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Vladimir [108]
3 years ago
9

Bryan has a prepaid cell phone with a balance of $75. He is charged the same rate per minute. Which function represents the bala

nce on the phone after (n) minutes?
Mathematics
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

36 minutes

Step-by-step explanation:

i have a different saying this so dont get mad

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The area of the base of the oblique pentagonal pyramid is 50 cm^2 and the distance from the apex to the center of the pentagon i
Verdich [7]
The complete question in the attached figure

Part a) 
we know that
ABC is a right triangle
∠ACB=45°
AC=hypotenuse------> 6√2 cm
sin 45=AB/AC-----> AB=AC*sin 45----> AB=6√2*√2/2----> AB=6 cm

the answer part a) is
AB=6 cm

Part b) 
we know that
volume of the pyramid=(1/3)*Area of the base*height
area of the base=50 cm²
height=6 cm
so
volume of the pyramid=(1/3)*50*6----> 100 cm³

the answer part b) is 
100 cm³

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A. 4 miles is equal to how many kilometer. B. 45 pound is equal to how many kilogram?​
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If a dish of ice cream holds 1/4 pound, how many dishes can you get from a 4 1/2 pound carton of Dan's ice cream?
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Hi! I don't really get these two questions, I would really really appreciate your help. Please if you don't mind include work in
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1. x intercept - (2,0)
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Match each set of vertices with the type of triangle they form.
Andrew [12]

Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

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3 years ago
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