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2 years ago

4(5x−3)+5x+5 = 118 solve for x

Mathematics

1 answer:

EleoNora [17]2 years ago

8 0

Answer:

x=1+2/15

Step-by-step explanation:

4(5x-3)=5x+5=118

We move all terms to the left:

4(5x-3)-(5x+5)=0

We multiply parentheses

20x-(5x+5)-12=0

We get rid of parentheses

20x-5x-5-12=0

We add all the numbers together, and all the variables

15x-17=0

We move all terms containing x to the left, all other terms to the right

15x=17

x=17/15

x=1+2/15

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2 years ago

In your internship with Lewis, Lee, & Taylor Inc. you have been asked to forecast the firm's additional funds needed (AFN) f

Allushta [10]

Answer:

Option B. -$22,000

Step-by-step explanation:

Last year data :

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Sales growth rate = 40%

Calculate total increase in total assets :

Increase in total assets = Total assets × Growth rate

= 120,000 × 40%

= $48,000

Calculate increase in spontaneous liabilities :

Increase in spontaneous liabilities = (50,000 + 20,000) × 40%

= $28,000

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calculate the increase sales :

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calculate profit margin for next year on increase sales :

Profit margin on increased sales = 280,000 × 20%

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Retained earning = profit - payout of profit

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2 years ago

Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes

Ivan

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2 $\times$ 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 $\times$ 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled $\triangle AOB$ .

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units$

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = $\dfrac{Total\ distance}{Total\ time}$

$\Rightarrow \dfrac{74.28}{5} = 14.86\ knots$

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2 years ago