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Blababa [14]
4 years ago
14

Describe all situation where vertical angles are also supplementary

Mathematics
2 answers:
timofeeve [1]4 years ago
6 0

Answer:

if  vertical angles are right angles  then the angles are supplementary.

Step-by-step explanation:

Vertical angles: When two lines are intersect to each other then the opposite angles made at the point of intersection are called vertical angles.

Vertical angles are always equal in measurement.

Supplementary angles: The angles whose sum is 180 degrees are called supplementary angle.

Let x and y are vertical angles

Then, x=y

If angles x and y are supplementary

x+y=180^{\circ}

x+x=180^{\circ}

2x=180^{\circ}

x=\frac{180}{2}=90^{\circ}

x=y=90^{\circ}

When x and y=90 degrees then the angle x and y are supplementary.

Therefore, we can say that if vertical angles are right angles  then the angles are supplementary.

Nataly_w [17]4 years ago
3 0
Vertical angles are always the same size. if they are supplementary the measures must add to 180. 180/2=90. vertical angles are supplementary only when they are right angles, our 90°
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B / -7 = 4<br><br> solve question correctly
Shkiper50 [21]

Answer:

b/-7=4

b=4*-7

b= - 28

Step-by-step explanation:

If this helps you mark as brainlist.!

4 0
3 years ago
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Antonio eats 1 slice of pizza in 3 minutes. How long will it take Antonio to eat 4 slices?​
GarryVolchara [31]

Answer:

12 minutes

4*3

That's the answer

8 0
3 years ago
A tortoise and a hare are competing in a 2000-meter race. The arrogant hare decides to let the tortoise have a 550-meter head st
andrew-mc [135]

Answer:

a) distance covered by hare d1 = 8t

b) distance covered by tortoise d2 = 5t + 550

c) ∆d = 550 - 3t

Step-by-step explanation:

Given;

Speed of hare u = 8m/s

Speed of tortoise v = 5 m/s

Initial distance of tortoise d0 = 550 m

a) using the equation of motion;

distance covered = speed × time + initial distance

d = vt + d0

For hare;

d0 = 0

Substituting the values;

d1 = 8t + 0

d1 = 8t

b)using the equation of motion;

distance covered = speed × time + initial distance

d2 = vt + d0

For tortoise;

d0 = 550m

Substituting the values;

d2 = 5t + 550

d2 = 5t + 550 m

c) the number of meters the tortoise is ahead of the hare.

∆d = distance covered by tortoise - distance covered by hare

∆d = d2 - d1

Substituting the values;

∆d = (5t + 550) - 8t

∆d = 550 - 3t

6 0
3 years ago
Which simplified fraction is equal to 0.53? 1) 24/45 2) 8/15 3) 48/90 4) 5/9
Shalnov [3]

To express a decimal number (with a finite decimal expression) in a fraction, you work like this: the numerator is the number without decimale separator (you simply pretend it's not there), while the denominator will be 10^n, where n is the number of digits after the decimal separator.

So, in your case, the number without decimal separator is 53, and you have two digits after the decimal separator, so the fraction will be

0.53 = \cfrac{53}{10^2} = \cfrac{53}{100}

which is irreducible.

6 0
3 years ago
Read 2 more answers
The set of the consecutive odd numbers 1, 3, 5, 7, ... , N has a sum of 400. How many numbers are in the set?
Amanda [17]

Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.

First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.

Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:

         1, 3, 5, 7, 9, 11, 13, 15 .

Number of items in the set . . . 8
Sum of all the items in the set . . . 64

Hmmm.  That's interesting.  64 happens to be the square of 8 . 
Do you think that might be all there is to it ?

Let's check it out:

Even-numbered lists of odd numbers:

1, 3                                   Items = 2, Sum = 4
1, 3, 5, 7                           Items = 4, Sum = 16
1, 3, 5, 7, 9, 11                 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .

Amazing !  The sum is always the square of the number of items in the set !

For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.

I slogged through it on my calculator, and it's true.

I never knew this before.  It seems to be something valuable
to keep in my tool-box (and cherish always).


3 0
3 years ago
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