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3 years ago

How were the numbers 1 10 100 and 1000 written by romans?

Mathematics

1 answer:

Arturiano [62]3 years ago

8 0

1 -- I
10 -- X
100 -- C
1000 -- M

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What is the simplest form of the ratio 11 : 16? A. 1 : 6 B. 5 : 8 C. 11 : 16 D. 22 : 32

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Answer:

Step-by-step explanation:

c because 11 and 16 have no GCF other than 1

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3 years ago

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Which expression(s) represent(s)®3 times the sum of sand 4"?

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Answer is : 3x+4 the second one

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2 years ago

adult tickets at a football game cost $6 and student tickets cost $2. A total of $1776 was collected on 372 tickets. How many ad

postnew [5]

Answer:

258

Step-by-step explanation:

Set up the two equations needed (x is the adult tickets and y is the student tickets)

x+y=372

6x+2y=1776

Solve by elimination

-2(x+y=372)

=-2x-2y=-744

Cancel like figures:

-2x-2y=-744

6x+2y=1776

= 4x=1032

x=258

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3 years ago

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Integer equivalent to 9^3/2

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364.5 because 9 to the third power is 729. 729 divided by 2 is 364.5

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4 years ago

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The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0

3 years ago