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PolarNik [594]
3 years ago
11

Repost this for help

Mathematics
1 answer:
inna [77]3 years ago
3 0

Answer:

I'm sorry I don't know if I do get an idea I'll come back and edit

Step-by-step explanation:

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For each sequence, determine whether it appears to be arithmetic, geometric, or neither.
Art [367]

Step-by-step explanation:

1. 2, 4, 6, 8, ...

This sequence has a common difference (+ 2).  So it is arithmetic.

2. 9, 16, 25, 36, ...

This sequence has neither a common difference nor a common ratio.

3. 64, 32, 16, 8, ...

This sequence has a common ratio (× ½).  So it is geometric.

7 0
3 years ago
Please help me solve each question! Make sure to show work so I can earn full credit.
omeli [17]

(1)

5x-5\geq10\,\,\mbox{or}\,\,-3x+1>13\\5x\geq15\,\,\mbox{or}\,\,-3x>12\\x\geq3\,\,\mbox{or}\,\,x

(2)

5x + 3\leq18 \,\,\mbox{and}\,\,4 - x < 6\\5x\leq15 \,\,\mbox{and}\,\,- x < 2\\x\leq3 \,\,\mbox{and}\,\,x > 2\\2

8 0
3 years ago
Read 2 more answers
A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2
alukav5142 [94]

Answer:

0.17 °/s

Step-by-step explanation:

Since the ladder is leaning against the wall and has a length, L and is at a distance, D from the wall. If θ is the angle between the ladder and the wall, then sinθ = D/L.

We differentiate the above expression with respect to time to have

dsinθ/dt = d(D/L)/dt

cosθdθ/dt = (1/L)dD/dt

dθ/dt = (1/Lcosθ)dD/dt where dD/dt = rate at which the ladder is being pulled away from the wall = 2 ft/s and dθ/dt = rate at which angle between wall and ladder is increasing.

We now find dθ/dt when D = 16 ft, dD/dt = + 2 ft/s, and L = 20 ft

We know from trigonometric ratios, sin²θ + cos²θ = 1. So, cosθ = √(1 - sin²θ) = √[1 - (D/L)²]

dθ/dt = (1/Lcosθ)dD/dt

dθ/dt = (1/L√[1 - (D/L)²])dD/dt

dθ/dt = (1/√[L² - D²])dD/dt

Substituting the values of the variables, we have

dθ/dt = (1/√[20² - 16²]) 2 ft/s

dθ/dt = (1/√[400 - 256]) 2 ft/s

dθ/dt = (1/√144) 2 ft/s

dθ/dt = (1/12) 2 ft/s

dθ/dt = 1/6 °/s

dθ/dt = 0.17 °/s

8 0
3 years ago
If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then is the triangle isosceles?
Ne4ueva [31]

Answer:

True

Step-by-step explanation:

The perpendicular bisector of the opposite side to the vertex bisects the angle at the vertex into two equal parts and also bisects the triangle into two equal parts.

Let A be the angle at the vertex, then assume that the angle is an isosceles triangle with base angles B.

We need to show that A = 180 - 2B for an isoceles triangle

The perpendicular bisector bisects A into two so the new angle in the vertex one half of the bisected triangle is A/2.

Since this half triangle is a right-angled triangle, the third angle in it is 90.

So, A/2 + B +  90 = 180 (Sum of angles in a triangle)

subtracting 90 from both sides, we have

A/2 + B + 90 - 90 = 180 - 90

A/2 + B = 90

subtracting B from both sides, we have

A/2 + B = 90

A/2 = 90 - B

multiplying through by 2, we have

A = 2(90 - B)

A = 180 - 2B

Since A = 180 - 2B, then our triangle is an isosceles triangle.

7 0
3 years ago
I need help no ilnks please and thank you
Elza [17]

Answer:

yep

Step-by-step explanation:

nice

8 0
3 years ago
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