Answer:
720, 5040
Step-by-step explanation:
It's always times plus 1. So,
1 x 2 = 2
2 x 3 = 6
6 x 4 = 24
24 x 5 = 120
120 x 6 = 720
720 x 7 = 5040
Answer:
y-2=3(x+5)
Step-by-step explanation:
so point slope form is y-y1=m(x-x1)
x1 and y1 are your ordered pair: (-5,2)
M is the slope.
All you have to do is plug in the numbers!
Answer:
Step-by-step explanation:sin
(
θ
)
=
40
41
Use the definition of sine to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.
sin
(
θ
)
=
opposite
hypotenuse
Find the adjacent side of the unit circle triangle. Since the hypotenuse and opposite sides are known, use the Pythagorean theorem to find the remaining side.
Adjacent
=
√
hypotenuse
2
−
opposite
2
Replace the known values in the equation.
Adjacent
=
√
(
41
)
2
−
(
40
)
2
Simplify
√
(
41
)
2
−
(
40
)
2
.
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Adjacent
=
9
Find the value of cosine.
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cos
(
θ
)
=
9
41
Find the value of tangent.
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tan
(
θ
)
=
40
9
Find the value of cotangent.
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cot
(
θ
)
=
9
40
Find the value of secant.
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sec
(
θ
)
=
41
9
Find the value of cosecant.
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csc
(
θ
)
=
41
40
This is the solution to each trig value.
sin
(
θ
)
=
40
41
cos
(
θ
)
=
9
41
tan
(
θ
)
=
40
9
cot
(
θ
)
=
9
40
sec
(
θ
)
=
41
9
csc
(
θ
)
=
41
40
2x+4-x-3 = x +1
x + 1 = x + 1
0 = 0
your number has to be an infinite number of solutions. It can be any number.
Answer:
0.04,0.25.0.52
Step-by-step explanation:
Given that you throw a dart at a circular target of radius 10 inches.
Assuming that you hit the target and that the coordinates of the outcomes are chosen at random,
probability that the dart falls
(a) within 2 inches of the center
Here favourable region has area of a circle with radius 2 inches and sample space has area of 10 inches
Prob = 
(b) within 2 inches of the rim.
For within two inches from the rim we have to select area of the ring i.e. area of big circle with 10 inches - area of smaller circle with 10-2 inches
Prob= 
c) within I quadrant
area of I quadrant / area of circle=0.25
d) within I quadrant and within 2 inches of the rim
= I quadrant area + 2 inches ring area - common area
= 
