Please help. Ignore The answer I have selected. I'm not sure if it's correct

What is the measure of angle RSL if it is (9x+27)° and angle BSA is (6x+66)°

Shalnov [3]

Answer:

Assume:

(angle) RSB and angle (LSA) = D

and (line) ASR and (line) LSB is striaght then

(9x +27) + D = 180

(6x + 66) + D = 180

9x + D = 153

6x + D = 114

substitute

6x + D = 114

D = 114 - 6 x

9x + (114 - 6x) =153

3x = 39

x = 13

D = 114 - 6(13)

D=36

Step-by-step explanation:

8 0

3 years ago

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Gravity Hospital purchased a new item of equipment on January 1,20X1. The equipment has an estimated useful life of ten years an

qaws [65]

Answer:

$25000

Step-by-step explanation:

If the salvage value is 20% of the cost, then 80% of the cost will be depreciated over 10 years. Over the 5 years from Jan 1 20X1 to Dec 31 20X5, the $10,000 accumulated depreciation represents 5/10 of that 80%, or 40% of the initial cost.

$10,000 = 0.40 × cost

$10,000/0.40 = cost = $25,000

The acquisition cost of the equipment was $25,000.

8 0

2 years ago

Total profit P is the difference between total revenue R and total cost C. Given the following total-revenue and total-cost fu

gavmur [86]

Answer:

The profit function is:

$P(x)=-x^2+1280x-3300$

The maximum value is 406, 300 occurring when x = 640.

Step-by-step explanation:

The revenue function is:

$R(x)=1300x-x^2$

And the cost function is:

$C(x)=3300+20x$

Then the total profit function will be:

$P(x)=R(x)-C(x)=(1300x-x^2)-(3300+20x)=-x^2+1280x-3300$

This is a quadratic function.

Therefore, the maximum value of the total profit will occur at its vertex point.

The vertex of a quadratic is given by:

$\displaystyle \Big(-\frac{b}{2a}, f\Big(-\frac{b}{2a}\Big)\Big)$

In this case, a = -1, b = 1280, and c = -3300.

Then the point at which the maximum profit occurs is at:

$\displaystyle x=-\frac{1280}{2(-1)}=640$

And the maximum profit will be:

$P(640)=-(640)^2+1280(640)-3300=406300$

5 0

2 years ago

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larisa86 [58]

Distance formula

$\boxed{\sf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Now

#1

$\\ \tt\Rrightarrow AB=\sqrt{(3-3)^2+(-2-8)^2}=\sqrt{100}=10units$

#2

$\\ \tt\Rrightarrow OK=\sqrt{(5-7)^2+(-2+2)^2}=\sqrt{4}=2units$

#3

$\\ \tt\Rrightarrow LM=\sqrt{(-4-4)^2+(6+4)^2}=\sqrt{64+100}=\sqrt{164}\approx 13units$

#4

$\\ \tt\Rrightarrow PC=\sqrt{(5-2)^2+(1-3)^2}=\sqrt{9+4}=\sqrt{13}=3.2units$

#5

$\\ \tt\Rrightarrow QR=\sqrt{(-2+10)^2+(3-9)^2}=\sqrt{64+36}=\sqrt{100}=10units$

8 0

2 years ago

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A family has two cars. The first car has a fuel efficiency of 20 miles per gallon of gas and the second has a fuel efficiency of

Lubov Fominskaja [6]

<u>Answer:</u>

First car: 35 gallons

Second car: 15 gallons

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

Assuming $x$ to be the gas consumed by 1st car and $50 - x$ to be the gas consumed by 2nd car.

We know that:

Distance traveled = fuel efficiency × gas consumed

$20x+35(50-x)=1225$

$20x+1750-35x=1225$

$35x-20x=1750-1225$

$15x=525$

$x=35$

So $50 - x = 50-35=15$

Therefore, first car consumed 35 gallons while second car consumed 15 gallons.

7 0

2 years ago

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