEFT TO RIGHT

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Point Q is the centroid of △ABC. QE = _____

pychu [463]

That’s the answer to your question.

7 0

2 years ago

use the guide to construct a two column proof proving △aec≅△deb, given that ca is parallel to db and e is the midpoint of ad. gi

MrMuchimi

By using the AAA congruence property of Triangles, △aec≅△deb is proved

It is given two triangles, Δaec and Δdeb, where e is the common point known as the midpoint of ad

Also, it is given that,

ca ║db

We need to prove that, △aec≅△deb

Then we'll use AAA congruence property of Triangles to prove the situation

As ca ║db

then, ∠cae = ∠ebd (Alternate angles)

∠ace = ∠edb (Alternate angles)

and ∠aec = ∠deb (common angles)

Thus, by AAA congruence property, △aec≅△deb

Hence, proved

To learn more about, congruence property, here

brainly.com/question/2039214

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5 0

1 year ago

a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface ar

Margarita [4]

A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)

Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.

The lateral faces are equilateral triangles of side length x.

The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.

now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.

by the Pythagorean theorem:

$h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x$

thus,

$Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2$

thus:

$x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\ \sqrt{3} }{3}=24 \sqrt{3}$ (cm^2)

but $x^{2}$ is exactly the base area, since the base is a square of sidelength = x cm.

So, the total surface area = base area + lateral area = $24 \sqrt{3}+72$ cm^2

Answer: $24 \sqrt{3}+72$ cm^2