I think we're not supposed to do more than 3 problems in one answer; certainly not five pages. Does the 1/3 mean there are 10 more to come?
I'll do the first page.
6)
Right triangle, opposite side of 10, adjacent side of 21,
tan x = opp/adj = 10/21
x = arctan 10/21 = 25.46°
Answer: 25
7)
cos x = adj / hyp = 10/14
x = arccos 10/14 = 44.42°
Answer: 44
8)
tan x = opp/adj = 12/24 = 1/2
x = arctan 1/2 = 26.57°
Answer: 27
9)
tan x = 31/32
x = arctan 31/32 = 44.09°
Answer: 44
10)
x = arctan 10/27 = 20.32°
Answer: 20
Answer:
c
Step-by-step explanation:
this is because your adding 5 to f(x) which would be x not y (pretty sure) and g(x) is y. your not adding 5 to g(x) therefore not b. D and A would never be correct for this because that would be subtraction of 5
Hi There!
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Question 2:
Obviously the right equation for this question is Point Slope Form.
Point Slope Form: y - y1 = m(x - x1)
y1 = -1
x1 = 3
m = 2
Question 2 Answer: y + 1 = 2(x - 3)
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Question 3:
Again use point slope form.
Find Slope:
7 - 1 = 6
-2 - 1 = -3
6/-3 = -2
Question 3 Answer: y - 1 = -2(x - 1)
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Hope This Helps :)
Answer:
El área del cuadrado es de 3 centímetros cuadrados.
Step-by-step explanation:
Dado que se desea pintar un cuadrado inscrito en una circunferencia de radio R=3cm como se muestra en la figura, para calcular el área del cuadrado se debe realizar el siguiente cálculo:
Radio = diámetro / 2
3 = X/2
X = 6
Hipotenusa del triángulo interno: 6 cm
Aplicando teorema pitagórico: lado al cuadrado mas lado al cuadrado es igual a hipotenusa al cuadrado.
L^2 + L^2 = 6
2L^2 = 6
L^2 = 6/2
L = √ 3
L = 1.732
Área de un cuadrado = L x L = L^2 = 3
Por lo tanto, el área del cuadrado es de 3 centímetros cuadrados.
