Here ya goes brotherrrrrrrrrrrrrr

Patsy can decorate 3 1/2 cakes in 1/4 hour. At this rate, how many cakes can she decorate in 1 hour?

nika2105 [10]

The answer is 14 because 3 1/2 multiplied by 4 is 14. Hope this helps

7 0

3 years ago

Hello pls help I'm dumb thx it's pre calc

Bumek [7]

(f-g)(x)=f(x)-g(x)=
$\frac{2x+6}{3x}-\frac{\sqrt{x}-8}{3x}=$
$\frac{2x+6-(\sqrt{x}-8)}{3x}=$
$\frac{2x+6-\sqrt{x}+8}{3x}=$
$\frac{2x-\sqrt{x}+14}{3x}$
answer is A

3 0

3 years ago

Describe at least two differences between constructing an inscribed regular hexagon and constructing an inscribed square.

liberstina [14]

Correct answer is A.

The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction.

<h3>How to calculate the diameter?</h3>

If you know the radius of the circle, multiply by 2 to get the diameter. The radius is the distance from the center of the circle to the edge. For example, if the radius of a circle is 4 cm, then the diameter is 4 cm x 2 = 8 cm. If you know the circumference of the circle, divide by π to get the diameter.

<h3>Briefing:</h3>

The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction which is A.

The diameter of the circle is used for the square as l=2r therefore corresponds to the diameter, while for a hexagon r=r being the radius of the circle.

To know more about Diameter visit:

brainly.com/question/5501950

#SPJ4

I understand that the question you are looking for is:

Describe at least two differences between constructing an inscribed regular hexagon and constructing an inscribed square.

A) The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction.

B) The radius of the circle is used for the square construction, but the diameter of the circle is used for the regular hexagon construction.

C) The square will need two arcs along the circle, but the hexagon will need two arcs above and two arcs below the diameter of the circle.

D) The square will need six arcs along the circle, but the hexagon will need two arcs above and two arcs below the diameter of the circle.

7 0

2 years ago

Discrete Math

andrezito [222]

Answer:

Part c: Contained within the explanation

Part b: gcd(1200,560)=80

Part a: q=-6 r=1

Step-by-step explanation:

I will start with c and work my way up:

Part c:

Proof:

We want to shoe that bL=a+c for some integer L given:

bM=a for some integer M and bK=c for some integer K.

If a=bM and c=bK,

then a+c=bM+bK.

a+c=bM+bK

a+c=b(M+K) by factoring using distributive property

Now we have what we wanted to prove since integers are closed under addition. M+K is an integer since M and K are integers.

So L=M+K in bL=a+c.

We have shown b|(a+c) given b|a and b|c.

//

Part b:

We are going to use Euclidean's Algorithm.

Start with bigger number and see how much smaller number goes into it:

1200=2(560)+80

560=80(7)

This implies the remainder before the remainder is 0 is the greatest common factor of 1200 and 560. So the greatest common factor of 1200 and 560 is 80.

Part a:

Find q and r such that:

-65=q(11)+r

We want to find q and r such that they satisfy the division algorithm.

r is suppose to be a positive integer less than 11.

So q=-6 gives:

-65=(-6)(11)+r

-65=-66+r

So r=1 since r=-65+66.

So q=-6 while r=1.

3 0

3 years ago

Ship A receives a distress signal from the southwest, and ship B receives a distress from the same vessel from the north. At wha

Fiesta28 [93]

I've attached an image showing the coordinates of Ship A and Ship B.

Answer:

distressed vessel is located at (1, 1.5)

Step-by-step explanation:

From the image attached, we can see that;

Coordinate of ship A is (3, 4)

Coordinate of ship B is (1, 1)

Now, we are told that Ship A receives a distress signal from the southwest, and ship B receives a distress from the same vessel from the north.

This means that the distressed ship is somewhere in between ship A & B.

To solve this, we will extend the arrow line from Ship A in that same SW direction and we will do the same for ship B in the same North direction. Their point of intersection is where the distressed ship is located.

Now, if we do that, we will see that they will intersect at the point; (1, 1.5)

Thus,distressed vessel is located at (1, 1.5)

5 0

3 years ago