Just add all the values, that's your denominator, then the number of pink roses is your numerator. :)
The answer is
(simplified to:)
Hopefully this helped! If you have any more questions or don't understand, feel free to DM me!! :)
The answer:
the full question is as follow:
<span>Point F is on circle C. What is the length of line segment GF?
12.5 units
15.0 units
17.5 units
20.0 units
according to the image, GF= GC + CF
CF is the radius of the circle, so it is CF =CE = 7.5
all that we want to find is the value of GC
let's consider the triangle GEC. This is a right triangle, so for finding GC, we can apply Pythagorean theorem
that is, GE² + EC² = GC², and from this, we have GC = sqrt(</span>GE² + EC² )
GC = sqrt(10² + 7.5² )=sqrt(56.25)=12.5
<span>
therefore, </span>GF= GC + CF=12.5 + 7.5 = 20.0
the answer is 20.0 units
Answer: The distance from vertex B to vertex H is 6.9 inches.
Step-by-step explanation:
The length of the line BH can be thought of as the hypotenuse of a triangle rectangle where the catheti are the lines HD (whit a measure of 4in) and line DB.
The length of the line DB can be thought of as the hypotenuse of a triangle rectangle where the catheti are lines DA and AB (both are 4in long)
Then if we use the Pythagorean's theorem, the length of line DB is:
(DB)^2 = (DA)^2 + (AB)^2
(DB)^2 = (4in)^2 + (4in)^2 = 32in^2
(DB) = √(32in^2) = 5.66 in
Whit this, we can find the length of line HB as:
(HB)^2 = (HD)^2 + (DB)^2
(HB)^2 = (4in)^2 + (DB)^2 = 16in^2 + 32in^2 = 48in^2
HB = √(48in^2) = 6.93 in
If we round to the nearest thent, we get:
HB = 6.9 in
The distance from vertex B to vertex H is 6.9 inches.
Answer:
Let x be the number;
We solve the equation x ÷ 4 - 5 = 15;
Then x ÷ 4 = 20;
Finally x = 80;
Step-by-step explanation:
There are an infinite number of possibilities, and not enough information
to decide which possibility is really the one inside the function machine.
Here are a few. Each of these gives the result that you described,
and there are an infinite number of others:
f(x) = x
f(x) = 2x + 1
f(x) = 10x + 9
f(x) = x² - 2
f(x) = 7x² - 8
f(x) = 31x³ + 30
f(x) = log( |x| ) - 1
f(x) = ln( |x| ) - 1
f(x) = x tan(45°)
.
.
etc.