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3 years ago

He number line below represents the solution to which inequality?

Mathematics

2 answers:

Umnica [9.8K]3 years ago

4 0

Answer:

i think it is a hope it helps

Step-by-step explanation:

Leno4ka [110]3 years ago

3 0

B hope it’s right sorry

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The given line passes through the points (-2,-4) and (2,2) What is the equation, in point-slope form, of the line that is parall

Morgarella [4.7K]

Answer:

$(y-1)=\frac{3}{2}(x+3)$

Step-by-step explanation:

The point slope form of a line is $(y-y_1)=m(x-x_1)$ where $x_1=-3\\y_1=1$ . We write

$(y-1)=m(x--3)\\(y-1)=m(x+3)$

Since the line is parallel to the line shown it will have the same slope. To find m, count the slope from each marked point on the graph. Count the rise then the run and create a fraction rise/run. The slope is 6/4 which simplifies to 3/2. Now substitute it for m.

$(y-1)=\frac{3}{2}(x+3)$

3 0

3 years ago

Algebra 2 Please Help

hammer [34]

No sé lo que quieres decir

5 0

3 years ago

A cable television compan is laying cable in an area with underground utilities. Two subdivisions are located on opposite sides

Schach [20]

Answer:

Step-by-step explanation:

Given:

Point Q = Q is on the north bank 1200 m east of P $40/m to lay cable underground and $80/m way to lay the cable.

Lets denote T be a point on north bank, therefore, point T is t m east and 100m north of P.

Note that 0 < t < 1200

Distance from P to T using Pythagoras theorem:

PT^2 = PQ^2 + QT^2

PT = √(t²+100²) = √(t²+10,000)

Distance from T to Q:

TQ = 1200 - t

From above, Cable from P to T is underwater, and costs $80/m to lay.

Cable from T to Q is underground, and costs $40/m to lay.

Total cost of the cable:

C(t) = 80√(t²+10,000) + 40(1200 - t)

Cost is at its minimum: when dC/dt = C'(t) = 0 and d^2C/dt^2 = C''(t) > 0

dC/dt = C'(t)

= 80 * 1/2 (t² + 10,000)^(-1/2) × 2t + 40 × (-1)

dC/dt = C'(t)

= 80t/√(t² + 10,000) - 40 = 0

80t/√(t² + 10,000) = 40

80t = 40√(t² + 10,000)

2t = √(t² + 10,000)

square both sides:

4t² = t² + 10,000

3t² = 10,000

t² = 10,000/3

t = 100/√3

≈ 57.735

d^2C/dt^2 =

C''(t) = [80√(t² + 10,000) - 80t × t/√(t²+10,000)] /√(t²+10,000)

C''(t) = [(80(t²+10,000) - 80t²) / √(t²+10,000)] /√(t²+10,000)

C''(t) = 800,000 / (t² + 10,000)^(3/2)

C''(t) > 0 for all t

Therefore C(t) is minimized at t = 100/√3

C(100/√3) = 80√(10,000/3 + 10,000) + 40 × (1200 - 100/√3)

= 80 √(40,000/3) + 48,000 - 4000/√3

= 16000/√3 + 48,000 - 4000/√3

= 48,000 + 12,000/√3

= 48,000 + 4,000√3

= 4000 (12 + √3)

≈ 54,928.20

Minimum cost is $54,928.20

This occurs when cable is laid underwater from point P to point T(57.735 m east and 100m north of P) and then underground from point T to point Q (1200 - 57.735)

= 1142.265 m

5 0

3 years ago

You use about 15 gallons of water when you shower. If you take 20% less time when you shower, estimate how many gallons of water

PIT_PIT [208]

I would have to say 85 but thats just a guess.

8 0

3 years ago

Number of Boxes Cost 18 $7.38 27 $15.39 30 $13.50 36 $18.36 40 $18.80 An office is restocking office supplies. The table shows t

malfutka [58]

<u>Answer</u>:

18 clips for $7.38 is the best value deal.

<u>Step-by-step explanation:</u>

An office is restocking office supplies. The table below shows the cost for the number of boxes of paperclips. We need to find what is the best value is.

To find the best value for money we can check the price of each paperclip.

To check the price of each paper clip we can use the following formula:

$\text{Price of each paper clip} =\frac{\text { price of the box }} {{\text {number of clips in the box }}}$