PLZ HELPWhat is the radius and diameter of the following circle?

Mathematics

2 answers:

bearhunter [10]3 years ago

7 0

The radius of the circle is 7 cm because the radius is half of the diameter. In your case, the diameter is 14 cm. So half of 14 is 7. 7 cm is the answer.
Hope this helps.

babymother [125]3 years ago

4 0

Solution:
d = diameter
r = radius

d = 2r
r = d/2
In your case d = 14
So
r = 14/2
r = 7 answer

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Please do them step by step;

ELEN [110]

For the first one c-9=(2)(4)+6
You are gonna multiply 2 and 4 first, 2*4=8. Now you have c-9=8+6. 8+6=14. So now the equation is c-9=14. Now we are going to add 9 to both sides. So c=23
For the second equation 10+y=90. We are going to subtract 10 from both sides. So, y=80.
Third, add 4 to both sides. a=10.
Fourth, subtract 8 from both sides. d=-9
Fifth, 5+2(3+4)=-x.
First we are going to add 3 and 4. 3+4=7. 5+2(7)=-x. Next we are going to multiply 2 and 7. 2*7=14. 5+14=-x. 5+14=19. So our equation is 19=-x. Next we are going to multiply both sides by -1. Making our final equation x=-19.
c=23
y=80
a=10
d=-9
x=-19
I hope that helps, if you need any further explanation just ask.<span />

3 0

3 years ago

at the movies, la quinta paid for drinks and popcorn for herself and her two children. she spend twice as much on popcorn as on

jarptica [38.1K]

Answer:

Step-by-step explanation:

let the amount spent on popcorn be x and amount spent on drinks be y

If the total bill paid is $17.94, then x+y = $17.94

If she spend twice as much on popcorn as on drinks, then y = 2x

Substitute y = 2x into the original equation

x+y = $17.94

x + 2x = $17.94

3x = $17.94

x = $17.94/3

x = $5.98

Since y = 2x

y = 2($5.98)

y = $11.96

Therefore she spent $5.98 on popcorn and $11.96 on drinks

4 0

3 years ago

A plane lamina with constant density p(x, y) = p occupies a region in the xy-plane bounded by a simple closed path C. Show that

stira [4]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.