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3 years ago

Which region did the Visigoths most likely settle?

Mathematics

2 answers:

Vlada [557]3 years ago

8 0

Answer:

Gallic i believe

Step-by-step explanation:

Naddika [18.5K]3 years ago

6 0

Answer:

It was Romania, so Spain.

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Can you help me with this please

krek1111 [17]

Answflasvkdiuihrytiy\ih\j

5 0

3 years ago

Given the function f(x)=3(x+2)-4 solve for the inverse function when x=2

Mars2501 [29]

The inverse is (x-2)/3 so the answer is f(x)=0

5 0

3 years ago

How many litres of milk can be put in six hemispherical bowl each of diameter 35cm

Lesechka [4]

Answer:

Approximately 67.348 litres can be put in six hemispherical bowl with a diameter of 35 centimetres.

Step-by-step explanation:

The volume of a hemisphere ( $V$ ), measured in cubic centimetres, is obtained from this formula:

$V = \frac{2\pi}{3}\cdot R^{3}$

Where $R$ is the radius of the hemisphere, measured in centimetres.

We know that radius is the half of the diameter ( $D$ ), measured in centimetres, then:

$R = 0.5\cdot D$

( $D = 35\,cm$ )

$R = 0.5\cdot (35\,cm)$

$R = 17.5\,cm$

Now, we get the volume of each hemispherical bowl:

$V = \frac{2\pi}{3} \cdot (17.5\,cm)^{3}$

$V \approx 11224.649\,cm^{3}$

The total volume of six hemispherical bowl is:

$V_{T} = 6\cdot V$

$V_{T}= 6\cdot (11224.649\,cm^{3})$

$V_{T} = 67347.894\,cm^{3}$

From Physics we know that 1 litre equals 1000 cubic centimetres. Then:

$V_{T} = 67.348\,L$

Approximately 67.348 litres can be put in six hemispherical bowl with a diameter of 35 centimetres.

3 0

3 years ago

What is this sum about like 3×3 what is that pls tell me

mars1129 [50]

3x3=9

hope this helps

7 0

3 years ago

Read 2 more answers

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

7 0

3 years ago