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3 years ago

What are the 3 cross section that are parallel to the base of the figure are congruent

Mathematics

1 answer:

puteri [66]3 years ago

8 0

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How many blocks with dimensions of One-third times 1 times 1 can fit in a unit cube?

gtnhenbr [62]

Answer: 3

Step-by-step explanation:

Given

The dimension of the small block is

$\dfrac{1}{3}\times 1\times 1$

The Volume of the cube having unit length is

$V=1\times 1\times 1\\V=1\ unit^3$

The number of blocks that can be in the unit cube is the division of the volume of the unit cube and block.

$\Rightarrow \dfrac{1\times 1\times 1}{\dfrac{1}{3}\times 1\times 1}\\\\\Rightarrow \dfrac{3}{1}=3$

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3 years ago

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If two friends invested in a ratio of 2:3 and earned a profit of Rs. 25,000 on their investments, then the smaller share of the

podryga [215]

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3 years ago

A pack of paper costs $3.75 including tax. Mr. Cooper wants to purchase packs of paper for his class and has a $20 budget. Write

Anastaziya [24]

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3 years ago

If y = x+2 were arranged to y= x+1, how would you graph the new function compare with the first one

Kobotan [32]

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3 years ago

In the circle below, DB = 22 cm, and m<DBC = 60°. Find BC. Ignore my handwriting.

Karo-lina-s [1.5K]

Answer:

$BC=11\ cm$

Step-by-step explanation:

step 1

Find the measure of the arc DC

we know that

The inscribed angle measures half of the arc comprising

$m\angle DBC=\frac{1}{2}[arc\ DC]$

substitute the values

$60\°=\frac{1}{2}[arc\ DC]$

$120\°=arc\ DC$

$arc\ DC=120\°$

step 2

Find the measure of arc BC

we know that

$arc\ DC+arc\ BC=180\°$ ----> because the diameter BD divide the circle into two equal parts

$120\°+arc\ BC=180\°$

$arc\ BC=180\°-120\°=60\°$

step 3

Find the measure of angle BDC

we know that

The inscribed angle measures half of the arc comprising

$m\angle BDC=\frac{1}{2}[arc\ BC]$

substitute the values

$m\angle BDC=\frac{1}{2}[60\°]$

$m\angle BDC=30\°$

therefore

The triangle DBC is a right triangle ---> 60°-30°-90°

step 4

Find the measure of BC

we know that

In the right triangle DBC

$sin(\angle BDC)=BC/BD$

$BC=(BD)sin(\angle BDC)$

substitute the values

$BC=(22)sin(30\°)=11\ cm$

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3 years ago