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sweet-ann [11.9K]
3 years ago
10

HELP ME PLEASE ASAP!

Mathematics
2 answers:
ladessa [460]3 years ago
7 0

Answer:

31

Step-by-step explanation:

dexar [7]3 years ago
7 0

Answer:

that's the answer

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Solve each equation 12x - 51 = 3(x + 7)
irinina [24]

Answer:

x=8      

Step-by-step explanation:

3 0
3 years ago
Plz<br> help<br> now<br> im<br> dew<br> in<br> a<br> hour
forsale [732]

Answer:

D. x<2

Step-by-step explanation:

-1+ 6-1-3x) >- 39 - 2x

Multiply the parentheses by 6

- 1-6 - 18x > - 39 - 2x

Calculate

7- 18x > - 39 - 2x

Move the terms

18x + 2x > -39 +7

Collect like terms Calculate

- 16x>-32

Divide both sides by - 16

x < 2

8 0
3 years ago
One counting number is 4 times great as a counting number.The product of the two number is 36.What is the sum of the two number
VARVARA [1.3K]

Answer:

The sum of the two numbers is 15

Step-by-step explanation:

Let the counting numbers be x and y.

Then we can write the following equations

x=4y.......1 (one is four times greater )

x(y)=36.....2 (the product of the two)

Putting equation 1 into equation 2, we get

4y(y)=36

{4y}^{2} = 36

Dividing through by 4,we get

{y}^{2} =  \frac{36}{4}

{y}^{2} = 9

Taking square root of both side, we obtain

y=3

Putting the value of y into equation , we get

x=4(3)=12

Hence the numbers are 3 and 12

Therefore the sum of the two numbers is 3+12=15

3 0
3 years ago
Need help with this math
Rudiy27

10,171.07 pesos

To every US dollar, it equal 18.84 pesos.

7 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
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